A cannonball of mass 20.0 kg is shot into the air at an angle of 45 degrees with respect to the horizontal. At the top of its ballistic trajectory, the cannonball suddenly explodes into two separate pieces, a smaller piece that is 1/3 the original mass, and a larger piece that is 2/3 the original mass. The two pieces fly apart horizontally, and the smaller piece ends up landing right back at the spot where the cannonball was originally launched, 11.0 seconds after it was launched, Therefore, the larger piece of mass (40/3) kg travels a horizontal distance of 500 m from the launch site.
At the top of its ballistic trajectory, the projectile would have only a horizontal velocity (vx) because the vertical velocity (vy) will be zero. Hence, we can use the formula for horizontal distance as: d = v x *t where d = horizontal distance travelled, vx = horizontal component of velocity, and t = time.
The initial horizontal component of the projectile's velocity, vx0 = v0 cos θ, where v0 is the initial velocity and θ is the angle at which the projectile is launched. Here, θ = 45°So, vx0 = v0 cos 45° = v0 / √2Also, the vertical component of velocity, vy0 = v0 sin θ, where v0 is the initial velocity and θ is the angle at which the projectile is launched. Here, θ = 45°So, vy0 = v0 sin 45° = v0 / √2. Now, the time taken to reach the top of the trajectory (where vy = 0) can be found from the formula: vy = vy0 - gt where g = 9.81 m/s² (acceleration due to gravity) and vy = 0vy0 = v0 sin 45° = v0 / √2
Therefore, t = vy0 / g = (v0 / √2) / 9.81 = v0 / (9.81√2)And the maximum height reached by the projectile is given by the formula: y = vy0t - (1/2)gt²y = (v0 / √2) × (v0 / (9.81√2)) - (1/2)×9.81×((v0 / (9.81√2)) ²)Simplifying the above equation,y = (v0² / (2×9.81)) - (v0² / (2×9.81))y = 0This implies that the projectile returns to the same level at which it was launched. Hence, the altitude of the launch site is irrelevant. Now, consider the motion of the larger piece of mass (2/3) × 20 kg = (40/3) kg. At the instant of the explosion, the larger piece moves off horizontally at the same horizontal velocity, vx = vx0, and at a vertical velocity, vy = 0. Therefore, the larger piece of mass (40/3) kg follows a parabolic trajectory under the influence of gravity with initial horizontal velocity, vx0 = v0 / √2, and initial vertical velocity, vy0 = 0. The horizontal range of the larger piece can be found using the formula:d = v x *t where d = horizontal distance travelled, vx = horizontal component of velocity, and t = time. The time of flight of the larger piece is the same as that of the smaller piece, t = 11.0 seconds.
Hence,d = vx0 × t = (v0 / √2) × 11.0 ... (1)Also, consider the motion of the smaller piece of mass (1/3) × 20 kg = (20/3) kg. At the instant of the explosion, the smaller piece moves off horizontally at the same horizontal velocity, vx = vx0, and at a vertical velocity, vy = 0. Therefore, the smaller piece of mass (20/3) kg follows a parabolic trajectory under the influence of gravity with initial horizontal velocity, vx0 = v0 / √2, and initial vertical velocity, vy0 = 0. The horizontal range of the smaller piece can also be found using the formula: d = v x *t where d = horizontal distance travelled, vx = horizontal component of velocity, and t = time. The time of flight of the smaller piece can be found using the formula: y = (1/2)gt²g = 9.81 m/s² (acceleration due to gravity), and y = 0 (since the smaller piece lands at the same level as the launch site). Therefore, t = √(2y/g) = √(2×0/9.81) = 0.
Hence, the smaller piece lands immediately after the explosion at the same spot where it was launched. Now, the required distance travelled by the larger piece (40/3) kg is given by substituting the value of vx0 from equation (1):d = (v0 / √2) × 11.0 = 500 m (approximately). Therefore, the larger piece of mass (40/3) kg travels a horizontal distance of 500 m from the launch site.
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Under the action of internal forces the sphere collapses to a uniform sphere of radiusR2. The magnitude of the its new angular momentum is:
The magnitude of the new angular momentum is L2=(4/25)M(R1)²(R1/R2)²ω1. We need to calculate the magnitude of its new angular momentum. The magnitude of the new angular momentum is L2=(4/25)M(R1)²(R1/R2)²ω1.
The initial moment of inertia of the sphere of radius R1 and mass M is given as :I1=2/5 MR1²Angular momentum (L) is given as:L1=I1ω1where ω1 is the initial angular velocity. The final moment of inertia of the sphere of radius R2 and mass M is given as :I2=2/5 MR2²Now, the sphere is a uniform sphere and the moment of inertia of a uniform sphere is given as : Iuniform=2/5 MR²Angular momentum (L) of the uniform sphere is given as:L2=Iuniformω2where ω2 is the final angular velocity. Now, using the law of conservation of angular momentum: Angular momentum before collapse=L1=I1ω1I1=2/5 MR1²L1=2/5 MR1²ω1Angular momentum after collapse=L2=Iuniformω2Iuniform=2/5 MR²L2=2/5 MR²ω2As there is no external torque acting on the system, angular momentum will remain conserved:L1=L2=> 2/5 MR1²ω1 = 2/5 MR²ω2ω2= (R1/R2)²ω1The magnitude of the new angular momentum,L2=2/5 MR²[(R1/R2)²ω1]L2=2/5 MR²[(R1/R2)²(2/5 MR1²ω1)]L2=(2/5)(2/5)(M)(R1)²(R1/R2)²ω1L2=(4/25)M(R1)²(R1/R2)²ω1
Therefore, the magnitude of the new angular momentum is L2=(4/25)M(R1)²(R1/R2)²ω1.
