You push on a refrigerator with a force of 20 N and cause the refrigerator to accelerate at 2 m/s/s. What is the refrigerator's mass in kg

Distinguishing and mastering different motions is to differentiate individual movements, while linking individual motions into a coherent, coordinated whole is to integrate movements.

Distinguishing and mastering different motions involves the ability to identify and understand separate movements or actions. It requires recognizing the unique characteristics and qualities of each motion and developing proficiency in performing them individually. This process helps in refining specific skills and techniques.

On the other hand, linking individual motions into a coherent, coordinated whole refers to the integration of these separate movements into a smooth and connected sequence. It involves transitioning seamlessly between different motions, combining them in a harmonious manner, and maintaining continuity and flow.

This integration is crucial for achieving fluidity, efficiency, and effectiveness in complex activities or performances that require multiple movements.

In summary, distinguishing and mastering different motions focuses on recognizing and developing proficiency in individual movements, while linking individual motions into a coherent, coordinated whole emphasizes the integration and seamless combination of these separate movements to create a unified and synchronized performance.

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The time of motion will be longer when the body ascends along the inclined plane with an initial velocity such that it stops at the same height h.

When the body moves down the inclined plane from a height h, it experiences the force of gravity acting in the direction of motion, which helps to increase its speed. This results in a shorter time of motion compared to the ascent.

On the other hand, when the body ascends along the inclined plane, it needs to work against the force of gravity. The initial velocity is crucial in determining whether the body can reach the same height h or not. If the initial velocity is not sufficient, the body will slow down and eventually come to a stop before reaching the height h. In this case, the time of motion will be longer as the body needs to work against gravity and overcome the height difference.

Therefore, the time of motion is longer when the body ascends along the inclined plane with an initial velocity such that it stops at the same height h.

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The resistance of each resistor in the circuit is 13.75 ohms, and the potential difference across each resistor is 2.1875 V.

To find the resistance of each resistor, we can use Ohm's Law, which states that resistance (R) is equal to the voltage (V) divided by the current (I).

Since the resistors are connected in series, the total resistance of the circuit is equal to the sum of the resistances of the individual resistors. Therefore, we divide the total voltage (12.0 V) by the total current (0.873 A) to find the resistance of each resistor.

Next, to determine the potential difference across each resistor, we can apply Ohm's Law again. Since the resistors are in series, the potential difference across each resistor is the same as the total potential difference.

Therefore, we divide the total voltage (12.0 V) by the number of resistors (5) to find the potential difference across each resistor.

By performing the calculations, we find that the resistance of each resistor is 13.75 ohms and the potential difference across each resistor is 2.1875 V.

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When the light from a galaxy 10 billion light years away leaves our galaxy, the universe is 4 billion years old.

What age was the universe when the light we see departed that galaxy?

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The potential difference decreases because as plate separation increases, the electric field between the plates decreases. The correct answer is option C.

As the plates of a capacitor are slowly pulled apart, the distance between them increases. This leads to a decrease in the electric field strength between the plates. Since the potential difference is directly related to the electric field strength, a decrease in the electric field results in a decrease in the potential difference between the plates. The charges on the plates remain constant, so the change in plate separation does not directly affect the amount of charge stored. Therefore, option C correctly indicates the change in the potential difference between the plates.

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The velocity of the wood and the bullet as they leave the target area is approximately 28.78 m/s, with both objects moving together at the same velocity.

To determine the velocity of the wood and the bullet as it leaves the target area, we can apply the principle of conservation of momentum. According to this principle, the total momentum before the collision is equal to the total momentum after the collision.
Given: Mass of the bullet (m_bullet) = 0.05 kg

Velocity of the bullet before collision (v_bullet) = 1180 m/s

Mass of the wood block (m_wood) = 2 kg

Velocity of the wood block before collision (v_wood) = 0 m/s (assuming it is initially at rest)

Let's denote the velocity of the bullet and wood after the collision as v_bullet' and v_wood', respectively. According to the conservation of momentum: Initial momentum = Final momentum
The initial momentum is given by:

Initial momentum = (mass of the bullet) × (velocity of the bullet before collision) + (mass of the wood block) × (velocity of the wood block before collision)

