Answer:
d = 52.20 miles
Explanation:
Given that,
Devon drove 50 miles west and then drove 15 miles north.
We need to find Devon's displacement. Let it is d. It can be calculated as follows :
[tex]d=\sqrt{50^2+15^2} \\\\=52.20\ \text{miles}[/tex]
Hence, Devon's displacement is 52.20 miles.
Which of the following is a definition of motion?
Answer:
sorry I don't get it so pls can u repeat the question .
Movement of any object from one position to another position with respect to the observer is called as Motion. Motion Along a Straight Line: When an object moves along a straight line, the motion of the object is called rectilinear motion. For example; motion of a car on highway.
QUESTION:
Which of the following is a definition of motion?
ANSWER:
In physics, motion is the phenomenon in which an object changes its position over time. Motion is mathematically described in terms of displacement, distance, velocity, acceleration, speed, and time.
What is the magnitude of the velocity after it hits the
ground?
9.3 m/s
12 m/s
41 m/s
73 m/s
Answer: 9.3m/s
Explanation:
Your question isn't complete but let me help out:
A 0.060 kg ball hits the ground with a speed of –32 m/s. The ball is in contact with the ground for 45 milliseconds and the ground exerts a +55 N force on the ball. What is the magnitude of the velocity after it hits the ground?
We would use Newton's law of motion to solve this which goes thus:
F = ma
f = m(v-u)/t
Cross multiply
ft = m(v-u)
where,
f = 55
t = 45/1000 = 0.045
m = 0.0060
u = -32
v = Unknown
Therefore,
55 × 45/1000 = 0.060(v - -32)
55 × 0.045 = 0.060(v + 32)
2.475 = 0.06(v + 32)
2.475 = 0.06v + 1.92
0.06v = 2.475 - 1.92
0.06v = 0.555
v = 0.555/0.06
v = 9.25m/s
v = 9.3m/s Approximately
Answer:
A.9.3 m/s
Explanation:
Two balls with masses of 2.0 kg and 6.0 kg travel toward each other at speeds of 12 m/s and 4.0 m/s, respectively. If the balls have a head-on, inelastic collision and the 2.0-kg ball recoils with a speed of 8.0 m/s, how much kinetic energy is lost in the collision
Answer:
The kinetic energy lost in the collision is 48 J
Explanation:
Given;
mass of the first ball, m₁ = 2.0 kg
mass of the second ball, m₂ = 6.0 kg
initial speed of the first ball, u₁ = 12 m/s
initial speed of the second ball, u₂ = 4 m/s
let v be the final velocity of the two balls after the inelastic collision
Apply the principle of conservation of linear momentum;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
2 x 12 + 6 x 4 = v(2 + 6)
48 = 8v
48 / 8 = v
v = 6 m/s
The initial kinetic energy of the balls is calculated as;
K.E₁ = ¹/₂m₁u₁² + ¹/₂m₂u₂²
K.E₁ = ¹/₂(2)(12²) + ¹/₂(6)(4)²
K.E₁ = 144 + 48
K.E₁ = 192 J
The final kinetic of the balls is calculated as;
K.E₂ = ¹/₂(m₁ + m₂)(v²)
K.E₂ = ¹/₂(2 + 6)(6²)
K.E₂ = ¹/₂(8)(6²)
K.E₂ = 144 J
The lost in kinetic energy of the balls is K.E₂ - K.E₁ = 144 J - 192 J = -48 J
Therefore, the kinetic energy lost in the collision is 48 J
The sound level at a distance of 1.48 m from a source is 120 dB. At what distance will the sound level be 70.7 dB?
Answer:
The second distance of the sound from the source is 431.78 m..
Explanation:
Given;
first distance of the sound from the source, r₁ = 1.48 m
first sound intensity level, I₁ = 120 dB
second sound intensity level, I₂ = 70.7 dB
second distance of the sound from the source, r₂ = ?