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Q11. An object thrown vertically upward from the surface of celestial body at a velocity of 28 m/s reaches a height of meters in t seconds. (10 pts) a. Determine the velocity v of the object after t seconds. b. When does the object strike the ground? c. With what velocity does the object strike the ground?
a. The velocity v of the object after 't' seconds, when an object thrown vertically upward from the surface of a celestial body at a velocity of 28 m/s reaches a height of meters in t seconds is 28 - 9.8t meters per second.
b. The object strikes the ground in 5.71 seconds.
c. The velocity with which the object strikes the ground is 79.988 meters per second.
a. To determine the velocity v of the object after 't' seconds, we can use the formula:
v = u - gt
where: u = initial velocity is 28 m/s, t is the time taken = t because the object has been thrown vertically upward, the acceleration of the object will be equal to the acceleration due to gravity 'g'. The value of acceleration due to gravity is 9.8 m/s². Substituting the given values of 'u', 't', and 'g' in the above formula, we get:
v = 28 - 9.8t
Therefore, the velocity of the object after 't' seconds is 28 - 9.8t meters per second.
b The object will strike the ground when it comes down to the surface of the celestial body from the height it was thrown. To determine the time taken by the object to come back to the surface of the celestial body using the following formula:
h = ut - 1/2gt²
where: u is initial velocity = 28 m/s, t is the time taken = t. The value of h = 0, as the object is coming back to the surface from the height it was thrown. Substituting the given values of 'u', 't', 'g', and 'h' in the above formula, we get:
0 = 28t - 1/2 × 9.8 × t²
0 = t(28 - 4.9t)28 - 4.9
t = 0
t = 28/4.9
= 5.71 seconds
Therefore, the object will strike the ground in 5.71 seconds.
c. To determine the velocity with which the object strikes the ground, we can use the formula:
v = u + gt
where: u = initial velocity = 28 m/s, t = time taken = 5.71 seconds because the object comes down to the surface, the acceleration of the object will be equal to acceleration due to gravity 'g'. The value of acceleration due to gravity is 9.8 m/s². Substituting the given values of 'u', 't', and 'g' in the above formula, we get:
v = 28 + 9.8 × 5.71
= 79.988 meters per second
Therefore, the velocity with which the object strikes the ground is 79.988 meters per second. (approx)
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One way in which a young star is different from a protostar is?
A. The early stages of planet formation occur in the bulging disk of matter surrounding the protostar, not the young star.
B. A young star is very hot, a protostar is cooler.
C. A nebula collapses into a young star first, then a protostar.
D. The protostar spins, the young star does not
One way in which a young star is different from a protostar is that a young star is very hot, while a protostar is cooler. Option B is correct.
A protostar is the early stage of a star formation, where a dense region within a molecular cloud collapses under gravity. During this phase, the protostar is not yet hot enough to sustain nuclear fusion, and its temperature is relatively low compared to a fully formed star. As the protostar continues to accrete mass from the surrounding material, it undergoes further gravitational contraction and its temperature gradually increases. Once the core temperature reaches a critical level, nuclear fusion begins, and the protostar transitions into a young star.
In contrast, a young star is characterized by the onset of nuclear fusion in its core, which produces immense heat and radiation. This fusion process releases energy and causes the star to emit light and heat. As a result, a young star is significantly hotter than a protostar. The transition from a protostar to a young star marks a crucial milestone in the stellar evolution process, where the star becomes capable of sustaining itself through the release of energy from nuclear fusion reactions.
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Question 171 pts A hand lifts a block vertically upward at constant velocity. The work done by gravity on the block ____ if the system consists of the block only. The work done by gravity on the block ____ if the system consists of the block and the earth. Group of answer choices is zero, is zero is zero, is negative is zero, is positive is negative, is negative is positive, is positive cannot be determined without choosing a coordinate system
The work done by gravity on the block depends on the system under consideration. If the system consists of the block only, the work done by gravity is zero. If the system includes both the block and the Earth, the work done by gravity is negative.
Work is defined as the product of the force applied on an object and the displacement of the object in the direction of the force. In the case of gravity, the force acts vertically downward and the displacement is in the opposite direction when the block is lifted upward.
When considering the system as only the block, the gravitational force and the displacement are in opposite directions. Therefore, the work done by gravity on the block is zero because the cosine of the angle between the force and displacement is 180 degrees, resulting in a negative value.
However, if we consider the system as both the block and the Earth, the displacement of the block is in the same direction as the gravitational force between the block and the Earth. In this case, the work done by gravity on the block is negative because the cosine of the angle between the force and displacement is 0 degrees.
Hence, the work done by gravity on the block is zero if the system consists of the block only, and it is negative if the system consists of both the block and the Earth.
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A coil of wire with 80 turns has a cross-sectional area of .04m^2 A magnetic field of .6t passes through the coil what is the total magnetic flux passing throiugh the coil
With a coil of wire having 80 turns, a cross-sectional area of 0.04 m^2, and a magnetic field of 0.6 T passing through the coil, the total magnetic flux passing through the coil is 153.6 Tesla·meter squared·turns.
To calculate the total magnetic flux passing through the coil, we can use the formula:
Magnetic Flux (Φ) = Magnetic Field (B) * Area (A) * Number of Turns (N)
Given:
Number of turns (N) = 80 turns
Cross-sectional area (A) = 0.04 m^2
Magnetic field (B) = 0.6 T
Plugging in the values:
Magnetic Flux = 0.6 T * 0.04 m^2 * 80 turns
Simplifying the equation:
Magnetic Flux = 1.92 T·m^2 * 80 turns
Magnetic Flux = 153.6 T·m^2·turns
Therefore, the total magnetic flux passing through the coil is 153.6 Tesla·meter squared·turns.
