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ArbitrLikvidat [17]
3 years ago
5

After the box comes to rest at position x1, a person starts pushing the box, giving it a speed v1. When the box reaches position

x2 (where x2>x1), how much work Wp has the person done on the box
Physics
1 answer:
Olin [163]3 years ago
7 0

Answer:

0.5m(v1)²

Explanation:

We know that work done on an object is equal to the change in the kinetic energy of that object. Hence,

W. D = ΔK. E.

ΔK. E. = K. E.(final) - K. E.(initial)

Initial Kinetic energy, K. E.(initial), is:

K. E.(initial) = 0 (the box was at West)

Final K. E., K. E.(final) is:

K. E.(final) = 0.5*m*(v1)²

Hence, work done on the box will be:

W. D. = 0.5m(v1)² - 0

W. D. = 0.5m(v1)²

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tectonic plates rubbing against each other.


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3 years ago
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30) A force produces power P by doing work W in a time T. What power will be produced by a force that does six times as much wor
schepotkina [342]

Answer:

A) 12P

Explanation:

The power produced by a force is given by the equation

P=\frac{W}{T}

where

W is the work done by the force

T is the time in which the work is done

At the beginning in this problem, we have:

W = work done by the force

T = time taken

So the power produced is

P=\frac{W}{T}

Later, the force does six times more work, so the work done now is

W'=6W

And this work is done in half the time, so the new time is

T'=\frac{T}{2}

Substituting into the equation of the power, we find the new power produced:

P'=\frac{W'}{T'}=\frac{6W}{T/2}=12\frac{W}{T}=12P

So, 12 times more power.

4 0
3 years ago
A compound machine is sued to lift a car. If you apply a force of 60 N to the machine, it lifts the car with a force of 550 N.
mrs_skeptik [129]

a) MA=550/60=9.17

b) Wu=1150*0.35=402.5 J

3 0
3 years ago
Read 2 more answers
A person walks the path shown below. The total trip consists of four straight-line paths.
dmitriy555 [2]

At the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

<h3>What is resultant displacement?</h3>

The resultant displacement of an object is the change in position of the object. It can be described as the shortest distance connecting the final position of the object to the initial position of the object.

<h3>Net horizontal displacement </h3>

Path 1 = 40 m

Path 2 = 0 m

Path 3 = 110 m x cos(30) = 95.26 m

Path 4 = 180 m x cos(60) = 90 m

Total horizontal displacement, X = 40 m + 0 m + 95.26 m + 90 m = 225.26 m

<h3>Net vertical displacement </h3>

Path 1 = 0 m

Path 2 = 230 m

Path 3 = 110 m x sin(30) = 55 m

Path 4 = 180 m x sin(60) = 155.885 m

Total horizontal displacement, Y = 0 m + 230 m + 55 m + 155.885 m = 440.885 m

<h3>Resultant displacement</h3>

R = √(X² + Y²)

R = √(225.26² + 440.885²)

R = 495.1 m

<h3>Direction of the displacement</h3>

θ = arc tan (Y/X)

θ = arc tan (440.885/225.26)

θ =  63⁰

Thus, at the end of the walk, the person's resultant displacement is 495.1 m at 63⁰ south of west.

Learn more about resultant displacement here: brainly.com/question/13309193

#SPJ1

3 0
1 year ago
A team of eight dogs pulls a sled with waxed wood runners on wet snow (mush!). The dogs have average masses of 18.5 kg, and the
meriva

Answer:

a.2.86 m/s^2

b.1058 N

Explanation:

We are given that

Mass of each dog,M=18.5 kg

Mass of sled with rider,m=250 kg

a.Average force,F=185 N

\mu_s=0.14

g=9.8 m/s^2

By Newton's second law

8F-f=(8M+m)a

a=\frac{8F-f}{8M+m}=\frac{8(185)-(0.14)(9.8)(250)}{8(18.5)+250}

a=2.86 m/s^2

b.By Newton's second law

T=ma+\mu_s mg

Substitute the values

T=250\times 2.86+0.14(250)(9.8)=1058 N

Hence, the force in the coupling between the dogs and the sled=1058 N

8 0
3 years ago
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