The intensity of sound in W/m² is given as;
[tex]dB = 10 Log[\frac{I}{I_o} ]\\\\For \ 120 dB\\\\120 = 10Log[\frac{I}{1*10^{-12}}]\\\\12 = Log[\frac{I}{1*10^{-12}}]\\\\10^{12} = \frac{I}{1*10^{-12}}\\\\I = 10^{12} \ \times \ 10^{-12}\\\\I = 1 \ W/m^2[/tex]
[tex]For \ 70.7 dB\\\\70.7 = 10Log[\frac{I}{1*10^{-12}}]\\\\7.07 = Log[\frac{I}{1*10^{-12}}]\\\\10^{7.07} = \frac{I}{1*10^{-12}}\\\\I = 10^{7.07} \ \times \ 10^{-12}\\\\I = 1 \times \ 10^{-4.93} \ W/m^2[/tex]
The second distance, r₂, can be determined from sound intensity formula given as;
[tex]I = \frac{P}{A}\\\\I = \frac{P}{\pi r^2}\\\\Ir^2 = \frac{P}{\pi }\\\\I_1r_1^2 = I_2r_2^2\\\\r_2^2 = \frac{I_1r_1^2}{I_2} \\\\r_2 = \sqrt{\frac{I_1r_1^2}{I_2}} \\\\r_2 = \sqrt{\frac{(1)(1.48^2)}{(1 \times \ 10^{-4.93})}}\\\\r_2 = 431.78 \ m[/tex]
Therefore, the second distance of the sound from the source is 431.78 m.
When landing from a jump, a basketball player of mass 82 kg has a velocity of 1.2 m/s right before they hit the ground. The player then lands on the floor with their feet and is quickly brought to a stop. If the the floor deforms by 0.025 m while the player lands on it, what was the average force under the player's feet during the landing
Answer:
2361.6N
Explanation:
Mass of player = 82kg
Velocity = 1.2m/s
Kinetic energy of player:
= 1/2mv²
= 1/2*82*1.2²
= 41x1.44
= 59.04J
Final kinetic energy = 0
Change in kinetic energy
|∆k| = |0-59.04|
= 59.04
Workdone by the feet = fd
d = 0.025
Fd = 59.04
F = 59.04/0.025
= 2361.6N
This is his average force.
These steps are involved in the production of work by a four-stroke heat engine. Place the steps in the correct order. Piston moves up, compressing gas and causing gas to increase its temperature: Piston moves up, pushing burned gases through exhaust valve: 1 Spark plug fires, causing combustion as gas is still compressed; increase in energy from combustion pushes piston down: Piston moves down, allowing fuel-air mixture to enter intake valve:
Answer:
correct order 2,4,3,1
Explanation: correct on edge
Which type of circuit would be best to use for lights used for decorations? Question 1 options: Series circuit. One bulb could go out and the strand will stay on. Series circuit. One bulb could go out and the rest go out. Parallel circuit. One bulb goes out and the rest go out. Parallel circuit. One bulb could go out and the strand will stay on.
Answer:
One bulb could go out and the strand will stay on.
Explanation:
In series circuit, there is only one path provided for the current to flow. So, all the lights are required to be in working condition, for the others to work. And if anyone light bulb goes out, the circuit will become incomplete and the rest of the strand will also go out. Because there is only one path for current flow which is now broken.
On the other hand, in parallel circuits, each light bulb has a separate connection with the source. Current path to each bulb is independent of the others. Therefore, if one bulb goes out, the rest of the strand will stay on.
So, the correct option is:
One bulb could go out and the strand will stay on.
Question 1 of 25
Two asteroids with masses 3.71 x 10 kg and 1.88 x 104 kg are separated by
a distance of 1,300 m. What is the gravitational force between the asteroids?
Newton's law of gravitation is F gravity
Gm, 2 The gravitational
constant Gis 6.67 x 10-11 Nm²/kg?