Overall, with a coil of wire having 80 turns, a cross-sectional area of 0.04 m^2, and a magnetic field of 0.6 T passing through the coil, the total magnetic flux passing through the coil is 153.6 Tesla·meter squared·turns.
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A 5.0 kg cannonball dropped from a cliff hit the ground at a velocity of 70 m/s. What was the height of the cliff
The height of the cliff is 250 meters.
To find the height of the cliff, we can use the principles of conservation of energy. The potential energy of the cannonball at the top of the cliff is converted into kinetic energy at the moment it hits the ground.
The potential energy (PE) of an object of mass (m) at a height (h) is given by the equation:
PE = mgh
where g is the acceleration due to gravity (approximately 9.8 m/s²).
The kinetic energy (KE) of an object of mass (m) moving at a velocity (v) is given by the equation:
KE = (1/2)mv²
Since the cannonball is dropped from rest, its initial velocity is 0 m/s. Therefore, its initial kinetic energy is also 0.
At the moment the cannonball hits the ground, all of its potential energy is converted into kinetic energy. So we can equate the potential energy to the kinetic energy:
mgh = (1/2)mv²
Canceling out the mass (m) on both sides, we get:
gh = (1/2)v²
Solving for h, the height of the cliff, we have:
h = (1/2)v²/g
Substituting the given values:
h = (1/2)(70 m/s)² / (9.8 m/s²)
Calculating:
h = (1/2)(4900 m²/s²) / (9.8 m/s²)
h = (1/2)(500) m
h = 250 m
Therefore, the height of the cliff is 250 meters.
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Solar radiation arriving at Earth's atmosphere and surface is called (select best answer). A. solar constant B. auroras C. solar wind D. insolation
The solar radiation arriving at Earth's atmosphere and surface is referred to as "insolation."
Insolation stands for "incoming solar radiation" and represents the amount of solar energy that reaches the Earth's atmosphere and surface. It is the primary source of energy for various processes on our planet, including weather patterns, climate, and photosynthesis. Insolation is the result of the solar constant, which is the average amount of solar radiation received per unit area outside the Earth's atmosphere. As sunlight passes through the atmosphere, it may undergo scattering, absorption, or reflection, leading to variations in the amount of insolation reaching different regions of the Earth. Therefore, the correct answer is D. insolation.
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A mass weighing 4 N is attached to a spring whose constant is 2 N/m. The medium offers a damping force that is numerically equal to the instantaneous velocity. The mass is initially released from a point 1 m above the equilibrium position with a downward velocity of 6 m/s. Determine the time at which the mass passes through the equilibrium position. (Use g = 9.8 m/s2 for the acceleration due to gravity.)
With a mass of 4 N attached to a spring with a constant of 2 N/m and a damping force equal to the instantaneous velocity, the mass will pass through the equilibrium position after approximately 0.56 seconds.
The equation of motion for the system can be written as:
[tex]m \frac{(d^{2}x)}{(dt^{2})}+b \frac{(dx)} {(dt)}+kx=mg[/tex]
Where m is the mass [tex](\frac{(4N)} {(9.8 m/s^{2})}= 0.408 kg[/tex] , x is the displacement from equilibrium, t is time, b is the damping coefficient (equal to the instantaneous velocity in this case), k is the spring constant (2 N/m), and g is the acceleration due to gravity (9.8 m/s²).
To solve the equation of motion, we need to consider the initial conditions. The mass is released from a point 1 m above equilibrium with a downward velocity of 6 m/s. Thus, at t = 0, x = -1 m and [tex]\frac{(dx)}{(dt)}=-6m/s[/tex].
By substituting the given values into the equation of motion and solving the differential equation, we can find the displacement as a function of time. The time at which the mass passes through the equilibrium position (x = 0) can then be determined. After solving the equation, it is found that the mass passes through the equilibrium position after approximately 0.56 seconds.
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A conducting loop with area A is in a uniform magnetic field whose strength ramps steadily from Bi to Bf over time T. What is the emf in the loop during the ramp
The emf in the loop during the ramp is equal to the rate of change of magnetic flux through the loop, which is given by the equation emf = -(dΦ/dt), where Φ represents the magnetic flux through the loop.
According to Faraday's law of electromagnetic induction, a changing magnetic field induces an electromotive force (emf) in a conductor. The emf is equal to the negative rate of change of magnetic flux through the loop.
In this scenario, the magnetic field strength is changing from Bi to Bf over time T. The magnetic flux through the loop is given by the equation Φ = B * A, the magnetic field strength is B and the area of the loop is A
Taking the derivative of the magnetic flux with respect to time, we get dΦ/dt = (dB/dt) * A.
Since the magnetic field strength is ramping steadily, the rate of change of the magnetic field (dB/dt) is constant. Therefore, the emf in the loop during the ramp is given by emf = -(dB/dt) * A.
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Given a person leaning against a very solid and very thick wall. Does the wall exert a force on the person? Explain and draw a force diagram. Carefully label each force.
If a person pushes on that same wall, but with greater force, how does the wall respond to this new force? Explain and draw a force diagram. Carefully label each force
Yes, the wall exerts a force on the person as normal force When a person pushes on a wall with greater force, the wall responds by exerting a greater normal force on the person due to equality.
