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Elenna [48]
3 years ago
10

10 points) A rubber ball and a lump of putty of the same mass m = 0.05 kg are thrown with the same speed v=10.0 m/s in positive

x-direction against a vertical wall. The ball bounces back with the same speed v’=10.0 m/s in negative x-direction. The putty sticks to the wall. Which of the two objects (rubber ball and a lump of putty) experiences the greater momentum change?
Physics
1 answer:
snow_lady [41]3 years ago
4 0

Answer:

the ball experiences the greater momentum change

Explanation:

You have to take into account that momentum change is given by

\Delta p=mv_f-mv_b

where vf and vb are the speed of the object after and before the impact.

In the case of the ball you have

\Delta p=(0.05kg)(-10.0\frac{m}{s})-(0.05kg)(10.0\frac{m}{s})=-1kg\frac{m}{s}

where the minus of vf is included due to the motion is in an opposite direction regarding with vb

And for the lump

\Delta p=(0.05kg)(10.0\frac{m}{s})-(0.05kg)(0\frac{m}{s})=0.5kg\frac{m}{s}

Hence, the ball experiences the greater change

hope this helps!!

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Can a goalkeeper at his goal kick a soccer ball into the opponent’s goal without the ball touching the ground? The distance will
zmey [24]

The goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

Explanation:

Consider the vertical motion of ball,

We have equation of motion v = u + at

     Initial velocity, u  = u sin θ

     Final velocity, v =  0 m/s    

     Acceleration = -g

     Substituting

                      v = u + at  

                      0 = u sin θ - g t

                      t=\frac{usin\theta }{g}

This is the time of flight.

Consider the horizontal motion of ball,

        Initial velocity, u =  u cos θ

        Acceleration, a =0 m/s²  

        Time, t=\frac{usin\theta }{g}  

     Substituting

                      s = ut + 0.5 at²

                      s=ucos\theta \times \frac{usin\theta }{g}+0.5\times 0\times (\frac{usin\theta }{g})^2\\\\s=\frac{u^2sin\theta cos\theta}{g}\\\\s=\frac{u^2sin2\theta}{2g}

This is the range.

In this problem

              u = 30 m/s

              g = 9.81 m/s²

              θ = 45° - For maximum range

Substituting

               s=\frac{30^2\times sin(2\times 45)}{2\times 9.81}=45.87m

Maximum horizontal distance traveled by ball without touching ground is 45.87 m, which is less than 95 m.

So the goalkeeper at his goal cannot kick a soccer ball into the opponent’s goal without the ball touching the ground

6 0
3 years ago
A force of 45 N is applied tangentially to the rim of a solid disk of radius 0.12 m. The disk rotates about an axis through its
bearhunter [10]

Answer:

Mass of the disk will be 2.976 kg

Explanation:

We have given force F = 45 N

Radius of the disk r = 0.12 m

Angular acceleration \alpha =140rad/sec^2

We know that torque \tau =I\alpha

And \tau =Fr

So Fr=I\alpha , here I is moment of inertia

So 50\times 0.12=I\times 140

I=0.0428kgm^2

We know that moment of inertia I=\frac{1}{2}mr^2

So 0.0428=\frac{1}{2}\times m\times 0.12^2

m = 2.976 kg

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Describe the motion for each segment below. Include start position, relative speed, and direction. Then calculate the speed of e
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Where is the picture?
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Which kind of cloud touches the ground making it hard to see
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A funnel cloud is a funnel-shaped cloud of condensed water droplets. They usually appear with a rotating column of air. These extend from the bottom of a cloud that does not touch the ground or a water surface.
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Why are we not crushed by the weight of the atmosphere on our shoulders?
antiseptic1488 [7]

Answer:

Due to equal pressure in all the direction at a particular level in a fluid medium (Pascal's Law)

Explanation:

We are not crushed by the weight of the atmosphere because atmosphere is a fluid and we are immersed into it. So, according to the Pascal's law the the pressure a each point in a horizontal level is equal in all the direction irrespective of the orientation of a body.

Variation of pressure in term of the height of a fluid medium is given as:

P=\rho.g.h

\rho=density of fluid

g = acceleration due to gravity

h = height of the free surface of the fluid from the immersed object.

  • And atmosphere has very less variation of pressure with change in height as it is a rare medium fluid and so for a human height there is very negligible variation of pressure at the heat of a human with respect to his toe.

5 0
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