A. 275 x 10"N
B. 4.13 x 10°N
C. 2.04 x 10°N
O D. 3.58 x 10-N
SUBMIT
Answer:
(A)
Explanation:
Answer:
275 x 10"N
Explanation:
A harmonic oscillator starts with an amplitude of 20.0 cm. After 10.0 s, the amplitude decreases to 15.0 cm. If the linear damping coefficient is 2.00 Ns/m, how much mass is oscillating
Answer: the amount of mass is oscillating is 34.8 kg
Explanation:
Given that;
amplitude A = 20.0 cm
time t = 10 s
amplitude decreases x = 15.0 cm
damping coefficient b = 2.00 N.s/m
amount of mass is oscillating = ?
we know that; amplitude can be expressed as;
x = Ae^-(∝t)
we substitute
15 = 20e^-∝(10)
∝ = 0.02877 s⁻¹
Hence mass m will be;
m = b/2∝
we substitute
m = (2 N.s/m) / ( 2 × 0.02877 s⁻¹)
m = 34.8 kg
Therefore the amount of mass is oscillating is 34.8 kg
A television camera is positioned 4 km from the base of a rocket launching pad. In order to keep the rocket in focus as it takes off, the camera must be programmed with the distance to the rocket and how that distance is changing. If we assume that the rocket rises vertically at a speed of 200 km/hr, how fast is the distance from the camera to the rocket changing when the rocket has risen 3 km
Answer:
120 km/hr
Explanation:
Let D be the distance between the rocket and the camera as the rocket is moving upwards. Let d be the distance the rocket moves and L be the distance between the camera and the base of the rocket = 4 km.
Now, at any instant, D² = d² + L²
= d² + 4²
= d² + 16 since the three distances form a right-angled triangle with the distance between the rocket and the camera as the rocket is moving upwards as the hypotenuse side.
differentiating the expression to find the rate of change of D with respect to time, dD/dt ,we have
d(D²)/dt = d(d² + 16)/dt
2DdD/dt = 2d[d(d)/dt]
dD/dt = 2d[d(d)/dt] ÷ 2D
Now d(d)/dt = vertical speed of rocket = 200 km/hr
dD/dt = 200d/D [D = √(d² + 16)]
dD/dt = 200d/[√d² + 16]
Now substituting d = 3 km, the distance the rocket has risen into the equation, we have
dD/dt = 200(3)/[√(3² + 16)]
dD/dt = 600/[√(9 + 16)]
dD/dt = 600/√25
dD/dt = 600/5
dD/dt = 120 km/hr
So, the speed at which the distance from the camera to the rocket changing when the rocket has risen 3 km is 120 km/hr.
Is anyone good at science I need help with 2 tests
Answer:
i am!
Explanation:
A wheel starts from rest and rotates with constant angular acceleration to reach an angular speed of 11.9 rad/s in 3.10 s. (a) Find the magnitude of the angular acceleration of the wheel.
Answer:
The value is [tex]\alpha = 3.84 \ rad/s^2[/tex]
Explanation:
From the question we are told that
The constant angular speed is [tex]w = 11.9 \ rad/s[/tex]
The time taken is [tex]t = 3.10 \ s[/tex]
Generally the magnitude of the angular acceleration is mathematically represented as
[tex]\alpha = \frac{w}{t}[/tex]
=> [tex]\alpha = \frac{11.9}{ 3.10 }[/tex]
=> [tex]\alpha = 3.84 \ rad/s^2[/tex]
An observer stands 150 meters from a fireworks display rocket, which is fired directly upward. When the rocket reaches a height of 200 meters, it is traveling at a speed of 12 meters/second. At what rate is the angle of elevation formed with the observer increasing at that instant
Answer:
[tex]-0.288\ \text{rad/s}[/tex]
Explanation:
x = Distance of observer from the initial location of the rocket = 150 m
y = Vertical displacement of the rocket from the ground = 200 m
r = Distance between observer and rocket
[tex]\dfrac{dy}{dt}[/tex] = Rate of change in height of rocket = 12 m/s
[tex]\dfrac{dx}{dt}[/tex] = Rate of change in x = 0
Distance between observer and rocket at y = 200 m
[tex]r=\sqrt{x^2+y^2}\\\Rightarrow r=\sqrt{150^2+200^2}\\\Rightarrow r=250\ \text{m}[/tex]
[tex]\tan\theta=\dfrac{y}{x}[/tex]
Differentiating with respect to time
[tex]\sec^2\theta\dfrac{d\theta}{dt}=\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{y\dfrac{dx}{dt}-x\dfrac{dy}{dt}}{x^2}}{\sec^2\theta}\\\Rightarrow \dfrac{d\theta}{dt}=\dfrac{\dfrac{0-150\times 12}{150^2}}{(\dfrac{250}{150})^2}\\\Rightarrow \dfrac{d\theta}{dt}=-0.288\ \text{rad/s}[/tex]
The rate of change of the angle of elevation is [tex]-0.288\ \text{rad/s}[/tex].
