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SpyIntel [72]
3 years ago
6

Ruby is out with her friends. Misfortune occurs and

Physics
1 answer:
zhannawk [14.2K]3 years ago
3 0

The work done is 2.35\cdot 10^5 J

Explanation:

The work done by a force on an object is given by:

W=Fd cos \theta

where

F is the magnitude of the force

d is the displacement of the object

\theta is the angle between the direction of the force and of the displacement

In this problem, we have

F = 1080 N is the force applied on the car

d = 218 m is the displacement of the car

And assuming the force is applied parallel to the motion of the car, \theta=0^{\circ}, and so the work done is

W=(1080)(218)(cos 0^{\circ})=2.35\cdot 10^5 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

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How much force would a 55kg person experience on a roller coaster if they travel through a loop with a radius of 55kg at a speed
Inessa [10]

Answer:

200 N

Explanation:

For a body moving in uniform circular motion, the force acting on it will be <em>centripetal force</em> and its direction is <em>radially inward</em> , pointing to the center.

The radially inward acceleration, or the centripetal acceleration is given by :

                                          a = v² / r

           where v is the speed at which the body is moving and r is the radius of the circle

Given-

m = 55kg

v = 14.1 m/s

r= 55m

We know that F = ma

⇒   F = m (  v²/ r )

⇒ F = 55 x 14.1 x 14.1 / 55

⇒ F =14.1 x 14.1 = 200 N

∴ <em>The force acting is 200 N</em>.

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Which of the following is an example of a conductor?
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A 1300-N crate rests on the floor. How much work is required to move it at constant speed (a)
kherson [118]

a) The work done is 920 J

b) The work done is 5200 J

Explanation:

a)

In this first part of the problem, the crate is moved horizontally at constant speed.

The work required in this case is given by

W=Fd cos \theta

where

F is the magnitude of the force applied

d is the displacement of the crate

\theta is the angle between the direction of the force and of the displacement

Here the crate is moved at constant speed: this means that the acceleration of the crate is zero, and so according to Newton's second law, the net force on the crate is zero: this means that the force applied, F, must be equal to the force of friction (but in opposite direction), so

F = 230 N

The displacement is

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied horizontally. Therefore, the work done is

W=(230)(4.0)(cos 0^{\circ})=920 J

b)

In this case, the crate is moved vertically. The force that must be applied to lift the crate must be equal to the weight of the crate (in order to move it a constant speed), therefore

F = W = 1300 N

The displacement this time is again

d = 4.0 m

And the angle is \theta=0^{\circ}, since the force is applied vertically, and the crate is moved also vertically. Therefore, the work done on the crate this time is

W=(1300)(4.0)(cos 0^{\circ})=5200 J

Learn more about work:

brainly.com/question/6763771

brainly.com/question/6443626

#LearnwithBrainly

4 0
3 years ago
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