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GREYUIT [131]
3 years ago
9

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

Physics
1 answer:
sveta [45]3 years ago
6 0

Answer:

the two ice skater have the same momentum but the are in different directions.

Paula will have a greater speed than Ricardo after the push-off.

Explanation:

Given that:

Two ice skaters, Paula and Ricardo, initially at rest, push off from each other. Ricardo weighs more than Paula.

A. Which skater, if either, has the greater momentum after the push-off? Explain.

The law of conservation of can be applied here in order to determine the skater that possess a greater momentum after the push -off

The law of conservation of momentum states that the total momentum of two  or more objects acting upon one another will not change, provided there are no external forces acting on them.

So if two objects in motion collide, their total momentum before the collision will be the same as the total momentum after the collision.

Momentum is the product of mass and velocity.

SO, from the information given:

Let represent the mass of Paula with m_{Pa} and its initial velocity with u_{Pa}

Let represent the mass of Ricardo with m_{Ri} and its initial velocity with u_{Ri}

At rest ;

their velocities will be zero, i.e

u_{Pa} = u_{Ri} = 0

The initial momentum for this process can be represented as :

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} = 0

after push off from each other then their final velocity will be v_{Pa} and v_{Ri}

The we can say their final momentum is:

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri} = 0

Using the law of conservation of momentum as states earlier.

Initial momentum = final momentum = 0

m_{Pa}u_{Pa} +  m_{Ri}u_{Ri} =  m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

Since the initial velocities are stating at rest then ; u = 0

m_{Pa}(0) + m_{Pa}(0) = m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}

m_{Pa}v_{Pa} +   m_{Ri}v_{Ri}  = 0

m_{Pa}v_{Pa} = - m_{Ri}v_{Ri}

Hence, we can conclude that the two ice skater have the same momentum but the are in different directions.

 B. Which skater, if either, has the greater speed after the push-off? Explain.

Given that Ricardo weighs more than Paula

So m_{Ri} > m_{Pa} ;

Then \mathsf{\dfrac{{m_{Ri}}}{m_{Pa} }= 1}

The magnitude of their momentum which is a product of mass and velocity can now be expressed as:

m_{Pa}v_{Pa} =  m_{Ri}v_{Ri}

The ratio is

\dfrac{v_{Pa}}{v_{Ri}} =\dfrac{m_{Ri}}{m_{Pa}} = 1

v_{Pa} >v_{Ri}

Therefore, Paula will have a greater speed than Ricardo after the push-off.

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Four charges with equal magnitudes of 10.6 × 10-12 C are placed at the corners of a rectangle. The lengths of the sides of the r
cricket20 [7]

Answer:

Figure a. E_net = 99.518 N/C

Figure b. E_net = 177.151 N / C

Explanation:

Given:

- Attachment for figures missing in the question.

- The dimensions for rectangle are = 7.79 x 3.99 cm

- All four charges have equal magnitude Q = 10.6*10^-12 C

Find:

Find the magnitude of the electric field at the center of the rectangle in Figures a and b.

Solution:

- The Electric field generated by an charged particle Q at a distance r is given by:

                                         E = k*Q / r^2

- Where, k is the coulomb's constant = 8.99 * 10^9

Part a)

- First we see that the charges +Q_1 and +Q_3 produce and electric field equal but opposite in nature. So the sum of Electric fields:

                                 E_1 + E_3 = 0

- For Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Hence, the net Electric Field at center of the rectangle can be given as:

                                  E_net = E_2 + E_4

                                  E_2 = E_4

                                  E_net = 2*E = 2*k*Q / r^2

- The distance r from each corner to mid-point of the rectangle is constant. It can be evaluated by Pythagoras Theorem as follows:

                                  r = sqrt ( (7.79/200)^2 + (3.99/200)^2 )

                                  r = sqrt ( 1.9151*10^-3 ) = 0.043762 m

- Plug the values in the E_net expression developed above:

                                  E_net = 2*(8.99*10^9)*(10.6*10^-12) / 1.9151*10^-3

                                 E_net = 99.518 N/C

Part b)

- Similarly for Figure b, for Charges -Q_2 and +Q_4, they are equal in nature but act in the same direction towards the negative charge -Q_2. Also, Charges -Q_1 and +Q_3, they are equal in nature but act in the same direction towards the negative charge -Q_1. These Electric fields are equal in magnitude to what we calculated in part a).

- To find the vector sum of two Electric Fields E_1,3 and E_2,4 we see the horizontal components of each cancels each other out. While the vertical components E_1,3 and E_2,4 are equal in magnitude and direction.

Hence,

                                  E_net = 2*E_part(a)*cos(Q)

- Where, Q is the angle between resultant, vertical in direction, and each of the electric field. We can calculate Q using trigonometry as follows:

                                  Q = arctan ( 3.99 / 7.79 ) = 27.12 degrees.

- Now, compute the net electric field E_net:

                                  E_net = 2*(99.518)*cos(27.12)

                                  E_net = 177.151 N / C

               

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Answer

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mass of the bar = 10 kg

smaller mass = 4 kg

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difference between two scale = 8 - 6

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