= m_bullet × v_bullet + m_wood × v_wood

= 0.05 kg × 1180 m/s + 2 kg × 0 m/s

= 59 kg·m/s
The final momentum is given by:

Final momentum = (mass of the bullet) × (velocity of the bullet after collision) + (mass of the wood block) × (velocity of the wood block after collision) = m_bullet × v_bullet' + m_wood × v_wood'
Since the bullet lodges into the wood block, they move together with the same final velocity v_common:

v_bullet' = v_wood' = v_common

Therefore, we have:

Final momentum = (mass of the bullet + mass of the wood block) × (velocity of the bullet and wood after collision)

= (m_bullet + m_wood) × v_common

Using the conservation of momentum principle, we can equate the initial and final momentum: 59 kg·m/s = (m_bullet + m_wood) × v_common

59 kg·m/s = (0.05 kg + 2 kg) × v_common

59 kg·m/s = 2.05 kg × v_common

Solving for v_common:

v_common = 59 kg·m/s / 2.05 kg

v_common ≈ 28.78 m/s

Therefore, the velocity of the wood and the bullet as they leave the target area is approximately 28.78 m/s, with both objects moving together at the same velocity.

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The magnitude of the impulse the ground exerts on the ball is equal to 2 times the product of the mass of the ball and its initial velocity: |J| = 2mv.

The impulse (J) experienced by an object is defined as the change in momentum (Δp) and can be calculated using the equation:

J = Δp = m * Δv

where m is the mass of the object and Δv is the change in velocity.

In this case, when the ball strikes the ground, the change in velocity is equal to the initial velocity of the ball (v) because it rebounds in the opposite direction.

Assuming the initial velocity of the ball before it strikes the ground is v and the final velocity after rebounding is -v (since it rebounds in the opposite direction), the change in velocity (Δv) is given by:

Δv = -v - v = -2v

Now, let's assume the mass of the ball is m.

Using the equation for impulse, we have:

J = m * Δv = m * (-2v) = -2mv

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The given statement "Hubble Space Telescope is the largest optical telescope ever made" is False because the Hubble Space Telescope is an iconic and remarkable telescope, it is not the largest optical telescope ever made

The Hubble Space Telescope (HST) is a renowned space-based observatory that has made significant contributions to our understanding of the universe. However, in terms of size, it is not the largest optical telescope. The HST has a primary mirror with a diameter of 2.4 meters (7.9 feet), which is relatively small compared to some ground-based optical telescopes.

The honor of being the largest optical telescope goes to several ground-based observatories. For example, the Gran Telescopio Canarias (GTC) located in Spain has a primary mirror with a diameter of 10.4 meters (34.1 feet). Additionally, the Keck Observatory in Hawaii consists of two telescopes, each with a primary mirror of 10 meters (32.8 feet) in diameter. These are just a couple of examples of larger optical telescopes on Earth.

While the Hubble Space Telescope is highly regarded for its scientific discoveries and the stunning images it has captured, it is not the largest optical telescope in terms of physical size.

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The magnitude of the force on the Q charge is now 16F. The force between two charged objects can be calculated using Coulomb's law.

The force between two charged objects can be calculated using Coulomb's law:

F = k * (|Q1| * |Q2|) / r²

Where F is the magnitude of the force, k is the electrostatic constant, |Q1| and |Q2| are the magnitudes of the charges on the two objects, and r is the distance between them.

Initially, both objects have a net charge of Q, and the force between them is F. Let's denote this force as F1.

Now, we replace one of the objects with another object that has a net charge of 4Q. The force between these two objects is F2.

Using Coulomb's law, we can write the following equation for F1:

F1 = k * (Q * Q) / r²

And for F2:

F2 = k * (Q * 4Q) / r²

Simplifying F1 and F2:

F1 = k * (Q²) / r²

F2 = k * (4Q²) / r²

Dividing F2 by F1 to find the ratio of the forces:

F2 / F1 = (k * (4Q²) / r²) / (k * (Q²) / r²)

= (4Q²) / (Q²)

= 4

Therefore, the magnitude of the force on the Q charge is now 4 times the original force. Since the original force was F, the magnitude of the force on Q is 4F.