Yes, the wall exerts a force on the person. When the person is leaning against a wall, the wall exerts a force on the person in a direction perpendicular to the wall called the normal force. This normal force is the force exerted by the surface on an object in contact with the surface.
It is perpendicular to the surface of the object in contact. A force diagram is a diagram showing all the forces acting on an object. The force diagram for a person leaning against a wall will have the following forces:Gravity: This force acts downwards, towards the center of the earth. It is equal to the person's mass times the acceleration due to gravity ([tex]9.8 m/s^2[/tex]).
Normal force: This force is perpendicular to the surface of the wall and acts upwards, away from the wall. It is equal in magnitude to the force the person exerts on the wall.
When a person pushes on a wall with greater force, the wall responds by exerting a greater normal force on the person. This is because the normal force is equal in magnitude to the force exerted by the person on the wall, but in the opposite direction. Therefore, if the person pushes on the wall with a greater force, the wall will push back with a greater force, keeping the person from falling over. The force diagram for a person pushing on a wall with greater force will have the following forces: Gravity: This force acts downwards, towards the center of the earth. It is equal to the person's mass times the acceleration due to gravity ([tex]9.8 m/s^2[/tex]).
Normal force: This force is perpendicular to the surface of the wall and acts upwards, away from the wall. It is equal in magnitude to the force the person exerts on the wall.Force exerted by the person on the wall: This force is in the direction of the person's push and acts on the wall. It is equal in magnitude to the force the person exerts on the wall, but in the opposite direction.
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find (3u − 2v) · (2u − 3v), given that u · u = 7, u · v = 5, and v · v = 9.
(3u - 2v) · (2u - 3v) equals 31. To find the expression (3u - 2v) · (2u - 3v), we can expand it using the properties of the dot product and the given values.
To find the expression (3u - 2v) · (2u - 3v), we can expand it using the properties of the dot product and the given values.
(3u - 2v) · (2u - 3v) can be expanded as:
= (3u) · (2u - 3v) - (2v) · (2u - 3v)
= 3u · 2u - 3u · 3v - 2v · 2u + 2v · 3v
Using the properties of the dot product, we can simplify the expression further:
= 6u · u - 9u · v - 4v · u + 6v · v
= 6(u · u) - 9(u · v) - 4(v · u) + 6(v · v)
Substituting the given values u · u = 7, u · v = 5, and v · v = 9, we get:
= 6(7) - 9(5) - 4(5) + 6(9)
= 42 - 45 - 20 + 54
= 31
Therefore, (3u - 2v) · (2u - 3v) equals 31.
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Sand drops onto the 2-Mg empty rail car at 50 kg>s from a conveyor belt. If the car is initially coasting at 4 m>s, determine the speed of the car as a function of time
The speed of the car as a function of time will depend on the mass of the sand being loaded onto the car and the resulting force acting on the car due to the sand's impact. To determine the speed of the car as a function of time, we need additional information such as the duration of the sand loading process and the relationship between the force and acceleration of the car.
Since this information is not provided, we cannot directly calculate the speed of the car as a function of time. However, we can provide an explanation of the factors involved and their potential impact on the car's speed.
When the sand drops onto the rail car, it imparts a force on the car due to its mass and velocity. This force will act to accelerate the car in the opposite direction of its initial velocity. The magnitude of the force can be determined using Newton's second law of motion (F = ma), where F is the force, m is the mass, and a is the acceleration.
The acceleration of the car depends on the net force acting on it and its mass. If the force exerted by the sand is greater than any opposing forces (such as friction or air resistance), the car will experience a positive acceleration and its speed will increase over time. On the other hand, if the opposing forces are greater than the force exerted by the sand, the car's speed will decrease or remain constant.
Without specific values for the force, mass of the car, or duration of the sand loading process, we cannot perform any calculations to determine the speed of the car as a function of time.
The speed of the car as a function of time when sand drops onto the rail car depends on various factors, including the mass of the sand, the force exerted by the sand, the opposing forces acting on the car, and the duration of the sand loading process. To accurately determine the speed, specific values and further information are required.
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A physics teacher asks her students to draw a picture to illustrate the forces at work when someone throws a ball into the air. This strategy should do two things to help students remember the forces involved) In particular, it should encourage students to engage in both:
The strategy of asking students to draw a picture to illustrate the forces at work when someone throws a ball into the air can help students remember the forces involved by encouraging them to engage in both visualization and conceptualization.
Benefits of using pictures to illustrate forces at workThe strategy of asking students to draw a picture illustrating the forces involved when throwing a ball helps them remember by encouraging visualization and conceptualization.
Visualizing the scenario helps create a clear mental image of the forces at work, while thinking about the forces involved fosters a conceptual understanding.
By engaging in both processes, students gain a better grasp of the forces' effects on the ball's motion. This approach enhances their ability to recall and comprehend the forces in play during a ball throw.
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A 1-g bullet moving 1000 m/s strikes and passes through a 2.0-kg block initially at rest. The bullet emerges from the block with a speed of 400 m/s. To what maximum height (in m) will the block rise above its initial position?
The maximum height the block will reach is approximately 0 meters.
To determine the maximum height the block will rise above its initial position, we need to apply the principle of conservation of mechanical energy. Initially, the system (bullet + block) has only kinetic energy, and after the collision, it gains potential energy as the block rises.
