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marysya [2.9K]
1 year ago
11

A juggler throws two balls to the same height so that one is at the halfway point going up when the other is at the halfway poin

t coming down. At that point:

Physics
1 answer:
horrorfan [7]1 year ago
4 0

We know that in a projectile motion the acceleration always points down and it has magnitude of 9.8 m/s^2 (the acceleration due to gravity). In this kind of motion we know that at equal heights the velocty has the same magnitude but opposite direction.

Since the juggler throws the ball to the same height this means that the balls will follow the same projectile motion then the properties mentioned above are satisfied.

Therefore we conclude that at that point:

Their accelerations are equal but their velocities are equal and opposite.

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Angelina_Jolie [31]

Answer:

E1 =  2996.667N/C E2 = 11237.5N/C

Explanation:

E1 = kQ1/r^2

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  = 2996.667N/C

E2 = kQ2/r^2

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A school bus moves slower and slower. Using what you have learned about forces, explain why the bus moves slower and slower.
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The water in a river flows uniformly at a constant speed of 2.50 m/s between parallel banks 80.0 m apart. You are to deliver a p
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Answer:

a)  The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) 133.33 m

c) 53.13°

d) 106.67 m

Explanation:

a) The swimmer should travel perpendicular to the bank to minimize the spent in getting to the other side.

b) velocity = distance * time

Let the velocity of the swimmer be v_{s} = 1.5 m/s

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t = 80/1.5

t = 53.33 s

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S = v_{r} t

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c) To minimize the distance traveled by the swimmer, his resultant velocity must be perpendicular to the velocity of the swimmer relative to water

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cos \theta = \frac{v_{s} }{v_{r} } \\cos \theta = 1.5/2.5\\cos \theta = 0.6\\\theta = cos^{-1} 0.6\\\theta = 53.13^{0}

d) Downstream velocity of the swimmer, v_{y} = v_{s} sin \theta\\

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the horizontal speed,

v_{x} = 2.5 - 1.5cos53.13\\v_{x} = 1.6 m/s

The downstream horizontal distance of the swimmer, x = v_{x} t

x = 1.6 * 66.67

x = 106.67 m

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