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3 years ago

What type of acceleration does an object moving in a circular path with constant speed experience?

Physics

2 answers:

Marysya12 [62]3 years ago

7 0

An object moving in a circular path has centripetal acceleration. (A)

AnnyKZ [126]3 years ago

5 0

An object moving in a circular path with constant speed experiences centripetal acceleration.

Centripetal Acceleration- “the rate of change of tangential velocity”

“The direction of the centripetal acceleration is always inwards along the radius vector of circular motion.”

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Describe a real-world example of the law of conservation of momentum.

mario62 [17]

Imagine a car crash. A car coming at a high speed has a head on collision with a car at rest. When the car makes impact, it will move the other car with it at a slower speed then it was travelling at. In this case, the velocity decreased since the car slowed down, but the mass increased since there are now two cars moving. Momentum was conserved because the change in mass accounts for the loss of velocity.

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3 years ago

Which of the following should a warm up NOT include?

Arlecino [84]

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3 years ago

A hiker walks 20.51 m at 33.16 degrees. What is the Y component of his displacement?

Serhud [2]

Answer:

The y component of his displacement is 11.22 meters

Explanation:

Components of the displacement

The displacement is a vector because it has a magnitude and a direction. Let's suppose a displacement has a magnitude r and a direction θ, measured with respect to the positive x-direction. The horizontal component of the displacement is calculated by:

$x=r\cos\theta$

The vertical component is calculated by:

$y=r\sin\theta$

The hiker has a displacement with magnitude r = 20.51 m at an angle of 33.16 degrees. Substituting in the above equation:

$y=20.51\sin(33.16^\circ)$

$y=11.22\ m$

The y component of his displacement is 11.22 meters

7 0

3 years ago

(a)

Marta_Voda [28]

a) The momentum of the coconut is 3 kg m/s

b) At first, the air resistance is negligible, so the coconut accelerates due to the force of gravity

c) The coconut reaches its terminal velocity

Explanation:

The momentum of an object is given by the equation

$p=mv$

where

m is the mass of the object

v is its velocity

For the coconut in this problem, we have:

m = 1.5 kg (mass)

v = 2 m/s (velocity)

Therefore, its momentum is

$p=(1.5)(2)=3 kg m/s$

There are only two forces acting on the coconut during its fall:

The force of gravity, of magnitude $mg$ (m= mass of the coconut, g = acceleration of gravity), acting downward
The air resistance, acting upward, whose magnitude is proportional to the speed of the coconut

During the first momentums of the fall, the speed of the coconut is still low, so the air resistance is mostly negligible, and therefore only the force of gravity is acting on the coconut. Since this force is constant, it means that the acceleration of the coconut is constant: therefore, its velocity keeps increasing during the fall, and the coconut speeds up.

If the tree is very tall, the fall of the coconut lasts long, and the speed of the coconut keeps increasing. Since the air resistance is proportional to the speed, this means that at some point, the air resistance is no longer negligible, and it starts to have some effect on the fall of the coconut. In particular, at a certain point, the air resistance will become equal (in magnitude) to the force of gravity (but opposite in direction): this means that from this point, the acceleration of the coconut will be zero, and therefore the coconut will continue its motion at constant velocity. This velocity is called terminal velocity, and it occurs when the force of gravity is equal to the air resistance:

$mg = F_r$