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vitfil [10]
3 years ago
8

At the start of a hockey game, the referee drops the puck between two players from opposing teams. Each player wants to push the

puck in the opposite direction. For several seconds the puck does not move even though both players are pushing on it with their hockey sticks. a.) Identify the forces acting on the puck. b.) Explain why the puck does not move. c.) Describe how one of the players could get the puck to move.
Physics
1 answer:
spin [16.1K]3 years ago
3 0

Answer:

a) puck is subjected to both the forces of the hockey sticks in a horizontal direction,

b)the puck does not move since the sum of the forces is zero

c) changing the magnitude or direction of its applied force

Explanation:

a) The puck is subjected to both the forces of the hockey sticks in a horizontal direction, these forces are of equal magnitude and opposite direction since the puck is at rest.

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.

c) the player has several ways to make the puck move

* slightly changing the angle of the club and therefore the direction of the force, in this case the disc comes out in the direction of this component

* 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.

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If the motor is operated at its full maximum rated capacity, then

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When 16.6 V is applied to a capacitor, it stores an energy of 8.15 x 10-4 J. How much charge is on the plates?
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The phenomenon of vehicle "tripping" is investigated here. The sport-utility vehicle is sliding sideways with speed v1 and no an
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Answer:

v_1  = 3.5 \ m/s

Explanation:

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moment of inertia about G , i.e I_G = 875 kg.m²

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H_1= H_2

mv_1 *0.765 = [I+m(0.765^2+0.895^2)] \omega_2

2140v_1*0.765 = [875+2140(0.765^2+0.895^2)] \omega_2

1637.1 v_1 = 3841.575 \omega_2

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\omega _2 = 0.4626 \ v_1

From the conservation of energy as well;we have :

T_2 +V_{2  \to 3} = T_3 \\ \\ \\  \frac{1}{2} I_A \omega_2^2 - mgh =0

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706.93 \ v_1^2 - 8657.49 =0

706.93 \ v_1^2  = 8657.49

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v_1 ^2 =  12.25

v_1  = \sqrt{ 12.25

v_1  = 3.5 \ m/s

6 0
3 years ago
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