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julsineya [31]
4 years ago
15

How many arrows of force would be needed in a free body diagram of a rocket taking off from the moon? Ignore any solar system ob

jects other than the moon itself.
One
Two
Four
Three

A person is trying to lift a crate that has a mass of 15.3 kg. The normal force of the floor is currently supplying 150 N of force. How much force does the person need to supply in order to pick up the crate?
150 N
144 N
294 N
130 N

A person is trying to lift a crate that has a mass of 30 kg. The normal force of the floor is 150 N of force. How much force is the person currently exerting?
114 N
150 N
294 N

Physics
1 answer:
Sophie [7]4 years ago
7 0

#1

When rocket is launched from the moon then there will be two forces on the rocket

1. Due to ejected gases the reaction force on rocket in forward direction

2. due to gravitational attraction of moon

So here we need to draw two arrow on its FBD in which first arrow is in the direction of motion of rocket which will be the reaction force on it while other arrow is opposite to its motion which is due to gravitational attraction force.

So answer would be TWO

#2

When crate is placed on the floor then in order to lift the crate upwards we always need an external force that will counter balance the normal force of crate.

Which mean if we make the normal force to be zero then crate will lift upwards and in order to make it zero we need to apply same force as that of normal force

So here the normal force is given as 150 N so we also need to apply 150 N force in order to lift it upwards


#3

If a person lift a box upwards by some force then we can say by force balance

Force applied by person + Normal force = Weight of the box

F_{app} + F_n = mg

we will plug in all values in it

F_{app} + 150 = 30 * 10

F_{app} = 300 - 150

F_{app} = 150 N

So we applied 150 N force on it

#4

In all cases we need to use force balance

like in case 3

T + T = W

2T = W

T = \frac{W}{2}

so here in 3rd case tension is half of the weight

which will be the least tension in all cases

so answer is CASE 3

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Answer:

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A grinding wheel 0.35 m in diameter rotates at 2600 rpm .
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It is calculated that a)The angular velocity of the wheel is 272.13 rad/s,

b)On the edge of the grinding wheel, the linear speed is 47.62 m/s,

and c) On the edge of the grinding wheel, the acceleration is 12958.08 m/s².

Calculation of angular velocity, linear speed & acceleration:

Provided that,

the diameter of the wheel = 0.35 m

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As 1 revolution = 2π rad

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So, the angular velocity is 272.13 rad/s.

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a = \frac{(47.62)^{2} }{0.175}

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Learn more about angular velocity here:

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