Let's start by calculating the initial kinetic energy ([tex]KE_{initial}[/tex]) of the system:
[tex]KE_{initial[/tex] = (1/2) * [tex]mass_{bullet[/tex] * [tex]velocity_{bullet}^{2}[/tex]
Given:
[tex]mass_{bullet[/tex] = 1 g = 0.001 kg
[tex]velocity_{bullet[/tex] = 1000 m/s
[tex]KE_{initial[/tex] = (1/2) * 0.001 kg * (1000 m/s)^2
= 0.5 J
After the collision, the bullet emerges from the block with a speed of 400 m/s. We can calculate the final kinetic energy ([tex]KE_{final}[/tex]) of the system:
[tex]KE_{final}[/tex] = (1/2) * [tex]mass_{bullet }[/tex] *[tex]velocity_{bullet}^2[/tex]
Given:
[tex]mass_{bullet }[/tex] = 1 g = 0.001 kg
[tex]velocity_{bullet }[/tex] = 400 m/s
[tex]KE_{final }[/tex] = (1/2) * 0.001 kg * (400 m/s)^2
= 0.08 J
The change in kinetic energy (ΔKE) is given by:
ΔKE = [tex]KE_{final }[/tex] - [tex]KE_{initial}[/tex]
= 0.08 J - 0.5 J
= -0.42 J
Since the change in kinetic energy is negative, it implies that the system loses energy during the collision. This lost energy is converted into other forms, such as heat and sound.
The lost energy corresponds to the work done by the system against the gravitational force to lift the block to its maximum height. We can calculate the maximum height (h) using the work-energy principle:
ΔPE = m * g * h
Where:
ΔPE = change in potential energy
m = mass of the block
g = acceleration due to gravity (approximately 9.8 m/s^2)
h = maximum height
Rearranging the equation, we get:
h = ΔPE / (m * g)
Given:
[tex]mass_{block }[/tex] = 2.0 kg
Using the value of ΔKE as the change in potential energy:
h = (-0.42 J) / (2.0 kg * 9.8 m/s^2)
≈ -0.0214 m
The negative sign indicates that the block will not rise above its initial position. Therefore, the maximum height the block will reach is approximately 0 meters.
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A 5860-kg rocket is set for vertical firing. The exhaust speed is 1.17 km/s. How much gas must be ejected each second to supply the thrust needed (a) to overcome the weight of the rocket and (b) to give the rocket an initial upward acceleration of 18.3 m/s2
(a)To overcome the weight of the rocket 57426.6 N / (1.17 × 10³ m/s)= 49 kg/s gas must be ejected each second to supply the thrust
(b)To give the rocket an initial upward acceleration of 18.3 m/s2 91.5 kg/s gas must be ejected each second to supply the thrust
Given: Mass of rocket (m) = 5860 kg Exhaust speed (v) = 1.17 km/s Initial upward acceleration (a) = 18.3 m/s²To calculate:
(a) Gas ejected each second to supply the thrust needed to overcome the weight of the rocket(b) Gas ejected each second to supply the thrust needed to give the rocket an initial upward acceleration of 18.3 m/s²(a) Gas ejected each second to supply the thrust needed to overcome the weight of the rocket:
Formula used:
Acceleration due to gravity (g) = 9.81 m/s²Thrust (T) = mg = mass of the rocket × acceleration due to gravity= 5860 kg × 9.81 m/s²= 57426.6 N (newton)
The thrust generated is equal to the force required to lift the rocket against the gravitational pull of the earth. Thus, the gas ejected each second will be equal to the thrust generated by the rocket divided by the exhaust speed of the rocket.= thrust generated / exhaust speed= 57426.6 N / (1.17 × 10³ m/s)= 49 kg/s
(b) Gas ejected each second to supply the thrust needed to give the rocket an initial upward acceleration of 18.3 m/s²:
Formula used:
Thrust (T) = ma Where ,m = mass of the rocket a = initial upward acceleration= 5860 kg × 18.3 m/s²= 107058 N
The gas ejected each second will be equal to the thrust generated by the rocket divided by the exhaust speed of the rocket.= thrust generated / exhaust speed= 107058 N / (1.17 × 10³ m/s)= 91.5 kg/s Hence, the gas ejected each second to supply the thrust needed to overcome the weight of the rocket is 49 kg/s and the gas ejected each second to supply the thrust needed to give the rocket an initial upward acceleration of 18.3 m/s² is 91.5 kg/s.
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A 85.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.538 meters off the ground. He hits the ground 0.0320 meters away from where he leapt. If the ball was moving horizontally when it was caught, how fast was the ball travelling?
_____m/s
A 85.0 kg football player leaps straight up into the air (with no horizontal velocity) to catch a pass. He catches the 0.430 kg ball precisely at the peak of his jump, when he is 0.538 meters off the ground. He hits the ground 0.0320 meters away from where he leapt.The ball was traveling approximately 0.097 m/s horizontally when it was caught.
To solve this problem, we can analyze the horizontal motion of the ball. Since the football player leaps straight up with no horizontal velocity, the horizontal distance traveled by the ball remains constant.
Given that the player hits the ground 0.0320 meters away from where he leapt and assuming the ball was caught at the same horizontal position, we can determine the horizontal speed of the ball.
Horizontal distance = Horizontal velocity * Time
The time it takes for the football player to reach the peak of his jump can be determined using the vertical motion equation:
Vertical distance = (1/2) * g * t^2
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and t is the time.
At the peak of the jump, the player's vertical velocity is momentarily zero. Using this information, we can solve for the time it takes for the player to reach the peak of his jump:
0.538 m = (1/2) * 9.8 m/s^2 * t^2
Simplifying:
t^2 = (2 * 0.538 m) / 9.8 m/s^2
t^2 ≈ 0.109
t ≈ √0.109
t ≈ 0.33 s (rounded to two decimal places)
Since the player catches the ball at the peak of his jump, the time it takes for the ball to reach that height is also approximately 0.33 seconds.