However, we also replaced one of the objects with another object with a net charge of 4Q. Thus, the new force is 4 times the original force multiplied by the charge ratio:

Magnitude of the force on Q = 4F * (|Q| / |4Q|)

= 4F * (1/4)

= 16F

The magnitude of the force on the Q charge is now 16F. Replacing one of the objects with another object having a net charge of 4Q leads to an increase in the force by a factor of 16.

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The kinetic energy of the bullet is 2K, twice the kinetic energy of bullet 1.

The kinetic energy of an object is the energy possessed due to its motion. The kinetic energy of an object of mass m moving with a velocity v is given by K = mv²/2 ( in joules)

Given: kinetic energy of bullet 1 = K =m₁×v²/2  

mass of bullet 2, m₂ = twice of mass of bullet 1, m₁

m₂ = 2 × m₁

the kinetic energy of bullet 2 is = m₂ × v²/2 , since speed are same.

in terms of m1,

kinetic energy of bullet 2 = 2m₁ × v²/2 = 2K

Therefore, the kinetic energy of the bullet is 2K, twice the kinetic energy of bullet 1.

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The SI unit for work is Joule (J) based on details given in the question.

Given that work is force x distance, and that the SI unit for force is Newton (symbol N), work will have units that equal Joule (J).Explanation:In physics, Work is defined as force applied on an object to move it through a distance. Work is a scalar quantity, and it's SI unit is the Joule (J), which is defined as the amount of work done by a force of one newton to move an object over a distance of one meter.

The mathematical expression for work is given by:Work = Force x DistanceWhere,Work = Work doneForce = Applied ForceDistance = Distance moved by the objectThe SI unit of force is newton (symbol N), which is the amount of force required to accelerate a mass of one kilogram at a rate of one meter per second squared.The relationship between work, force, and distance is given by the formula:

Work = Force x Distance (W = F x D)where,W = Work done (in Joule)F = Applied force (in Newton)D = Distance moved by the object (in meter)Therefore, the units of work are obtained by multiplying the units of force (N) with the units of distance (m), which gives us the unit of work as joules (J).

Thus, the SI unit for work is Joule (J).

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The value of the frequency of oscillation of the object is 1.67 Hz. The answer is most near to option C.

The number of waves that pass a fixed place in a unit of time is referred to as frequency. It also indicates how many cycles or vibrations a body in periodic motion experiences in a given unit of time.

The term time period refers to the amount of time that is required for one full oscillation to take place.

From the figure, it is shown that the object is completing an oscillation between the time durations of t = 0.2s and t = 0.8s.

So, the time period of oscillation of the object is,

T = 0.8 - 0.2

T = 0.6 s

Therefore, the frequency of oscillation of the object is given by,

f = 1/T

f = 1/0.6

f = 1.67 Hz

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False. The RC time constant for the filter capacitor in a power supply is typically set to be small, not large, so that the capacitor discharge appears linear.

The RC time constant is a measure of the time it takes for a capacitor to charge or discharge through a resistor in an RC circuit. In the case of a power supply filter capacitor, the RC time constant is typically set to be small rather than large.

When the RC time constant is small, it means that the product of the resistance (R) and the capacitance (C) in the circuit is small. This results in a faster charging and discharging process for the capacitor.

In a power supply, the filter capacitor is used to smooth out the voltage ripple and provide a steady DC output. By setting the RC time constant to be small, the capacitor can discharge quickly and respond rapidly to changes in the input voltage, minimizing the voltage ripple.

If the RC time constant were set to be large, it would result in a slower discharge of the capacitor. This would lead to a non-linear discharge curve, which is undesirable in a power supply where a steady and linear output voltage is desired. Therefore, the RC time constant for the filter capacitor is typically set to be small so that the discharge of the capacitor appears linear.

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The area of the largest circle at the surface of the water over which light from the ring could escape is approximately 554.45 square millimeters.

To calculate the area of the largest circle at the surface of the water over which light from the ring could escape, we need to consider the concept of critical angle and total internal reflection.

When light travels from a medium with a higher refractive index to a medium with a lower refractive index, it can be refracted or reflected depending on the angle of incidence.

When the angle of incidence is greater than the critical angle, total internal reflection occurs, and the light is reflected back into the original medium.

In this case, the ring is going into the water, which has a higher refractive index than the surrounding air. Therefore, we need to determine the critical angle for light to escape from the water into the air.

The critical angle (θc) can be calculated using Snell's law:

sin(θc) = n2 / n1

Where n2 is the refractive index of the air (approximately 1) and n1 is the refractive index of water.

The refractive index of water is around 1.33. Plugging in the values, we have:

sin(θc) = 1 / 1.33

θc ≈ 48.75 degrees

Now, we can consider the situation at the surface of the water. The light from the ring can escape into the air if the angle of incidence is less than the critical angle.

To determine the area of the largest circle over which light can escape, we consider a circle centered at the point where the ring enters the water.

The radius of this circle will be the distance from the center of the ring to the point where the light reaches the surface of the water.

Since we are treating the ring as a point, the radius of the circle will be equal to the depth of the water where the ring enters, which is 13.3 mm.

The area of the circle can be calculated using the formula:

Area = π * radius^2

Substituting the values:

Area = π * (13.3 mm)^2

Calculating the result:

Area ≈ 554.45 mm²

Therefore, the area of the largest circle at the surface of the water over which light from the ring could escape is approximately 554.45 square millimeters.

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We can calculate the potential energy of the microwave antenna:

Therefore, the potential energy of the microwave antenna is approximately 43,126 joules. The formula for calculating potential energy is PE = mgh

where:

Potential energy is a form of energy that an object possesses due to its position, shape, or condition. It is the energy that an object possesses because of its potential to do work. The amount of potential energy an object potential energy is the energy possessed by an object or a system due to its position or configuration in a field of force. It can be thought of as the stored energy that an object has due to its position relative to other objects.

This energy can be released and converted into other forms of energy, such as kinetic energy, when the object moves or changes its position. Examples of potential energy include gravitational potential energy, elastic potential energy, and chemical potential energy.

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A. The largest possible magnitude of the acceleration of the electron due to the magnetic field is 1.89 × 10⁶ m/s².

B. The smallest possible magnitude of the acceleration is 0 m/s².

C. If the actual acceleration is one-fourth of the largest magnitude, the angle between the electron velocity and the magnetic field is 60 degrees.

When an electron moves through a magnetic field, it experiences a force known as the magnetic force. The magnitude of this force can be calculated using the formula F = qvBsinθ, where F is the force, q is the charge of the electron, v is its velocity, B is the magnetic field magnitude, and θ is the angle between the velocity and the magnetic field.

A. To find the largest possible magnitude of acceleration, we need to consider the maximum value of sinθ, which is 1. Therefore, the maximum acceleration occurs when the electron's velocity is perpendicular to the magnetic field. Using the given values, we can calculate the largest acceleration as F/m = (qvBsinθ)/m = (1.6 × 10⁻¹⁹ C)(2.40 × 10⁶ m/s)(7.90 × 10^−2T)/9.11 × 10⁻³¹ kg ≈ 1.89 × 10⁶ m/s².

B. The smallest possible magnitude of acceleration occurs when the angle θ between the velocity and the magnetic field is 0 degrees, causing sinθ to be 0. In this case, the magnetic force is zero, resulting in no acceleration.

C. If the actual acceleration is one-fourth of the largest magnitude, it would be 1.89 × 10⁶ m/s² divided by 4, which is approximately 4.73 × 10⁵ m/s². To find the angle θ, we rearrange the formula as sinθ = (a/m)/(qvB) and substitute the known values, sinθ = (4.73 × 10⁵ m/s²)/(1.6 × 10^-19 C)(2.40 × 10⁶ m/s)(7.90 × 10^−2T) ≈ 0.156. Taking the inverse sine of 0.156, we find that the angle θ is approximately 60 degrees.

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The mechanism typically used to ensure that a pressure-sensitive switch latches into position regardless of the operator's speed is a latch or locking mechanism.

A latch or locking mechanism is designed to hold the pressure-sensitive switch in its actuated position once it has been triggered by the operator. This mechanism ensures that even if the operator releases pressure quickly, the switch remains latched in the activated state until it is intentionally reset or released.

