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vodomira [7]
3 years ago
15

Three people pull simultaneously on a stubborn donkey. Jack pulls directly ahead of the donkey with a force of 61.3 N, Jill pull

s with 83.9 N in a direction 45° to the left, and Jane pulls in a direction 45° to the right with 137 N. (Since the donkey is involved with such uncoordinated people, who can blame it for being stubborn?) Find the magnitude of the net force the people exert on the donkey.
Physics
1 answer:
miv72 [106K]3 years ago
7 0

Answer:

F_{net} = 220.8 N

Explanation:

It is pulled by three forces as given below

1. Jack pulls directly ahead of the donkey with a force of 61.3 N,

2. Jill pulls with 83.9 N in a direction 45° to the left, and

3. Jane pulls in a direction 45° to the right with 137 N.

Now net force directly in forward direction given as

F_x = 61.3 N + 83.9 cos45 + 137cos45

F_x = 217.5 N

Now similarly in perpendicular to this we have

F_y = 137 sin45 - 83 sin45

F_y = 38.2 N

Now net force is given by them

F_{net} = \sqrt{F_x^2 + F_y^2}

F_{net} = \sqrt{217.5^2 + 38.2^2}

F_{net} = 220.8 N

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Explanation:

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Answer:

Distance between slits, d=2.89\times 10^{-7}\ m  

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So, d=\dfrac{1/2\lambda}{sin\theta}

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Answer:

The magnitude of displacement is 0.082m

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Where M = combined mass of the platform and the two people.

V = velocity of the platform

m = mass of the ball

v = velocity of the ball

The distance that the platform moves is given by:

X = Vt ...eq2

Where t is the time that the ball is in the air.

The time the ball is in the air is given by:

L/(v-V) ...eq3

Where L is the length of the platform

The quantity(v-V) = velocity of the ball relative to the platform.

Combining eq2 and eq3

X = (V/(v - V))L

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X = - (3.36kg × 3.12m) /( 119kg + 3.36kg)

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Therefore, the distance moved by the platform is the magnitude of this displacement 0.082m.

7 0
3 years ago
Read 2 more answers
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