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Ostrovityanka [42]
3 years ago
7

Chuck and Jackie stand on separate carts, both of which can slide without friction. The combined mass of Chuck and his cart, mca

rt, is identical to the combined mass of Jackie and her cart. Initially, Chuck and Jackie and their carts are at rest. Chuck then picks up a ball of mass mball and throws it to Jackie, who catches it. Assume that the ball travels in a straight line parallel to the ground (ignore the effect of gravity). After Chuck throws the ball, his speed relative to the ground is vc. The speed of the thrown ball relative to the ground is vb. Jackie catches the ball when it reaches her, and she and her cart begin to move. Jackie's speed relative to the ground after she catches the ball is vj. When answering the questions in this problem, keep the following in mind: The original mass mcart of Chuck and his cart does not include the mass of the ball. The speed of an object is the magnitude of its velocity. An object's speed will always be a nonnegative quantity. Find the relative speed u between Chuck and the ball after Chuck has thrown theball.
Express the speed in terms ofv_c and v_b.
Physics
1 answer:
dmitriy555 [2]3 years ago
5 0

As per momentum conservation we can say that Chuck and ball as a system has no external force

So here momentum of them must be conserved

now here initially if they are at rest

so we can say

P_i = P_f

0 = m_{cart}v_c + m_{ball} v_b

now we have

v_b = \frac{m_{cart}v_c}{m_{ball}

now the relative velocity of the cart and ball is given as

v_r = v_c + \frac{m_{cart}v_c}{m_{ball}

also we can say since they are moving in opposite direction so relative speed will be

v_r = v_c + v_b

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Answer:

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Explanation:

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In the direction of the y-axis (perpendicular to the ice) you have the weight of the disk and the normal to this weight that are also in equilibrium.

b) the puck does not move since the sum of the forces is zero, which implies that the forces of the hockey sticks are of equal magnitude and opposite direction.

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* inclined the stick slightly so that the force has a vertical component and the puck jumps in this direction

* Increasing the magnitude of the force so that the puck comes out in the opposite direction to the player

* The worst case, decreasing its force to zero and the disk comes out in its direction by the other force that had the same magnitude.

3 0
3 years ago
A force of 20 N acts on a 4 kg cart for 10 s . How far will it go (using the second law) ?
Elden [556K]
This took me a short while to figure out, but I am still not entirely sure if this is correct, this is just from my basic understanding of Newtons Second Law of Motion.

You have a 4kg cart with a force of 20N acting on it.

The formula for working out the acceleration is.
a=Fnet÷mass

Substitute in the information.
a=20N÷4kg

Now you solve it to give you.
a=5m/s

So now what you should be able to do is figure out that after 10 seconds the cart travelling at 5m/s would have travelled 10 metres.

This is achieved by finding out how many 5's go into 10 which is 2.
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The 4kg cart has travelled 10 meters in 10 seconds with a force of 20N acting upon it.

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8 0
3 years ago
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Answer:

600 mC

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so number of electrons, n = (10 x 10-3)/(1.6 x 10-19) = 6.25 x 1016  so in a minute the charge is 10 * 60 = 600 mC

4 0
3 years ago
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