There is no specific calculation involved in the mechanism itself, as it is a physical mechanism designed to engage and hold the switch in position. The calculation would depend on the specific design and engineering requirements of the latch or locking mechanism.

By incorporating a latch or locking mechanism into the design of a pressure-sensitive switch, it can reliably maintain its actuated state regardless of the speed at which the operator applies or releases pressure. This ensures consistent and accurate operation of the switch, regardless of the operator's actions

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Isabela's acceleration was 0.8 m/s².

acceleration = (final velocity - initial velocity) / time

Given that Isabela's initial velocity was 3 m/s and her final velocity was 5.4 m/s, and the time elapsed was 5 seconds - 2 seconds = 3 seconds, we can calculate her acceleration:

acceleration = (5.4 m/s - 3 m/s) / 3 s

acceleration = 2.4 m/s / 3 s

acceleration = 0.8 m/s²

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Car 1 experiences a greater change in momentum compared to Car 2, implying that Car 1 exerts a stronger magnitude force on Car 2 during the collision. hence, the correct answer is Car 1.

To determine which car exerts a stronger magnitude force on the other during an elastic collision, we can analyze the situation using the principles of conservation of momentum.

Let's denote the mass of car 1 as [tex]$m_1$[/tex], the mass of car 2 as [tex]$m_2$[/tex], the initial velocity of car 1 as [tex]$v_{1i}$[/tex], and the initial velocity of car 2 as [tex]$v_{2i}$[/tex].

According to the conservation of momentum, the total momentum before the collision is equal to the total momentum after the collision:

[tex]$m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}$[/tex]

Now, based on the given information that car 1 has double the mass of car 2 ([tex]$m_1 = 2m_2$[/tex]) and car 2 has double the speed of car 1 [tex]($v_{2i} = 2v_{1i}$)[/tex], we can substitute these values into the equation:

[tex]$2m_2v_{1i} + m_2(2v_{1i}) = 2m_2v_{1f} + m_2v_{2f}$[/tex]

Simplifying the equation further:

[tex]$4m_2v_{1i} = 2m_2v_{1f} + m_2v_{2f}$[/tex]

Dividing both sides of the equation by [tex]$m_2$[/tex] (since it cancels out):

[tex]$4v_{1i} = 2v_{1f} + v_{2f}$[/tex]

From the equation, we can observe that the final velocities [tex]$v_{1f}$[/tex]  [tex]$v_{2f}$[/tex] represent the change in velocities of car 1 and car 2, respectively.

Since [tex]$v_{1i}$[/tex] and [tex]$v_{1f}$[/tex] are the velocities of car 1, and [tex]$v_{2f}$[/tex] is the velocity of car 2, the right-hand side of the equation represents the sum of the change in momentum for both cars.

Now, comparing the magnitudes of the velocities involved, we can see that [tex]$|v_{1i}| < |v_{1f}|$[/tex] (due to the collision), and [tex]$|v_{2f}| > |v_{1f}|$[/tex] (since car 2 has double the initial velocity of car 1).

Thus, [tex]$2v_{1f} + v_{2f}$[/tex] will have a larger magnitude compared to [tex]$4v_{1i}$[/tex]. Therefore, car 1 experiences a greater change in momentum compared to Car 2, implying that Car 1 exerts a stronger magnitude force on Car 2 during the collision.

Hence, the correct answer is Car 1.

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The coefficient of kinetic friction between the tires and the road can be determined with the given information. A car is moving at a speed of 14.3 m/s on a horizontal road. The brakes are applied and the car stops after traveling for 4.41 s.

Distance covered by the car before coming to a stop can be calculated as follows:

Now, we know that the kinetic energy of the car is converted into heat and sound energy due to frictional force acting between the tires and the road. The amount of work done by the frictional force can be calculated as follows:

The work done by the car to bring it to rest is given by:

We can find the change in kinetic energy of the car as follows:

The work done by the car to bring it to rest is equal to the change in kinetic energy of the car.

