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vivado [14]
2 years ago
7

In an experiment, a disk is set into motion such that it rotates with a constant angular speed. As the disk spins, a small spher

e of clay is dropped onto the disk, and the sphere sticks to the disk. All frictional forces are negligible. What would happened to the angular momentum and the total kinetic energy of the disk-sphere system immediately before and after the collision?
Physics
1 answer:
boyakko [2]2 years ago
6 0

Answer:

  L₀ = L_f ,  K_f < K₀

Explanation:

For this exercise we start as the angular momentum, with the friction force they are negligible and if we define the system as formed by the disk and the clay sphere, the forces during the collision are internal and therefore the angular momentum is conserved.

This means that the angular momentum before and after the collision changes.

Initial instant. Before the crash

        L₀ = I₀ w₀

Final moment. Right after the crash

        L_f = (I₀ + mr²) w

we treat the clay sphere as a point particle

how the angular momentum is conserved

       L₀ = L_f

       I₀ w₀ = (I₀ + mr²) w

       w = \frac{I_o}{I_o + m r^2}   w₀

having the angular velocities we can calculate the kinetic energy

       

starting point. Before the crash

        K₀ = ½ I₀ w₀²

final point. After the crash

        K_f = ½ (I₀ + mr²) w²

sustitute

        K_f = ½ (I₀ + mr²)  ( \frac{I_o}{I_o + m r^2}   w₀)²

        Kf = ½  \frac{I_o^2}{ I_o + m r^2}   w₀²

we look for the relationship between the kinetic energy

        \frac{K_f}{K_o}=   \frac{I_o}{I_o + m r^2}

       \frac{K_f}{K_o } < 1

      K_f < K₀          

we see that the kinetic energy is not constant in the process, this implies that part of the energy is transformed into potential energy during the collision

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Mario rdoll a coin up a slope at 2 m/s. It travels 2.7 m, comes to a stop and rolls back down. What is the coin's acceleration?
leva [86]

The coin's acceleration is <u>0.37 m/s²</u>

Acceleration is the rate of change of the velocity of an item with appreciation to time. Accelerations are vector portions. The orientation of an item's acceleration is given by the orientation of the net pressure appearing on that object.

<u>Calculation:-</u>

<u />

V² = U -2aS

a = U/2S

  = 2/2×2.7

   =  <u>0.37 m/s²</u>

Acceleration is the charge at which velocity modifications with time, in terms of each speed and route. A factor or an object moving in a straight line is accelerated if it quickens or slows down. movement on a circle is extended despite the fact that the rate is consistent because the course is continually changing.

Learn more about acceleration here:- brainly.com/question/29110429

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8 0
1 year ago
A meter stick balances horizontally on a knife-edge at the 51 cm mark. With two nickels stacked over the 6.0 cm mark, the stick
Oliga [24]

Answer:

65g

Explanation:

Two main conditions for equilibrium are:

I. The resultant force must be equal to zero. That is, sum of the forces acting in one direction about a point must be equal to the sum of the forces acting in the opposite direction about the same point.

II. The resultant moment must be equal to zero. That is, sum of the moments in one direction about a point must be equal to the sum of the moments in another direction about the same point.

For the above question,

the 51cm mark is the point where the resultant weight of the meter stick lies,

the pivot or point is the 45cm mark where the stick balanced when 2 nickels ( total mass (5.0g x 2) 10g were placed at the 6cm mark.

Using the conversion factor:

1000g(1kg) = 10N, we can convert mass to weight, calculate the weight of the meter stick then reconvert to mass.

That is,

mass of 2 nickels = 10g = 10/1000 = 0.01N.

Moment = Force x distance from line of force to pivot of rotation

Applying the principle of equilibrium,

Moment of left side = Moment of right side

0.01 x (45-6) = W x (51-45)

Where W = weight of the meter stick

W x 6 = 0.01 x 39

W x 6 = 0.39

W = 0.39/6

W= 0.065N

Therefore, mass of meter stick = 0.065 x 1000 = 65g.

4 0
2 years ago
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
3 years ago
a skier starts from rest and skis down a 82 meter tall hill labeled h1, into a valley and staught back up another 35 meter hill(
horrorfan [7]

Answer:

She is going at 30.4 m/s at the top of the 35-meter hill.    

Explanation:

We can find the velocity of the skier by energy conservation:

E_{1} = E_{2}

On the top of the hill 1 (h₁), she has only potential energy since she starts from rest. Now, on the top of the hill 2 (h₂), she has potential energy and kinetic energy.

mgh_{1} = mgh_{2} + \frac{1}{2}mv_{2}^{2}    (1)

Where:

m: is the mass of the skier

h₁: is the height 1 = 82 m

h₂: is the height 2 = 35 m

g: is the acceleration due to gravity = 9.81 m/s²  

v₂: is the speed of the skier at the top of h₂ =?

Now, by solving equation (1) for v₂ we have:

v_{2}^{2} = \frac{2mg(h_{1} - h_{2})}{m}  

v_{2} = \sqrt{2g(h_{1} - h_{2})} = \sqrt{2*9.81 m/s^{2}*(82 m - 35 m)} = 30.4 m/s    

Therefore, she is going at 30.4 m/s at the top of the 35-meter hill.

I hope it helps you!  

6 0
2 years ago
4.) The diagram to the right is the orbit of a comet:
Brilliant_brown [7]

Answer: This is what I found hope it helps

Explanation:

6 0
3 years ago
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