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Alina [70]
3 years ago
6

A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it st

ruck the wall. Has the momentum of the ball changed, and if so, by how much?
Physics
1 answer:
jeka57 [31]3 years ago
3 0

Answer:

The magnitude of change in momentum is (2mv).

Explanation:

The momentum of an object is given by the product of mass and velocity with which it is moving.

Let the mass of ball is m. A tennis player smashes a ball of mass m horizontally at a vertical wall. The ball rebounds at the same speed v with which it struck the wall.

Initial speed of the ball is v and final speed, when it rebounds, is (-v). The change in momentum is given by :

p = final momentum - initial momentum

p=-mv-mv\\\\p=-2mv

So, the magnitude of change in momentum is (2mv).

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Answer:

Think it is C

Explanation:

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3 years ago
Kendall has empty graduated cylinder with markings and an identical graduated cylinder partway filled with water. She also has a
Elis [28]
The answer is B.
She can measure the mass of the water, marble and the graduated cylinder with the balance.
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3 years ago
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A rock is thrown horizontally from a high building at 33.8 m/s. What is the magnitude of its velocity 4.25 s later?
Alex17521 [72]
<h2>Answer:53.63ms^{-2}</h2>

Explanation:

The equations of motion used in this question is v=u+at

When a object is projected horizontally from a sufficiently height,the x-component of acceleration remains zero because there is no force that drags the object in x direction.

But,due to gravity,the object accelerates downward at a rate of 9.8ms^{-2}.

In X-Direction,

Given that initial velocity=u_{x}=33.8ms^{-1}

Using v=u+at,

v_{x}=33.8+(0)4.25=33.8ms^{-1}

In Y-Direction,

Given that initial velocity=u_{x}=0ms^{-1}

Using v=u+at,

v_{y}=0+(9.8)4.25=41.65ms^{-1}

v=\sqrt{v_{x}^{2}+v_{y}^{2}}

v=\sqrt{1142.44+1734.72}=\sqrt{2877.163}=53.63ms^{-1}

7 0
3 years ago
A 0.3-kg object connected to a light spring with a force constant of 19.3 N/m oscillates on a frictionless horizontal surface. A
Ghella [55]

The total work <em>W</em> done by the spring on the object as it pushes the object from 6 cm from equilibrium to 1.9 cm from equilibrium is

<em>W</em> = 1/2 (19.3 N/m) ((0.060 m)² - (0.019 m)²) ≈ 0.031 J

That is,

• the spring would perform 1/2 (19.3 N/m) (0.060 m)² ≈ 0.035 J by pushing the object from the 6 cm position to the equilibrium point

• the spring would perform 1/2 (19.3 N/m) (0.019 m)² ≈ 0.0035 J by pushing the object from the 1.9 cm position to equilbrium

so the work done in pushing the object from the 6 cm position to the 1.9 cm position is the difference between these.

By the work-energy theorem,

<em>W</em> = ∆<em>K</em> = <em>K</em>

where <em>K</em> is the kinetic energy of the object at the 1.9 cm position. Initial kinetic energy is zero because the object starts at rest. So

<em>W</em> = 1/2 <em>mv</em> ²

where <em>m</em> is the mass of the object and <em>v</em> is the speed you want to find. Solving for <em>v</em>, you get

<em>v</em> = √(2<em>W</em>/<em>m</em>) ≈ 0.46 m/s

8 0
3 years ago
Two canoeists in identical canoes exert the same effort paddling and hence maintain the same speed relative to the water. One pa
Serga [27]

Answer:

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

Explanation:

v_r = Speed of river

v_c = Speed of canoe

v_r+v_c=2.8\ m/s

v_r-v_c=-1.9\ m/s

Adding the equations we get

2v_r=0.9\\\Rightarrow v_r=\frac{0.9}{2}\\\Rightarrow v_r=0.45\ m/s

0.42+v_c=2.8\ m/s\\\Rightarrow v_c=2.8-0.45\\\Rightarrow v_c=2.65\ m/s

Speed of river is 0.45 m/s

Speed of boat is 2.65 m/s

6 0
3 years ago
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