a horse moves a sleigh 1.00 kilometers by applying a horizontal 2000 Newton force on its harness for 45 minutes. what is the power of the horse?

Answers

Answer 1

Answer: -- If 2,000 newtons of force were applied through a distance of 1,000 meters,
then 2,000,000 newton-meters = 2,000,000 joules of work were done.

-- 45 minutes = (45 x 60) = 2,700 seconds

-- Power = (work) / (time) = (2,000,000 j) / (2,700 s) = 740.74 watts

Interestingly, that's almost exactly 1 horsepower. (0.99295... of 746 watts)

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Which of the following is an example of heat transfer through radiation?A. A pot of soup warming on a hot burner.

B. A crayon melting on a hot sidewalk.

C. Beach sand getting warmed by the sun.

D. An egg getting cooked in a pot of boiling water.

Answers

A). The hot burner is in contact wit the pot,
warming it with conduction.

B). The hot sidewalk is in contact with the crayon,
melting it with conduction.

C). The sun is not in contact with anything, and there is
no fluid to convect between it and the sand. The only way
for the sun to transfer heat to the sand is with radiation.

D). The boiling water is in contact with the egg,
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A 0.145-kg baseball pitched at 37.5 m/s is hit on a horizontal line drive straight back toward the pitcher at 57.0 m/s . If the contact time between bat and ball is 2.50×10−3 s , calculate the average force between the ball and bat during contact.

Answers

Answer:

The average force between the ball and bat during contact is 1131N.

Explanation:

The average force can be found employing Newton's second law:

$F = ma$ (1)

Where F is the force, m is the mass and a is the acceleration.

The acceleration can be determined employing the equations for a Uniformly Accelerated Rectilinear Motion:

$v_(f) = v_(i) + at$ (2)

Where $v_(f)$ is the final velocity, $v_(i)$ is the initial velocity and t is the time.

Then, a will be isolated from equation 2:

$a = (v_(f)-v_(i))/(t)$ (3)

Replacing equation (3) in equation (1) it is gotten:

$F = m(v_(f)-v_(i))/(t)$

$F = (0.145Kg)((57.0m/s-37.5m/s))/((2.50x10^(−3)s))$

$F = (0.145Kg)((19.5m/s))/((2.50x10^(−3)s))$

$F = 1131Kg.m/s{2}$

But $1N = 1Kg.m/s^(2)$ , therefore:

$F = 1131N$

So, the average force between the ball and bat during contact is 1131N.

You need to raise a heavy block by pulling it with a massless rope. You can either (a) pull the block straight up height h, or (b) pull it up a long, frictionless plane inclined at a 15∘ angle until its height has increased by h. Assume you will move the block at constant speed either way.

Answers

Final answer:

In both scenarios, the work done on the heavy block is the same, as it is determined by the change in the vertical height. However, pulling the block up the inclined plane may require less force because the work is distributed over a larger distance.

Explanation:

The subject of this question is based on the concept of work and energy in physics. When you pull the heavy block straight upwards (scenario a), the work done is equal to the force times the distance, or Work = mg*h, where m is the mass of the block, g is the acceleration due to gravity, and h is the height it needs to rise. For pulling the block up the inclined plane (scenario b), the work done still equals mg*h as the vertical distance it rises is the same.

This is because, according to the principle of work and energy, the work done on an object is equal to the change in its kinetic energy. Since the speed of the block remains constant in both scenarios, the kinetic energy does not change, meaning the work done on the block is the same in both scenarios.

However, pulling the block up the inclined plane may require less force because of the larger distance over which the work is done. But the overall work is the same in both cases.

Learn more about Work and Energy here:

brainly.com/question/16987255

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A 250 gram ball at the end of a string is revolving uniformly in a circle of radius 0.75 meters.The ball makes 2.0 revolutions per second. What is the centripetal acceleration?

Answers

The centripetal acceleration of the ball would be 88.44 m/ $s^2$ .

Centripetal acceleration

The centripetal acceleration (ac) of an object moving in a circle at a constant speed is given by the formula:

ac = (v^2) / r

where v is the speed of the object and r is the radius of the circle.

In this case, the ball is revolving uniformly in a circle of radius 0.75 meters, and it makes 2.0 revolutions per second. To find the speed of the ball (v), we need to convert the number of revolutions per second to the angular velocity (ω) in radians per second:

ω = 2π x (number of revolutions per second)

ω = 2π x 2.0 = 4π radians per second

The speed of the ball (v) is then given by: