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2 years ago

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diving to work one morning, you get a flat tire. when using the car jack, you apply 120 N of force to the jack and the jack in t

urns applies 2000 N of force to lift the car up. what is the mechanical advantage of the jack

1 answer:

dybincka [34]2 years ago

6 0

Mechanical advantage = (output force) / (input force)

MA = (2,000 N) / (120 N)

<em>MA = 16.67</em> (16 and 2/3)

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A 20-N force acts on an object with a mass of 2.0kg. what is the objects acceleration?

ale4655 [162]

Answer:

10 m/s^2

Explanation:

Equation: F = ma.

a = acceleration

m = mass

F = force

Because we are trying to find acceleration instead of force we want to rearrange the equation to solve for a which is F/m = a.

F = 20

m = 2

a = ?

a = F/m

a = 20/2

a = 10 m/s^2

7 0

3 years ago

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Football player 1 has a mass of 80 kg and a velocity of 2 m/s east while player 2 has a mass of 70 kg and a velocity of 3 m/s we

sukhopar [10]

The total momentum of the system is equal to 50 Kgm/s.

<u>Given the following data:</u>

Mass 1 = 80 kg

Velocity 1 = 2 m/s east.

Mass 2 = 70 kg

Velocity 2 = 3 m/s west.

To determine the total momentum of the system:

Mathematically, momentum is given by the formula;

$Momentum = mass \times velocity$

<u>For Football player 1:</u>

$Momentum = 80 \times 2$

Momentum 1 = 160 Kgm/s.

<u>For Football player 2:</u>

$Momentum = 70 \times 3$

Momentum 1 = 210 Kgm/s.

Now, we can calculate the total momentum of the system:

$Total\;momentum = momentum \;2 - momentum \;1\\\\Total\;momentum = 210-160$

Total momentum = 50 Kgm/s.

<u>Note:</u> We subtracted because the football players were moving in opposite directions.

Read more: brainly.com/question/15517471

6 0

2 years ago

PLEASE ANYONE SOLVE THIS NOW FAST PLEASE IM IN A HURRY

m_a_m_a [10]

Hello,

<u>Solution for A:</u>

Force = 3.00N

Mass = 0.50 Kgs

Time = 1.50 Seconds

According to newton's second law of motion;

Force = Mass times Acceleration(a)

3.00 = 0.50 * a

a = 3.00/0.50 = 6.00 m/s^2

We know that acceleration = Velocity / time

So Velocity = time * acceleration = 1.50 * 6 = 9.00 m/s^2

<u>Solution for B:</u>

The net force = 4.00N - 3.00N = 1.00N to the left

Force = 1.00N

Mass = 0.50Kg

Time = 3.00 Seconds

Again; F = MA (Where F is force, M is mass and A is acceleration)

1.00N = 0.5 * A

A = 1/0.5 = 2 m/s^2

Velocity = Acceleration * Time = 2 * 3 = 6 m/s

3 0

2 years ago

It is a general exercise that we do before practicing sports activities<br><br> worm-up or cool down

Montano1993 [528]

It is called a warm-up.

3 0

2 years ago

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Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

2 years ago