Find the resultant of the following displacement:
A = 20 Km 30° south of east;
B = 50 Km due west;
C = 40 Km north east;
D = 30 Km 60° south of west.
Answer:
Explanation:
Basically you just have to find the left vectors. To do so divide A, C and D into horizontal and vertical vector. A: 10km to south and 10root3 to east. Just sine and cosine of 30 at 20km. D: 15 km to west and 15root3 to south. Again sine and cosine of 60 at 30 km. C: 45 degrees so 20root2 to north and east. Add all these up with B. Then you have 7.696 km due south and 19.395 km due west. Resultant displacement magnitude = root(7.696^2+19.395^2)=20.866 south of west with angle=arctan(7.696/19.395)=21.644 degrees
An object that floats in water weighs 20 N in air.
a. What is the weight of the object in water?
b.What is the Upthrust acting on the object in water?
c. What is the weight of the water displaced by the object?
Answer:
a. Weight of Object in Water = 20 N
b. Up thrust = 20 N
c. Weight of Water Displaced = 20 N
Explanation:
a.
The weight of the object remains same in the water as well. Because, the same force of gravity is acting there as well. Hence,
Weight of Object in Water = 20 N
b.
Since, the object floats on the water. Therefore, according to Archimedes' principle the up thrust force acting on the object must be equal to the weight of object:
Up thrust = Weight of object
Up thrust = 20 N
c.
From Archimedes' Principle, we know that the up thrust or the Buoyant force is equal to the weight of the water displaced by the object. therefore:
Weight of Water Displaced = Up thrust
Weight of Water Displaced = 20 N
What is the significance of Nucleotides in Chromosomes?
Answer:
it comprises of the DNA/RNA bipolymer molecules
In order to model the motion of an extinct ape, scientists measure its hand and arm bones. From shoulder to wrist, the arm bones are 0.60 m long and their mass is 4.0 kg. From wrist to the tip of the fingers, the hand bones are 0.10 m long and their mass is 1.0 kg. In the model above, each bone is assumed to have a uniform density. When the arm and hand hang straight down, the distance from the shoulder to the center of mass of the arm-hand system is most nearly
Answer:
0.37 m
Explanation:
Let the shoulder be the origin.
The center of mass of the arm bones is 0.60 m/2 = 0.30 m and the center of mass of the hand bones is 0.10 m/2 = 0.05 m since they are modeled as straight rods with uniform density and the center of mass of a rod is x = L/2 where L is the length of the rod.
The center of mass y = (m₁y₁ + m₂y₂)/(m₁ + m₂) where m₁ = mass of arm bones = 4.0 kg, y₁ = distance center of mass of arm bones from shoulder = 0.30 m, m₂ = mass of hand bones = 1.0 kg and y₂ = distance of center of mass hand bones from shoulder = x₁ + distance of center of hand bones from wrist = 0.60 m + 0.05 m = 0.65 m
Substituting these into the equation for the center of mass, we have
y = (m₁y₁ + m₂y₂)/(m₁ + m₂)
y = (4.0 kg × 0.30 m + 1.0 kg × 0.65 m)/(4.0 kg + 1.0 kg)
y = (1.20 kgm + 0.65 kgm)/5.0 kg
y = 1.85 kgm/5.0 kg
y = 0.37 m
The distance of the center of mass from the shoulder is thus y = 0.37 m
Which is not true about waves?
