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Tamiku [17]
3 years ago
5

A steel ball and a piece of clay have equal mass. They are dropped from the same height on a horizontal steel platform. The ball

bounces back with nearly the same speed with which it hit. The clay sticks to the platform. Which object experiences the greater momentum change? (A) the ball (B) the clay change(C) Both experience the same momentum (D) there is no momentum change for either (E) more information is required
Physics
1 answer:
emmasim [6.3K]3 years ago
6 0

Answer: The ball (option A)

Explanation: change in momentum is defined by the formulae m(v - u) where m = mass of object, v = final velocity and u = initial velocity.

For the ball, it hits the ground and bounces back with the same speed, that's final velocity equals initials (v = - u)

Change in momentum = m( -u- u) = m(-2u) = m(-2u) = -2mu

For the clay, it final velocity is zero since it sticks to the floor, hence (v =0)

m(v - u) = m(0 - u) = - mu.

-2mu (change in momentum from the ball) is greater than - mu ( change in momentum of clay)

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Vlad [161]

Answer:

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Explanation:

4 0
2 years ago
Two golf balls are thrown with the same speed off the top of a large bridge at the same time. ball a is thrown straight downward
iren2701 [21]
I think that ball  a hit the ground because it says that it went straight down.
8 0
3 years ago
(4A) The mass of Earth is 5.972 * 10^24 kg, and the radius of Earth is 6,371 km.
faltersainse [42]

Answer:

x₁ = 345100 km

Explanation:

The direction of the attraction forces between the earth and the object, and between the moon and the object, are in opposite direction and  (along the straight line between the centers of earth and moon) and as gravity is always attractive, the net force will become zero when both forces are equal. According to this:

Let  call "x₁"  distance between center of the earth and the object, and

"x₂" the distance between center of the moon and the object, Mt mass of the earth, Ml mass of the moon, m₀ mass of the object

we can express:

F₁  ( force between earth and the object )

F₁ = K *  Mt * m₀/ ( x₁)²        K is a gravitational constant

F₂  (force between mn and the object)

F₂ = K * Ml * m₀ / (x₂)²

Then:

F₁ = F₂               K*Mt*m₀ / x₁²   =  K*Ml*m₀ /x₂²

Or  simplifying the expression

Mt/ x₁²  =  Ml/ x₂²

We know that   x₁   +  x₂  = 384000 Km then

x₁ =  384000 - x₂

Mt/( 384000 - x₂)²  =  Ml / x₂²

Mt *  x₂²  =  Ml *( 384000 - x₂)²

We need to solve for x₂

Mt *  x₂²  =  Ml *[ ( 384000)² + x₂² - 768000*x₂]

By substitution:

5.972*10∧24*x₂² = 7.348*10∧22 * [ 1.47*10∧11 ] + 7.348*10∧22*x₂² -

                                7.348*10∧22*768000*x₂

Simplifying by 10∧22

5.972*10²*x₂²  = 7.348* [ 1.47*10∧11 ] + 7.348*x₂²- 7.348*768000*x₂

Sorting out

5.972*10²*x₂²- 7.348*x₂² = 10.80*10∧11 - 56,43* 10∧5*x₂

(597,2 - 7,348 )* x₂²  = 10.80*10∧11 - 56.43*10∧5*x₂

590x₂²  + 56.43*10∧5*x₂ - 10.80*10∧11 = 0

Is a second degree equation

x₂  =  -56.43*10∧5 ± √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  = -56.43*10∧5 + √3184*10∧10 + 25488*10∧11  / 1160

x₂ ₁  =  -56.43*10∧5 + √3184*10∧10 + 254880*10∧10  / 1160

x₂ ₁  = -56.43*10∧5 + 10∧5 [ √3184 + 254880 ] /1160

x₂ ₁  =  -56.43*10∧5 + 508* 10∧5  / 1160

x₂ ₁  =  451.27*10∧5/1160

x₂ ₁  =  4512.7*10∧4 /1160

x₂ ₁  = 3.89*10∧4  km (distance between the moon  and the object)

x₂ ₁  = 38900 km

x₂ = 38900 km

We dismiss the other solution because is negative and there is not a negative distance

Then the distance between the earth and the object is:

x₁  = 384000 - x₂

x₁ = 384000 - 38900

x₁ = 345100 km

5 0
3 years ago
Why doesn’t a machine that increases force break the law of conservation of energy?
USPshnik [31]

Answer:

A machine in which work input equals work output. energy can be used to do work, work can be used to transfer energy. The change in the kinetic energy of an object is equal to the net work done on the object.

hope this helps

8 0
3 years ago
A 5 kilograms bowling ball is dropped out a window. It hits the ground, and bounces upward. The velocity change of the ball is n
ioda

Answer:

13.5

Explanation:

Mass: 5kg

Initial Velocity: -15

Final Velocity: 12

Force: 10

We can use the equation: Vf = Vi + at

We need to find acceleration, and we can use the equation, F=ma,

We have mass and the force so it would look like this, 10=5a, and 5 times 2 would equal 10, so acceleration would be 2.

Now we have all the variables to find time.

Back to Vf = Vi + at, plug the numbers in, 12 = -15 + 2(t)

Plugging them in into desmos gives 13.5 for time.

4 0
2 years ago
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