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OLga [1]
3 years ago
8

Come all ruj-jxgn-qua​

Physics
1 answer:
Shalnov [3]3 years ago
5 0

Answer:

this app is for solving doubts not sending links ok

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Can you hit the monkey at a fast speed with gravity on?
Ierofanga [76]

Answer:

It depends how far the monkey is and how hard you throw the object because If the monkey is close, then you will most likely hit it

7 0
3 years ago
One block rests upon a horizontal surface. A second identical block rests upon the first one. The coefficient of static friction
goblinko [34]

Answer:

The magnitud of the force is 124.8N.

Explanation:

First we have to find the value of the static friction coefficient, when the external force F is applied to upper block (i will call it A Block) we have a free body diagram as the one shown in the figure i attached, so since this block has no aceleration in any direction the force F should be equal to the friction force between A and B block, one we noticed this we can use the equation for the Friction force to find the coefficient:

0=F-FrictionAB

F=FrictionAB=Nab*μs

and again, since the block has no acceleration the normal between A and B block should be equal to the weigth of the first block, so we have:

0=Nab-W

Nab=W=mg

replacing this we have:

F=μs*Nab=μs*mg=41.6N

and  μs=41.6N/(mg)

now it's time to see the free body diagram for the b block, if we now apply the F force to the B block the diagram should look like in the figure.

the color of the arrow gives you an idea of where the force comes from, the blue ones comes from the B block, the red ones from the A block and the brown ones from the ground.

now for the B block you can see two friction forces, one for the ground and one for the A block, both of these directed bacwards, and two normal forces, again one for the ground and one for the A block but the normal force for the A block is aiming downwards.

again we use the fact that the block is not accelerating in any direction so the sum of the forces in x and y direction have to be 0, so:

F-Friction1(ground)-Friction2(AB)=0

This is the new external F force that we are looking for:

F=Friction1(ground)+Friction2(AB)

we know Friction2(AB) because we found that in the previous block so:

F=Friction1(ground)+mg*μs

for the other friction we have to use the equation:

Friction(ground)=N(ground)*μs

from y axis we have:

N(ground)-w-Normal(AB)=0

N(ground)=w+Normal(AB)

we found the value of Normal(AB) with the previous block so:

N(ground)=mg+mg=2mg

and:

Friction(ground)=2mg*μs

F=Friction(ground)+mg*μs

F=2mg*μs+μs*mg=3mg*μs

and since: μs*mg=41.6N

the new F force would be:

F=3mg*μs=41.6*3=124.8N

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