Hence, we can write:

The force of friction acting on the car can be expressed in terms of the coefficient of kinetic friction (μ) as follows:

We can substitute the expression for the force of friction into the equation for work done by friction and equate it to the expression for the change in kinetic energy. We get:

We can express this as follows:

Hence, we get:

Now, we can substitute the given values into this expression to calculate the coefficient of kinetic friction.μ = (1/2) × (14.3 m/s)^2 / (9.81 m/s^2 × 63.063 m)μ = 0.4119 (approx)

Therefore, the coefficient of kinetic friction between the tires and the road is approximately 0.4119.

Kinetic energy is the energy that an object possesses due to its motion. It depends on the mass of the object and its velocity. The formula for kinetic energy is KE = 1/2 mv^2, where m is the mass of the object and v is its velocity. The unit of kinetic energy is joules (J).

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The initial speed of the bullet is approximately 632.45 m/s. This calculation is based on the principles of conservation of linear momentum and conservation of mechanical energy in the system consisting of the bullet, block, and spring.

We can solve this problem using the principle of conservation of linear momentum and the principle of conservation of mechanical energy.

1. Conservation of linear momentum:

Before the collision, the bullet is moving horizontally with an unknown speed (let's call it v_bullet), and the block is initially at rest. After the collision, the bullet embeds itself in the block, so the combined objects move together.

Using the conservation of linear momentum, we have:

(m_bullet)(v_bullet) = (m_block + m_bullet)(v_combined)

(0.01 kg)(v_bullet) = (0.99 kg + 0.01 kg)(v_combined)

2. Conservation of mechanical energy:

After the collision, the block and bullet slide along the surface until they encounter the spring. At this point, the mechanical energy is conserved.

The potential energy stored in the compressed spring equals the initial kinetic energy of the block-bullet system.

(1/2)(k)(x^2) = (1/2)(m_combined)(v_combined)^2

Given:

m_bullet = 0.01 kg (mass of the bullet)

m_block = 0.99 kg (mass of the block)

k = 400 N/m (spring constant)

x = 0.1 m (compression of the spring)

We can now solve the equations simultaneously to find the initial speed of the bullet:

From the conservation of linear momentum:

(0.01 kg)(v_bullet) = (0.99 kg + 0.01 kg)(v_combined)

0.01 v_bullet = 1.00 v_combined

v_combined = 0.01 v_bullet

Substituting this into the conservation of mechanical energy equation:

(1/2)(400 N/m)(0.1 m)^2 = (1/2)(1.00 kg)(0.01 v_bullet)^2

20 J = 0.00005 v_bullet^2

Solving for v_bullet:

v_bullet^2 = (20 J) / (0.00005 kg)

v_bullet^2 = 400,000 m^2/s^2

v_bullet = √(400,000) m/s

v_bullet ≈ 632.45 m/s

the initial speed of the bullet is approximately 632.45 m/s.

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If one follows the arc of the handle of the Big Dipper away from the dipper, the first moderately bright star one comes to is Polaris.

The Big Dipper is a prominent asterism in the northern sky that is located in the constellation Ursa Major. The Big Dipper is made up of seven stars that seem to create a shape similar to a dipper or scoop. It is often used as a pointer for finding the North Star, Polaris.

The Big Dipper has been used in cultures all over the world to track the seasons and agricultural cycles, as well as for navigation purposes.

Polaris is the brightest star in the constellation Ursa Minor and is also known as the North Star or Pole Star. Polaris is located at the end of the handle of the Little Dipper and is the closest bright star to the north celestial pole.

This means that it appears to be stationary in the night sky while all other stars appear to move around it. Polaris is frequently used in navigation, especially in the northern hemisphere.

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The tension in a transverse wave on a string can be calculated using the equation y = (F/μ) sin(kx-wt), where F is the tension force, μ is the linear density of the string, k is the wave number, x is the position of the particle, w is the angular frequency, and t is time.

Using the given equation y = 0.021 sin (8πt + 2x) and the linear density of the string (16 x 10-2 g/cm), the tension in the string is calculated to be 2.53 N.

To find the tension in the string, the equation for a transverse wave on a string is used: y = (F/μ) sin(kx-wt), where F is the tension force, μ is the linear density of the string, k is the wave number, x is the position of the particle, w is the angular frequency, and t is time.

Rearranging this equation to solve for F, we get F = μwk^2A, where A is the amplitude of the wave.

