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Vedmedyk [2.9K]

3 years ago

12

A football is kicked at ground level with the speed of 32 m/s at an angle of 40 degrees to the horizontal. how much later does i

t hit the ground?

1 answer:

LUCKY_DIMON [66]3 years ago

4 0

the question was a football is kicked at ground level with a speed of 32 m/s the an angle of 40 degrees to the horizontal how much later does it hit the ground probably like 30 more or 40 sense it was around 32

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How do you find the velocity of an object?

Scorpion4ik [409]

<span>Basically just divide the change in position by the change in time. </span>

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A student, standing on a scale in an elevator at rest, sees that his weight is 840 n. as the elevator rises, his weight increase

Firlakuza [10]

<span>1) at rest his weight is 840 N

=> 840N = mass * g => mass = 840 N / g = 840 N / 9.8 m/s^2 = 85.7 kg

2) as the elevator rises, his weight increases to 1050 N,

The reading of the scale is the norma force of it over the body of the person.

And the equation for the force is: Net force = mass * acceleration = normal force - weight at rest

=> mass * acceleration = 1050 N - 840 N = 210 N

acceleration = 210 N / mass = 210 N / 85.7 kg = 2.45 m/s^2 (upward)

3) when the elevator slows to a stop at the 10th floor, his weight drops to 588 N

=> mass * acceleration = 588 N - 840 N = - 252 N

=> acceleration = - 252 N / 85.71 kg = - 2.94 m / s^2 (downward)

Answer:

Acceleration at the beginning of the trip 2.45 m/s^2 upward
Acceleration at the end of the trip 2.94 m/s^2 downward
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The tire on this drag racer is severely twisted: The force of the road on the tire is quite large(most likely several times the

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Answer: Speeding Up

Explanation:

Force is proportional to the acceleration of an object. The greater the force, the more acceleration will be produced.

When the force on the tire is equal to the weight of the car, the car is reaching a stability as a result of increase in motion.

But when the force of the load on the tire is quite large(most likely several times the weight of the car) and is directed forward, then, the car is at high speed.

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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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