Now, we can calculate the horizontal velocity of the ball:
Horizontal distance = Horizontal velocity * Time
0.0320 m = Horizontal velocity * 0.33 s
Solving for the horizontal velocity:
Horizontal velocity = 0.0320 m / 0.33 s
Horizontal velocity ≈ 0.097 m/s (rounded to three decimal places)
Therefore, the ball was traveling approximately 0.097 m/s horizontally when it was caught.
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The maximum sustainable electric field strength in air is 3.0×106 V/m.
What is the magnitude, in volts, of the maximum potential difference between two parallel conducting plates separated by 0.59 cm of air?
The magnitude of the maximum potential difference between the two parallel conducting plates is 1770 volts.
What is the voltage of the maximum potential difference between two parallel conducting plates separated by 0.59 cm of air?The maximum sustainable electric field strength in air is given as 3.0×10⁶V/m. To calculate the maximum potential difference between the two conducting plates, we need to determine the electric field strength between them and then multiply it by the distance.
Given that the plates are separated by 0.59 cm (or 0.0059 m) of air, we can calculate the electric field strength using the equation E = V/d, where E represents the electric field strength, V represents the potential difference, and d represents the distance between the plates.
Rearranging the equation, we have V = E × d. Plugging in the values, the potential difference V is equal to (3.0×10⁶ V/m) × (0.0059 m), resulting in 1770 volts.
Therefore, the magnitude of the maximum potential difference between the two parallel conducting plates is 1770 volts.
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To keep the forces on the riders within allowable limits, many loop-the-loop roller coaster rides are designed so that the loop is not a perfect circle but instead has a larger radius of curvature at the bottom than at the top. Explain why this is so.
the centripetal force acting on the rider at the top is less than at the bottom, which means that the forces on the rider are kept within allowable limits.
Loop-the-loop roller coaster rides are designed in such a way that the loop is not a perfect circle but has a larger radius of curvature at the bottom than at the top in order to keep the forces on the riders within allowable limits. However, to understand the reason behind this design, we need to first understand what happens to the rider as he passes through the loop.
When the rider is at the top of the loop, the gravitational force acting on him is less than the centripetal force required to maintain the circular motion, so the net force acts upward. Hence, the rider feels lighter, or weightless. However, when he is at the bottom of the loop, the gravitational force acting on him is greater than the centripetal force, and hence, the net force acts downward, adding to the weight of the rider.
Therefore, to keep the forces acting on the rider within allowable limits, the radius of curvature at the bottom is made larger than at the top. This means that the speed of the roller coaster at the top of the loop is slower than at the bottom of the loop, as the radius of curvature is smaller.
Therefore, the centripetal force acting on the rider at the top is less than at the bottom, which means that the forces on the rider are kept within allowable limits.
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What is the minimum angular velocity (in rpmrpm ) for swinging a bucket of water in a vertical circle without spilling any
The minimum angular velocity (in rpm) required for this is given by the above formula. Let's put the values of g and r.g = 9.81 m/s² (acceleration due to gravity)
r = 0.5 m (radius of the circle)ω=√g/r = √(9.81/0.5) = 6.26 rad/sNow, we need to convert rad/s to rpm.
1 rad/s = 60/(2π) rpm6.26 rad/s = 60 × (6.26/2π) rpm≈ 60 × 1.00 rpm= 63 rpm (approx.)Hence, the minimum angular velocity (in rpm) required for swinging a bucket of water in a vertical circle without spilling any is approximately 63 rpm.
The minimum angular velocity (in rpm) for swinging a bucket of water in a vertical circle without spilling any can be calculated using the formula given below:
ω=√g/rWhere,
ω - minimum angular velocityg - acceleration due to gravityr - radius of the circleFor swinging a bucket of water in a vertical circle without spilling any, the centrifugal force must be equal to the gravitational force acting on the water in the bucket.
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what kinds of systems are most commonly described by schematics?
The systems that are most commonly described by schematics are electrical and mechanical systems. A schematic is a type of diagram that illustrates how a system works by showing the interconnections between components. It provides a visual representation of the system's architecture that is easy to read and interpret.
The schematic diagram displays the components of the system and their connections. Schematic diagrams can be used to explain complex systems in a simple way. They are particularly useful for illustrating how electrical and mechanical systems work.A mechanical schematic is a diagram that shows the mechanical components of a system, such as gears, belts, and pulleys. It provides a visual representation of how the system is designed to work, allowing engineers to identify potential problems and optimize the design. Electrical schematics, on the other hand, show the wiring and electronic components of a system. They provide a clear and concise representation of how electrical signals flow through the system, making it easy to identify potential problems and troubleshoot issues.
Schematic diagrams are used in a wide range of fields, from automotive and aerospace engineering to robotics and computer hardware design. They are an essential tool for engineers and designers who need to understand the inner workings of complex systems.
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Suppose a project with a negative net present value would provide intangible benefits. To estimate the annual value of intangible benefits needed to accept the project, ______ the negative net present value excluding intangible benefits by the ______.
A. multiply, present value of $1 factor
B. multiply, present value factor for an annuity
C. divide, present value of $1 factor
D. divide, present value factor for an annuity
By dividing the negative net present value by the present value factor for an annuity, we can estimate the annual value of intangible benefits needed to offset the negative net present value and make the project financially acceptable. Therefore, the correct answer is option D: divide, present value factor for an annuity.
To estimate the annual value of intangible benefits needed to accept a project with a negative net present value, we need to adjust the negative net present value by a certain factor. The factor used for this adjustment depends on the nature of the intangible benefits.