O Compression and rarefaction are found in longitudinal waves.
O Sound waves are transverse waves.
O Amplitude, wavelength, and troughs all describe parts of a wave.
O Electromagnetic waves are transverse waves.
Answer:
sound waves are transverse waves
a guitar string is 0.620 m long, and oscillates at 234 Hz. if a player uses his finger to shorten the string to 0.480 m, what is the new frequency?
Answer:
the new frequency is : 302.25Hz
using the formula F2 = [tex]\frac{F1 L1}{L2}[/tex]
Explanation:
meaning of frequencyfrequency of a string is the number of vibrations of a plucked string per second. it is measured in Hertz.
the frequency of a string is inversely proportional to twice the length of the string. which means the longer the string, the smaller the freqency and the higher the string the higher the frequency.
f ∝ 1/2L.
f = k/2L
where f = frequency
L = length of string
k = constant
k = 2fL is a constant
given data
L1 = 0.62m
f1 = 234Hz
L2 = 0.48m
2f1L1 = 2f2L2
f1L1 = f2L2
f2 = f1L1/L2
f2 = [tex]\frac{234 x 0.62}{0.48}[/tex] = 302.25Hz
in conclusion, the new frequency is 302.25Hz
learn more about frequency of a vibrating string: brainly.in/question/1149252
SPJ2
An object that is initially traveling at 17.6 m/s has
1,743 J of work performed on it. If the mass of the
object is 283 kg, its final kinetic energy will be
J.
1. What is the total distance traveled?
A 3.0m
B 4.0m
C 5.0m
D 6.0m
Answer:
c
Explanation:
If vector A= aj^ and vector B = bj^, then vector A×B is equal to
Answer:
0 because j×j =0
Explanation:
so a×b is 0
As a freely falling object speeds up, what is happening to its acceleration - does it increase, decrease, or stay the same? (a) Ignore air resistance. (b) Consider air resistance.
Essay ANSWER
Answer:
(a)When ignoring air resistance its accelerating increases steadily .
(b)When considering air resistance then its acceleration decreases this could either be uniformly or unevenly.
Hope this helped.
In Space, an astronaut releases a wrench from his hand. The wrench has a mass of 4 grams and is traveling with a velocity of -15m/s. The Astronaut’s mass is 70kg. What is his Velocity?
Answer:
[tex]v_=-8.5\times 10^{-4}\ m/s[/tex]
Explanation:
Given that,
Mass of a wrench, m₁ = 4 g = 0.004 kg
Speed of wrench, v₁ = -15 m/s
Mass of the Astronaut, m₂ = 70 kg
We need to find Astronaut's velocity. Let it is v₂. Using the conservation of linear momentum to find it.
[tex]m_1v_1=m_2v_2\\\\v_2=\dfrac{m_1v_1}{m_2}\\\\v_2=\dfrac{0.004\times (-15)}{70}\\\\=-8.5\times 10^{-4}\ m/s[/tex]
So, the speed of Astronaut is [tex]8.5\times 10^{-4}\ m/s[/tex].
How much work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth?
s = 40m - 4m = 36m
W = F × s
= 10N × 36m = 360J
A bit of explanation :W = Work (J)
F = Force / weight (N)
s = distance (m)
Work done in physics is the product of force and displacement. The displacement for the object is 36 m and force acts on it is 10 N. Then the work done is 360 J.
What is work done?Work done is the dot product of force acting on a body and the resultant displacement. When a force applied on an object results in a displacement from its position, the force is said to be work done.
Work done is a vector quantity thus, characterised by a magnitude and direction. The common unit of work done is joule.
Given that force applied on the weight = 10 N
displacement occurred = 40 m - 4 m = 36 m
Work done = F . ds
ds = 36 m and f = 10 N
Then W = 10 N × 36m
= 360 J.