Using the given equation y = 0.021 sin (8πt + 2x),

we can see that the amplitude of the wave is A = 0.021 m. To find k and w, we can compare this equation to the standard form of a transverse wave: y = A sin(kx-wt). We see that k = 2, and w = 8π.

Finally, substituting these values for k, w, and A into the equation for tension, we get F = (16 x 10-2 g/cm)(8π)^2(2)^2(0.021 m), which simplifies to 2.53 N.

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The temperature of water, t = 50°C. The pump is at a height of 3 m above the open tank level. The velocity in the pipe is v = 0.9 m/s. The available (NPSH) is 2.7m.

We need to calculate the available NPSH for the pump (NPSHA).

NPSHA can be calculated using the following formula:

NPSHA = Ha + PATM/Pg - hf - Hf

where

Ha is the height of the fluid above the pump centerline and

Pg is the density of the fluid.

Hf is the suction head and PATM is the atmospheric pressure.

Pg can be calculated using the following formula:

Pg = ρg

where ρ is the density of water and g is the acceleration due to gravity.

We know that:

ρ = 998 kg/m³ (density of water at 50°C)g = 9.81 m/s² (acceleration due to gravity)

PATM = 101.3 kPa (atmospheric pressure at sea level)

Ha = 3 m - 0.5 m = 2.5 m (assuming the centerline of the pump is at a height of 0.5 m from the top of the pump)

NPSHA = Ha + PATM/Pg - hf - Hf= 2.5 + (101.3/9.81*10³) / 998 - 1 - 0= 2.7 m

Therefore, the available NPSH for the pump is 2.7 m.

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If the resistance in the circuit is reduced by half, then the current flowing through the circuit will increase to 6.0 amps.

Ohm's Law states that the current flowing through a conductor between two points is directly proportional to the voltage across the two points, given that the temperature and other physical conditions remain constant.

Mathematically, Ohm's Law can be expressed as:

V = I × R

Where:

V represents the voltage (in volts, V)

I represents the current (in amperes, A)

R represents the resistance (in ohms, Ω)

Given:

Current (I₁) = 3.0 A

Voltage (V) = 12 V

Resistance (R₁) = ?

Make R₁  the subject of the formula:

R₁ = V / I₁

R₁ = 12 V / 3.0 A

R₁ = 4.0 Ω

Now, let's calculate the new current (I₂) when the resistance is reduced by half:

Resistance (R₂) = R₁ / 2

R₂ = 4.0 Ω / 2

R₂ = 2.0 Ω

Recall Ohm's Law:

I₂ = V / R₂

I₂ = 12 V / 2.0 Ω

I₂ = 6.0 A

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When examining a piston crown that has vertical cracks through the edges of the piston's combustion chamber bowl, Technician A suggests the piston crown was not getting properly cooled due to low mass airflow in the combustion chamber during the valve overlap period. The correct suggestion is Technician A because the piston crown to not get properly cooled and thus causes cracks

While examining a piston crown that has vertical cracks through the edges of the piston's combustion chamber bowl, it was found that the technician A suggested that the piston crown was not getting properly cooled due to low mass airflow in the combustion chamber during the valve overlap period. While technician B suggests the cracks were caused by inadequate cooling by the oil cooler nozzles.

In this situation, technician A is correct. During valve overlap, a low-mass airflow occurs, which leads to an increased temperature of the combustion chamber. This causes the piston crown to not get properly cooled and thus causes cracks. Therefore, technician A's suggestion is correct.

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The velocity of the projectile is 13.86 m/s (approx) to the leave the barrel of the cannon.

The mass of the rubber ball, m = 5.34 g = 0.00534 kg

The force constant of the spring, k = 8.05 N/m

The displacement of the spring, x = 5.05 cm = 0.0505 m

The distance travelled by the ball, d = 15.9 cm = 0.159 m

The force of friction acting on the ball, f = 0.0326 N

To find: The velocity of the projectile, v

The potential energy of the compressed spring is given as,

U = 1/2 k x²= 1/2 × 8.05 × 0.0505²= 0.0102 J

When the ball is fired, the potential energy of the spring is converted into the kinetic energy of the ball and the work done against the friction force.