In this case, the question suggests that the intangible benefits can be estimated as an annual value. To adjust the negative net present value, we need to divide it by a certain factor.
The correct answer is option D: divide, present value factor for an annuity.
By dividing the negative net present value by the present value factor for an annuity, we can estimate the annual value of intangible benefits needed to offset the negative net present value and make the project financially acceptable.
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Two 2.0 kg bodies, A and B, collide. The velocities before the collision are and m/s.After the collision, What are (a) the final velocity of B and (b) the change in the total kinetic energy (including sign)
(a) The final velocity of body B after the collision is m/s.
(b) The change in the total kinetic energy is 0 J.
Step 1:
(a) The final velocity of body B after the collision can be determined.
(b) The change in the total kinetic energy, including the sign, can be calculated.
Step 2:
(a) To find the final velocity of body B after the collision, we need more information about the velocities before the collision. Please provide the velocities so that a specific answer can be given.
(b) The change in the total kinetic energy can be determined by calculating the difference between the initial kinetic energy and the final kinetic energy. However, without knowing the velocities before and after the collision, it is not possible to determine the change in kinetic energy.
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A boat is traveling upstream at 10 km/h with respect to the water of a river. The water is flowing at 9.0 km/h with respect to the ground. What are the (a) magnitude and (b) direction of the boat’s velocity with respect to the ground? A child on the boat walks from front to rear at 3.0 km/h with respect to the boat.What are the (c) magnitude and (d) direction of the child’s velocity with respect to the ground?
(a) the magnitude of the boat’s velocity with respect to the ground is 1 km/s, (b) the direction of the boat’s velocity with respect to the ground is opposite to the flow of the river, (c) the magnitude of the child’s velocity with respect to the ground is 2km/s and (d) direction of the child’s velocity with respect to the ground is along the flow of the river.
The relative speed of a body w.r.t to another body is the rate of change of separation between the two bodies with time.
Vab = Va- Vb, where Vab = relative speed of a w.r.t b, Va = speed of a, and Vb = speed of b.
Given: Speed of the boat upstream w.r.t river Vbr = 10 km/h
Speed of water of the river, Vr = - 9.0 km/h (negative sign shows direction opposite to boat)
speed of child w.r.t boat, Vcb = - 3.0 km/h
For upstream, the direction of the boat is opposite to the direction of the water of the river.
(a) Vbr = Vb - Vr, where Vb is the velocity of the boat w.r.t ground.
10 = Vb - (-9)
Vb = 1 km/s
(b) the direction of the boat w.r.t ground is opposite of the flow of the river.
(c) Vcb = Vc - Vb, where Vc is the velocity of child w.r.t ground
Vc = Vcb + Vb = - 3 + 1
Vc = -2 km/s
(d) Direction of the child's velocity is along the current of the river.
Therefore, (a) the magnitude of the boat’s velocity with respect to the ground is 1 km/s, (b) the direction of the boat’s velocity with respect to the ground is opposite to the flow of the river, (c) the magnitude of the child’s velocity with respect to the ground is 2km/s and (d) direction of the child’s velocity with respect to the ground is along the flow of the river.
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a certain jet is capable of a steady 20 degree climb. how much altitude does the jet gain when it moves 1 km through the air
The jet gains approximately 342 meters in altitude when it moves 1 km through the air at a steady 20-degree climb.
To determine the altitude gained by the jet when it moves 1 km through the air, we need to calculate the vertical displacement using trigonometry.
Given that the jet climbs at a steady 20 degrees, we can use the sine function to find the vertical displacement.
Vertical displacement = Distance * sin(angle)
In this case, the distance is 1 km (1000 m), and the angle is 20 degrees.
Vertical displacement = 1000 m * sin(20 degrees)
Using a calculator, we find that sin(20 degrees) is approximately 0.342.
Vertical displacement ≈ 1000 m * 0.342 ≈ 342 m
Therefore, the jet gains approximately 342 meters in altitude when it moves 1 km through the air at a steady 20-degree climb.
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Use Newton's method to approximate the value of
52−−√3523
as follows:
Let x1=3x1=3 be the initial approximation.
The second approximation x2x2 is
and the third approximation x3x3 is
The second approximation x2 is approximately 588.67, and the third approximation, x3, is approximately 297.01.
To use Newton's method to approximate the value of √(3523), we start with an initial approximation x1 = 3.
The formula for the next approximation, x2, is given by:
x2 = x1 - f(x1)/f'(x1)
where f(x) = x^2 - 3523 and f'(x) is the derivative of f(x).
Let's calculate the second approximation, x2:
f(x) = x^2 - 3523
f'(x) = 2x
Plugging in x1 = 3:
x2 = x1 - f(x1)/f'(x1)
= 3 - (3^2 - 3523)/(2*3)
= 3 - (9 - 3523)/6
= 3 - (-3514)/6
= 3 + 585.67
= 588.67
So, the second approximation x2 is approximately 588.67.
To find the third approximation, x3, we repeat the process using x2 as the new initial approximation:
x3 = x2 - f(x2)/f'(x2)
Plugging in x2 = 588.67:
x3 = x2 - f(x2)/f'(x2)
= 588.67 - (588.67^2 - 3523)/(2*588.67)
= 588.67 - (346780.8489 - 3523)/(1177.34)
= 588.67 - (343257.8489)/1177.34
= 588.67 - 291.66
= 297.01
Therefore, the third approximation, x3, is approximately 297.01.
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A diver of mass m drops from a board 10 m above the water surface. Neglect air resistance. What is the gain of kinetic energy at 5 m above ground with respect to the starting point at 10m above ground?
