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Bess [88]
2 years ago
15

Change the following sentences into Passive Voice: (5M)

Physics
1 answer:
geniusboy [140]2 years ago
7 0

Answer:

passive voice

many messengers all over the world was sent by emperor Ashoka to preach Buddhism.

Which pen was liked by you?

Has your passport size photo been taken for the application form?

A beautiful bicycle was given to me on my birthday by my father.

The plants is being watered by the gardener.

Paul said that he will never leave you and he will always be with you.

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A 41 kg girl and a 5.0 kg sled are on the frictionless ice of a frozen lake, 15 m apart but connected by a rope of negligible ma
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(a) 0.8 m/s^2

The force exerted on the sled is F = 4.0 N. We can calculate the acceleration of the sled by using Newton's second law:

F=ma

where

m = 5.0 kg is the mass of the sled

a is the acceleration of the sled

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{5.0 kg}=0.8 m/s^2

(b) 0.098 m/s^2

According to Newton's third law (action-reaction law), since the girl exerts a force on the sled, then the sled exerts an equal and opposite force on the girl as well. This means that the force exerted on the girl is also F = 4.0 N. As before, we can calculate the acceleration of the girl by using Newton's second law:

F=ma

where

m = 41 kg is the mass of the girl

a is the acceleration of the girl

Solving the equation for a, we find:

a=\frac{F}{m}=\frac{4.0 N}{41 kg}=0.098 m/s^2

(c) 5.8 s

Taking the initial position of the girl as x = 0, the position at time t of the girl is given by:

x(t)=\frac{1}{2}a_g t^2

where a_g = 0.098 m/s^2 is the acceleration of the girl.

The sled starts instead its motion from x = 15 m, so its position at time t is given by

x'(t)=15-\frac{1}{2}a_s t^2

where a_s=0.8 m/s^2 is the acceleration of the sled, and the negative sign is due to the fact that the sled accelerates in opposite direction to the girl's acceleration.

The girl and the sled meet when x(t) = x'(t). So, we find:

\frac{1}{2}a_g t^2=15-\frac{1}{2}a_s t^2\\(a_g+a_s) t^2=30 m\\t=\sqrt{\frac{30 m}{a_g+a_s}}=\sqrt{\frac{30 m}{0.8 m/s^2+0.098 m/s^2}}=5.8 s

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You are driving at the speed of 27.7 m/s (61.9764 mph) when suddenly the car in front of you (previously traveling at the same s
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1) Acceleration of the car in front: -7.89 m/s^2

The only data we need for this part of the problem is:

u = 27.7 m/s --> initial velocity of the car

\mu=0.804 --> coefficient of friction between the car wheels and the road

From the coefficient of friction, we can find the deceleration of the car. In fact, the force of friction is given by

F=-\mu mg

where m is the car's mass and g=9.81 m/s^2 is the acceleration due to gravity. We can find the car's acceleration by using Newton's second law:

a=\frac{F}{m}=\frac{-\mu mg}{m}=\mu g=(0.804)(9.81 m/s^2)=-7.89 m/s^2

And the negative sign means it is a deceleration.


2) Braking distance for the car in front: 48.6 m

This can be found by using the following SUVAT equation:

v^2 - u^2 = 2aS

where

v=0 is the final velocity of the car

u=27.7 m/s is the initial velocity of the car

a=-7.89 m/s^2 is the acceleration of the car

S is the braking distance

By re-arranging the formula, we find S:

S=\frac{v^2-u^2}{2a}=\frac{0-(27.7 m/s)^2}{2(-7.89 m/s^2)}=48.6 m


3) Minimum safe distance at which you can follow the car: 15.0 m

In this case, we must calculate the thinking distance, which is the distance you travel before hitting the brakes. During this time, the speed of your car is constant, so the thinking distance is given by

d_t = ut=(27.7 m/s)(0.543 s)=15.0 m

After hitting the brakes, your car decelerates at the same rate of the car in front of you, so the braking distance is the same of the other car:

d_b=48.6 m

So the total distance your car covers is

S'=d_t+d_b=15.0 m +48.6 m=63.6 m

At the same time, the car in front of you just covered a distance of 48.6 m. So, in order to avoid the collision, you should travel at a distance equal to

d=S'-S=63.6 m-48.6 m=15.0 m


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