W = U - f × d= 0.0102 - 0.0326 × 0.159= 0.00512 J

The kinetic energy of the ball is given as,

K.E = 1/2 m v²

Here, m = 0.00534 kg

Substituting the given values of kinetic energy and mass,

0.00512 = 1/2 × 0.00534 × v²

v² = 0.00512 / (0.5 × 0.00534)

v² = 191.7

v = √191.7 = 13.86 m/s (approx)

Therefore, the velocity of the projectile is 13.86 m/s (approx).

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Based on the observation that the radio stations come in more clearly in your friend's car compared to your own, it can be concluded that the radio damping in your friend's vehicle is lower than in yours.

Radio damping refers to the attenuation or reduction of radio signals as they propagate through various materials, such as the body of a vehicle. Higher radio damping means that the radio signals are more attenuated, resulting in weaker reception and potentially poorer audio quality.

If the radio stations come in more clearly in your friend's car, it suggests that her vehicle has lower radio damping compared to yours. This means that the materials or construction of her car allow for better transmission and reception of radio signals, resulting in stronger and clearer reception of the stations.

The observation that the radio stations come in more clearly in your friend's car indicates that her vehicle has lower radio damping compared to yours. This suggests that her car's design or construction allows for better transmission and reception of radio signals, resulting in improved reception quality.

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The upper arm bone shortens by approximately 0.51 mm when the wrestler stands on one hand during the maneuver.

To calculate the amount by which the upper arm bone shortens, we can use the concept of strain and the Young's modulus of the material.

The strain (ε) is defined as the ratio of the change in length (ΔL) to the original length (L), which can be expressed as ε = ΔL / L.

The Young's modulus (Y) represents the stiffness of the material and relates the stress (σ) to the strain (ε) through the equation σ = Y * ε. In this case, we have the original length (L) of the bone, which is 38.0 cm. The radius (r) of the bone is 2.10 cm.

The change in length (ΔL) can be calculated using the equation ΔL = (σ / Y) * L.

To find the stress (σ), we can consider the weight of the wrestler as the force acting on the bone. The weight (W) is given by W = m * g, where m is the mass of the wrestler (150 kg) and g is the acceleration due to gravity (9.8 m/s^2).

The stress (σ) is then calculated as σ = W / A, where A is the cross-sectional area of the bone.

Using the formula for the cross-sectional area of a rod (A = π * r^2), we can substitute the values into the equations and find that the upper arm bone shortens by approximately 0.51 mm.

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After the collision, the truck and the car will move with different speeds. The truck's speed will decrease, while the car's speed will increase.

The final speeds of the vehicles can be calculated using the principles of conservation of momentum and kinetic energy. The speed of the truck after the collision is approximately 4.62 m/s, while the speed of the car is approximately 27.38 m/s.

In an elastic collision, both momentum and kinetic energy are conserved. Initially, the truck has momentum given by the product of its mass and velocity (1510 kg * 16.0 m/s), while the car has zero momentum as it is stationary.

After the collision, the truck and the car move with different speeds. Let the final velocity of the truck be Vt and the final velocity of the car be Vc. Applying the conservation of momentum, the total initial momentum must equal the total final momentum.

(1510 kg * 16.0 m/s) + (729 kg * 0 m/s) = (1510 kg * Vt) + (729 kg * Vc)

Simplifying the equation gives:

24160 kg·m/s = 1510 kg·Vt + 729 kg·Vc

The final speeds, we also consider the conservation of kinetic energy. The total initial kinetic energy is zero (car at rest), and the total final kinetic energy is given by the sum of the kinetic energies of the truck and the car.

(1/2) * (1510 kg) * (Vt)^2 + (1/2) * (729 kg) * (Vc)^2 = (1/2) * (1510 kg) * (16.0 m/s)^2 + (1/2) * (729 kg) * (0 m/s)^2

Simplifying the equation gives:

(1/2) * (1510 kg) * (Vt)^2 + (1/2) * (729 kg) * (Vc)^2 = 963,680 J

Now, we have a system of equations to solve simultaneously. Solving these equations will give us the values of Vt and Vc. The resulting speeds are approximately Vt = 4.62 m/s for the truck and Vc = 27.38 m/s for the car.

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