The gain of kinetic energy at 5 m above ground will be mg(h - H) joules.
Given that:
Initial height, h = 10 m
Final height, H = 5 m
Initial velocity, u = 0 m/s
The mechanical energy is the sum of potential energy (PE) and kinetic energy (KE). At the starting point, when the diver is 10 m above the water surface, all the energy is in the form of potential energy since the diver is at rest. Then the equation is given as,
0.5mu² + mgh = 0.5mv² + mgH
mgh = KE(gain) + mgH
KE = mgh - mgH
KE = mg(h - H)
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Your coworker is supposed to grab the packages as they arrive at the top of the ramp. But she misses one and it slides back down. What is its speed when it returns to you
The speed of the package when it returns to you will depend on the specific conditions of the ramp and any external forces. Without additional information, we cannot determine the exact speed of the package when it returns.
When the package is released from the top of the ramp, it has an initial speed that depends on factors such as the height of the ramp and the angle of the incline. However, since the question does not provide specific values for these factors, we cannot calculate the exact speed.
In general, when an object slides down a ramp without any external forces acting on it (such as friction or air resistance), it will conserve mechanical energy. This means that the total mechanical energy of the object (kinetic energy + potential energy) remains constant.
As the package slides down the ramp, it converts its initial potential energy into kinetic energy. When it reaches the bottom and starts moving back up, it will convert its kinetic energy back into potential energy. At the highest point of its return, it will momentarily come to rest before starting to slide back down.
At this highest point of return, the package's speed is zero since it has momentarily stopped. As it starts sliding back down, it will regain speed, and its speed at any given point will depend on the specific conditions of the ramp and any external forces.
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true false If two different lenses photograph the same scene, the perspective relations in the images they capture will likely be the same.
The given statement, "If two different lenses photograph the same scene, the perspective relations in the images they capture will likely be the same." is false. Different lens will have different perspective and the image will not be similar.
The connections between the items in a picture, especially their relative sizes, locations, and distances between them, are all part of perspective in an image. Photographers may alter perspective in a variety of ways to affect how the picture appears to have depth and space as well as to give the image a feeling of scale.
A common misconception among photographers is that wide-angle lenses offer a different viewpoint than, example, telephoto lenses. However, in reality, the perspective itself is not affected by the focal length of the lens; rather, the perspective varies as the camera location or viewpoint does. Even though it drastically modifies the size of objects in the image, adjusting the lens focal length without also changing the camera location has no effect on the perspective in the scene.
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At a distance of 4.3 cm from the center of a very long uniformly charged wire, the electric field has magnitude 2000 N/C and is directed toward the wire. What is the charge on a 1.0 cm length of wire near the center
The sign of charge density is negative and the magnitude is [tex]4.78\times10^{-9}\,C/m[/tex].
The charge near the center of the wire is [tex]q=\lambda[/tex]
[tex]l=-4.78\times10^{-9}\times(0.01)[/tex]
=[tex]-4.78\times10^{-11}\,C[/tex]
Electric field due to long uniformly charged wire is[tex]E=\frac{\lambda}{2\pi\epsilon_0 r}[/tex]
{\lambda} is charge density, and r is the distance from the center of the wire.
[tex]E=2000\,N/C , r=4.3\,cm=0.043\,m ,[/tex]
A) Direction of the electric field is towards the wire, which means the line charge is negatively charged.
[tex]E=\frac{\lambda}{2\pi\epsilon_0 r}[/tex]
[tex]\lambda=2\pi\epsilon_0 r[/tex]
[tex]E=2\pi(8.85\times10^{-12})(0.043)(2000)[/tex]
=[tex]4.78\times10^{-9}\,C/m[/tex]
Charge density is [tex]\lambda[/tex][tex]=-4.78\times10^{-9}\,C/m[/tex]
The sign of charge density is negative and the magnitude is [tex]4.78\times10^{-9}\,C/m[/tex]
B) Charge density = charge/ length
[tex]\lambda=\frac{q}{l}[/tex]
The charge near the center of the wire is [tex]q=\lambda[/tex]
[tex]l=-4.78\times10^{-9}\times(0.01)[/tex]
=[tex]-4.78\times10^{-11}\,C[/tex]
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A 1,500 kg car sliding on frictionless ice at 15 m/s hits a stationary, 2,500 kg minivan. The two vehicles are locked together after impact on the ice. What is their common speed after impact
The common speed of the car and minivan after impact on the frictionless ice is 6 m/s.
How can we calculate the common speed of the car and minivan after impact?When the car and minivan collide, the principle of conservation of momentum can be applied to determine their common speed after impact. According to this principle, the total momentum before the collision is equal to the total momentum after the collision, assuming there are no external forces acting on the system.
The momentum (p) of an object is given by the product of its mass (m) and velocity (v): p = m * v.
Initially, the car has a mass of 1,500 kg and a velocity of 15 m/s, while the minivan has a mass of 2,500 kg and is stationary (0 m/s). After the collision, the two vehicles lock together and move with a common velocity (v').
Applying the conservation of momentum equation:
Initial momentum of the car = Final momentum of the car and minivan
(1,500 kg * 15 m/s) = (1,500 kg + 2,500 kg) * v'
Simplifying the equation:
22,500 kg⋅m/s = 4,000 kg * v'
Dividing both sides by 4,000 kg:
v' = 22,500 kg⋅m/s / 4,000 kg
v' = 5.625 m/s
Therefore, the common speed of the car and minivan after impact is 5.625 m/s, which can be rounded to 6 m/s.
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