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3 years ago

Newton’s third law in football?

1 answer:

6 0

Newtons third law of motion applies to football as when the ball is thrown, force is applied to the football thrown forward and also force back to the player. Therefore it states that every action has a reaction. The player does not get a huge force back from throwing the football due to the player having a bigger mass than the football being thrown.

In short, when the football is throw, there is a reaction with the ball and the player accordingly with the force put into the ball.

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A 1100 kgkg safe is 2.4 mm above a heavy-duty spring when the rope holding the safe breaks. The safe hits the spring and compres

Drupady [299]

Answer:

191.36 N/m

Explanation:

From the question,

The Potential Energy of the safe = Energy of the spring when it was compressed.

mgh = 1/2ke²............... Equation 1

Where m = mass of the safe, g = acceleration due to gravity, h = height of the save above the heavy duty spring , k = spring constant, e = compression

Making k the subject of the equation,

k =2mgh/e²................ Equation 2

Given: m = 1100 kg, h = 2.4 mm = 0.0024 m, e = 0.52 m

Constant: g = 9.8 m/s²

Substitute into equation 2

k = 2(1100)(9.8)(0.0024)/0.52²

k = 51.744/0.2704

k = 191.36 N/m

Hence the spring constant of the heavy-duty spring = 191.36 N/m

3 0

2 years ago

How do you find the normal force here? I forgot

kakasveta [241]

Normal force is mass x gravity, so mass x 9.81

6 0

3 years ago

Read 2 more answers

what is the acceleration of an object if the object has an initial speed of 230 m/s and speeds up to 650 m/s. The time it takes

adell [148]

Answer:

explain this

Explanation:

7 0

2 years ago

The main purpose of an air bag is to stop a passenger during a car accident in a greater amount of time than if the air bag were

Simora [160]

Answer:

a) 45571 N

b) 22786 N

c) 4557 N

Explanation:

Since the goal of the airbag is helping the person to stop after the collision in a greater time, this means that the change in momentum must finish when this is just zero.
In other words, the change in momentum, must be equal to the initial one, but with opposite sign.

$\Delta p = - p_{o} = -m*v = -55 kg*29m/s = -1595 kgm/s (1)$

Now, just applying the original form of Newton's 2nd Law, we know that this change in momentum must be equal to the impulse needed to stop the person:

$\Delta p = F* \Delta t (2)$

So, as we know the magnitude of Δp from (1) and we have different Δt as givens, we can get the different values of F (in magnitude) required to stop the person for each one of them, as follows:

$F_{1} = \frac{\Delta p}{\Delta t_{1}} = \frac{1595kgm/s}{0.035s} = 45571 N (3)$

$F_{2} = \frac{\Delta p}{\Delta t_{2}} = \frac{1595kgm/s}{0.07s} = 22786 N (4)$

$F_{3} = \frac{\Delta p}{\Delta t_{3}} = \frac{1595kgm/s}{0.35s} = 4557 N (5)$

4 0

3 years ago

Why might the 1990 earthquake and volcanic eruption in Luzon be related?

Art [367]

I am not sure but i think the answer is C

6 0

3 years ago