calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.

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Answer 1

The force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s is 7000 N in the opposite direction to the car's motion.

Calculate the force required to stop a car of mass 1400 kg in 2 seconds if it is moving with a velocity of 10 m/s.

To solve the given problem, we can use the equation:

F = (m * Δv) / Δt

where F = force

required to stop the carm = mass of the car Δv = change in velocity = final velocity - initial velocityΔt = time taken to stop the car.

Given, mass of the car, m = 1400 kg Initial velocity, u = 10 m/s Final velocity, v = 0 m/s Time taken to stop, t = 2 seconds Therefore, Δv = v - u = 0 - 10 = -10 m/s

Substituting the given values in the above equation, we get:

F = (m * Δv) / Δt = (1400 kg * (-10 m/s)) / (2 s) = -7000 N

Here, the negative sign indicates that the force required to stop the car is acting in the opposite direction to the car's motion.

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Related Questions

an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2. after 5 seconds, the object will have a speed of

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Answer : If an object falls freely from rest on a planet where the acceleration due to gravity is 20 m/s2 then after 5 seconds, the object will have a speed of 100 m/s

This can be calculated using the equation v = a*t, where v is the velocity, a is the acceleration due to gravity, and t is the time elapsed. Therefore, in this case, v = 20 m/s2 * 5 s = 100 m/s. These values are given in question, so we just have to put them in equation.

Since the object is falling freely, its acceleration remains constant and it follows a uniform acceleration motion. Therefore, the velocity of the object will increase linearly with time. After 10 seconds, the velocity will double to 200 m/s, and so on.

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pry on the power steering reservoir to adjust the tension of the power steering belt. true or false?

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The statement "pry on the power steering reservoir to adjust the tension of the power steering belt" is: false.

The tension of the power steering belt is adjusted by adjusting the position of the power steering pump. There is a tension adjustment bolt on the power steering pump that is used to adjust the tension of the power steering belt. The adjustment bolt should be turned clockwise or counterclockwise to adjust the tension of the belt.

A belt tension gauge may be used to ensure that the belt is properly tensioned. A pry bar should not be used on the power steering reservoir to adjust the tension of the power steering belt. This could cause damage to the reservoir or other components of the power steering system. The reservoir should be inspected for damage or leaks, but it should not be used to adjust the tension of the belt.

In summary, the tension of the power steering belt should be adjusted by adjusting the position of the power steering pump, not by prying on the power steering reservoir.

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measurements show a certain star has a very high luminosity (100,000 x the sun's) while its temperature is quite cool (3500 k). how can this be?

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The star might be quite large in size, with a much larger surface area than the sun. This would increase its luminosity despite its cooler temperature.

The star has a high luminosity (100,000 x the sun's) and a cool temperature (3500 K) because of its size.

A star's luminosity is proportional to its size, so if a star is very large, it can have a high luminosity even if it is relatively cool.

Another possibility is that the star is in a phase of its life cycle where it has expanded and cooled, such as a red giant or supergiant, but still retains a high luminosity due to its large size.

These stars have relatively low surface temperatures, but their large sizes give them very high luminosities.

Therefore, this star is likely very large and thus has a very high luminosity despite its low temperature.

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