Therefore, the work is required to lift a 10-newton weight from 4.0 meters to 40 meters above the surface of Earth is 360 J.
To find more on work done, refer here:
https://brainly.com/question/13662169
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PLEASE HELP!!! Which process in living things evolved as oxygen levels increased in Earth’s atmosphere?
glycolysis
photosynthesis
fermentation
aerobic respiration
Answer: Areobic Resporation
Explanation; -Aerobic respiration is the process in living things that evolved as oxygen levels increased in earth's atmosphere.
True or False: Solids always have a higher density than liquids and gases.
Answer:
TRUE THE ANSWER IS TRUE
draw simple vector diagram and resolve 60N at an angle of 30° from the horizontal.. plz help guys
Explanation:
The attatched figure shows the vector diagram for a force that has magnitude of 60 N and it is acting at an angle of 30° from the horizontal.
When it is resolved, the horizontal and vertical components are given by :
[tex]F_x=F\cos\theta\\\\=60\times \cos30\\\\=51.96\ N[/tex]
And
[tex]F_y=F\sin\theta\\\\=60\times \sin30\\\\=30\ N[/tex]
Hence, this is the required solution.
A jet aircraft is traveling at 219 m/s in horizontal flight. The engine takes in air at a rate of 77.9 kg/s and burns fuel at a rate of 4.19 kg/s. The exhaust gases are ejected at 934 m/s relative to the aircraft. Find the thrust of the jet engine. Answer in units of N. 019 (part 2 of 2) 10.0 points Find the delivered power. Answer in units of W
Answer: thrust [tex]-85905.24N[/tex]
delivered power [tex]= -18813247.56W[/tex]
Explanation:
given data:
horizontal speed = 219 m/s
rate of air intake = 77.9 kg/s
fuel combustion rate = 4.19 kg/s.
exhaust gases ejected = 934 m/s
solutuon:
Thrust of the engine
[tex]Fth = Ve * ( Mf - Ma ) - Vj *Ma[/tex]
[tex]= 934 * ( 4.19 - 77.9 ) - 219 *77.9[/tex]
[tex]= -68845.14 - 17060.1[/tex]
=[tex]-85905.24N[/tex]
delivered power
[tex]= Fth * Vj[/tex]
[tex]= -85905.24N * 219[/tex]
[tex]= -18813247.56W[/tex]
Given:
Horizontal speed = 219 m/sRate of air = 44.9 kg/sFuel combustion rate = 4.19 kg/sExhaust gas ejected = 934 m/sThrust of the engine will be:
→ [tex]F^{th} = Ve(Mf-Ms)-Vj\times Ma[/tex]
By putting the values, we get
[tex]= 934(4.19-77.9)-219\times 77.9[/tex]
[tex]= -68845.14-17060.1[/tex]
[tex]= -85905.24 \ N[/tex]
hence,
The delivered power will be:
= [tex]F^{th}\times Vj[/tex]
= [tex]-85905.24\times 219[/tex]
= [tex]-18813247.56 \ W[/tex]
Thus the solution above is right.
Learn more about power here:
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Question 1 of 10
A wave meets a large barrier that has a small opening in it. The part of the
wave that meets the opening bends as it passes through. Which statement
best describes what has happened to the wave?
OA. The opening in the barrier absorbed all of the wave's energy.
OB. The wave was reflected as it passed through the opening in the
barrier.
OC. The opening in the barrier caused the wave to change speed and
refract.
OD. The wave diffracted as it passed through the opening in the
barrier.
Answer:
OD. The wave diffracted as it passed through the opening in the barrier.
Explanation:
A progressive wave (i.e waves in motion) has the capacity to bend around an obstacle on its path. This is one of the general properties of waves called diffraction. Others are: reflection, refraction, interference. Note that only transverse waves undergo polarization.
Diffraction of waves is the ability of waves to bend around an obstacle on its path during progression.
Thus, the bending of the part of waves as it passes through the barrier implies that the wave diffracted as it passed through the opening in the barrier.
Answer:
The wave diffrected as it passed through the opening
Explanation:
