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3 years ago

Zorn and Porsha are ice skating. Porsha has a mass of 60 kg, and Zorn has a mass of 40 kg. As they face each

1 answer:

algol133 years ago

3 0

Assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Missing part of the question: determine the magnitude of Porsha's acceleration.

Given the data in the question;

Mass of Porsha; $m_{porsha} = 60kg$

Mass of Zorn; $m_{zorn} = 40kg$

Force of Porsha push; $F_{porsha} = 168N$

Magnitude of Porsha's acceleration; $a = \ ?$

To determine the magnitude of Porsha's acceleration, we use Newton's second laws of motion:

$F = m*a$

Where m is the mass of the object and a is the acceleration.

We substitute the mass of Porsha and the force he used into the equation

$168N = 60kg * a\\\\a = \frac{168kg.m/s^2}{60kg}\\\\a = 2.8m/s^2$

Therefore, assuming the friction between the skaters and the ice is negligible, the magnitude of Porsha's acceleration is 2.8m/s².

Learn more: brainly.com/question/25125444

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2 years ago

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4 years ago

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3 years ago

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Ezra (m = 20.0 kg) has a tire swing and wants to swing as high as possible. He thinks that his best option is to run as fast as

Dmitriy789 [7]

Answer:

a) $v=5.6725\,m.s^{-1}$

b) $h= 1.6420\,m$

Explanation:

Given:

mass of the body, $M=20\,kg$

mass of the tyre, $m=10\,kg$

length of hanging of tyre, $l=3.5m$

distance run by the body, $d=10m$

acceleration of the body, $a=3.62m.s^{-2}$

(a)

Using the equation of motion :

$v^2=u^2+2a.d$ ..............................(1)

where:

v=final velocity of the body

u=initial velocity of the body

here, since the body starts from rest state:

$u=0m.s^{-1}$

putting the values in eq. (1)

$v^2=0^2+2\times 3.62 \times 10$

$v=8.5088\,m.s^{-1}$

Now, the momentum of the body just before the jump onto the tyre will be:

$p=M.v$

$p=20\times 8.5088$

$p=170.1764\,kg.m.s^{-1}$

Now using the conservation on momentum, the momentum just before climbing on the tyre will be equal to the momentum just after climbing on it.

$(M+m)\times v'=p$

$(20+10)\times v'=170.1764$

$v'=5.6725\,m.s^{-1}$

(b)

Now, from the case of a swinging pendulum we know that the kinetic energy which is maximum at the vertical position of the pendulum gets completely converted into the potential energy at the maximum height.

So,

$\frac{1}{2} (M+m).v'^2=(M+m).g.h$

$\frac{1}{2} (20+10)\times 5.6725^2=(20+10)\times 9.8\times h$

$h\approx 1.6420\,m$

above the normal hanging position.

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3 years ago

A car moving at 10 m/s crashes into a large bush and stops in 1.3 m. Using the Work-Energy theorem, calculate the average force

ad-work [718]

Answer:

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

It takes 0.26 s to bring the passenger in the car to a halt.

Explanation:

Hi there!

The negative work (W) needed to bring a moving object to stop is equal to its kinetic energy (KE):

W = KE

F · s = 1/2 · m · v²

Where:

F = applied force on the passenger.

s = displacement of the passenger.

m = mass of the passenger.

v = velocity of the passenger.

Solving the equation for F:

F = 1/2 · m · v² / s

Replacing with the data:

F = 1/2 · 70 kg · (10 m/s)² / 1.3 m

F = 2692.3 N

The magnitude of the average force the seat belt exerts on the passenger is 2692.3 N.

According to the second law of Newton:

F = m · a

Where "a" is the acceleration of the passenger.

We also know from kinematics that the velocity of an object can ve calculated as follows:

v = v0 + a · t

Where:

v = velocity of the passenger at time t.

v0 = initial velocity.

t = time.

a = acceleration.

When the passenger stops, its velocity is zero. So replacing a = F/m, let´s solve the equation for the time it takes the passenger to stop:

v = v0 + a · t

0 = 10 m/s + (-2692.3 N/ 70 kg) · t

-10 m/s / (-2692.3 N/ 70 kg) = t

t =0.26 s

It takes 0.26 s to bring the passenger in the car to a halt.

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3 years ago