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velikii [3]
3 years ago
13

You throw a bouncy rubber ball and a wet lump of clay, both of mass m, at a wall. Both strike the wall at speed v, but while the

ball bounces off with no loss of speed, the clay sticks. What is the change in momentum of the clay and ball, respectively, assuming that toward the wall is the positive direction?
Physics
2 answers:
shutvik [7]3 years ago
8 0

Answer:

Change in momentum of the wet lump of clay = -mv

Change in momentum of the bouncy rubber ball = -2mv

Explanation:

Momentum, p = mv

Change in momentum of the wet lump of clay

The initial velocity of the wet lump of clay of mass, m is v, so its initial momentum is p₁ = mv. Since it sticks to the wall, its final velocity is zero. So its final momentum is p₂ = m × 0 = 0. Its change in momentum Δp = p₂ - p₁ = 0 - mv = -mv

Change in momentum of the bouncy rubber ball

The initial velocity of the rubber ball of mas m is v, so its initial momentum is p₃ = mv. Since it bounces of the wall with no loss of speed and thus moves in the opposite direction to its initial direction, its final velocity is -v. So its final momentum is p₄ = m × -v = -mv. Its change in momentum Δp = p₄ - p₃ = -mv - mv = -2mv

nadezda [96]3 years ago
6 0

Your answer


<h2>-mv; -2mv</h2>

Good luck!

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A rock falls off a cliff. How fast will it be going after falling for 4.33 seconds?
bixtya [17]

Answer:42.43m/s

Explanation:According to vf=vi+at, we  can calculate it since v0 equals to 0. vf=0+9.8m/s^2*4.33s= 42.434m/s

4 0
1 year ago
An applied force of 50 N is used to accelerate an object, that weighs 73 N, to the right across a frictional surface. The object
Hunter-Best [27]

Answer:

5.38 m/s^2

Explanation:

NET force causing the object to accelerate  =  50 -10 = 40 N

Mass of the object =  73 N / 9.81 m/s^2 = 7.44 kg

F = ma

40 = 7.44 * a         a = 5.38 m/s^2

6 0
1 year ago
Problem 1: Spherical mirrorConsider a spherical mirror of radius 2 m, and rays which go parallel to the optic axis. What is thep
SIZIF [17.4K]

Answer:

1) iii i= 1m, 2)  iii and iv, 3)  i = f₂ (L-f₁) / (L - (f₁ + f₂))

Explanation:

Problem 1

For this problem we use two equations the equations of the focal distance in mirrors

              f = r / 2

              f = 2/2

             f = 1 m

The builder's equation

           1 / f = 1 / o + 1 / i

Where f is the focal length, "o and i" are the distance to the object and the image respectively.

For a ray to arrive parallel to the surface it must come from infinity, whereby o = ∞ and 1 / o = 0

              1 / f = 0 + 1 / i

              i = f

              i = 1 m

The image is formed at the focal point

The correct answer is iii

Problem 2

For this problem we have two possibilities the lens is convergent or divergent, in both cases the back face (R₂) must be flat

Case 1 Flat lens - convex (convergent)

              R₂ = infinity

              R₁ > 0

Cas2 Flat-concave (divergent) lens

             R₂ = infinity

              R₁ <0

Why the correct answers are iii and iv

Problem 3

For a thick lens the rays parallel to the first surface fall in their focal length (f₁), this is the exit point for the second surface whereby the distance to the object is o = L –f₁, let's apply the constructor equation to this second surface

          1 / f₂ = 1 / (L-f₁) + 1 / i

          1 / i = 1 / f₂ - 1 / (L-f₁)

           1 / i = (L-f₁-f₂) / f₂ (L-f₁)

           i = f₂ (L-f₁) / (L - (f₁ + f₂))

This is the image of the rays that enter parallel to the first surface

6 0
4 years ago
Suppose you increase your walking speed from 4 m/s to 13 m/s in a period of 3 s. What is your acceleration?
iogann1982 [59]

The acceleration formula goes like this: a= (vf-vi)/t so it would be (13-4)/3 Thus the answer is 3m/s^2

7 0
4 years ago
Select all that apply.
slega [8]
A chemist is likely to:
<span>1. analyze the ingredients in ice cream
</span><span>2. determine how to separate gasoline from other substances in petroleum</span>
7 0
3 years ago
Read 2 more answers
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