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mr_godi [17]
1 year ago
9

Calculate work when the force exerted to push an object is 75 n and the distance is 1.2 kilometers.

Physics
1 answer:
grigory [225]1 year ago
7 0

The magnitude of the work done to push the object is 90,000 J

<h3>When is Work done on a body ?</h3>

Work is done on a body when the direction of the force applied is parallel to the direction of the displacement of the body.

The following parameters are given;

  • Force F = 75 N
  • Distance S = 1.2 Km
  • Work W = ?

First convert the distance in kilometer to meter by multiplying it by 1000

S = 1.2 x 1000 = 1200 m

From the definition of work,

W = F x S

Substitute the parameters into the formula

W = 75 x 1200

W = 90,000 J

Therefore, the magnitude of the work done to push the object is 90,000 J

Learn more about Work here: brainly.com/question/25573309

#SPJ1

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Answer:

41.4496148484\ m/s

Explanation:

\epsilon_0 = Permittivity of free space = 8.85\times 10^{-12}\ F/m

\sigma = Surface charge density = 5.9\times 10^{-8}\ C/m^2

\Delta x = 0.57-0.26

q = Charge = 6.9\times 10^{-9}\ C

m = Mass of object = 8.3\times 10^{-9}\ kg

Electric field due to a sheet is given by

E=\dfrac{\sigma}{2\epsilon_0}\\\Rightarrow E=\dfrac{5.9\times 10^{-8}}{2\times 8.85\times 10^{-12}}\\\Rightarrow E=3333.33\ V/m

Electric field is given by

E=\dfrac{V}{d}

Voltage is given by

V=E\Delta x

Kinetic energy is given by

K=qV

\dfrac{1}{2}mv^2=qE\Delta x\\\Rightarrow v=\sqrt{\dfrac{2qE\Delta x}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 6.9\times 10^{-9}\times 3333.33\times (0.57-0.26)}{8.3\times 10^{-9}}}\\\Rightarrow v=41.4496148484\ m/s

The initial speed of the object is 41.4496148484\ m/s

7 0
3 years ago
A school bus transporting students to school is an example of
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A 1,400 kg car accelerates from rest to 30 m/s in 6.0 seconds. what is the net force on the car?
Allushta [10]
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When calculating the power bill, power companies use _____.
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I think the correct answer from the choices listed above is option B. When calculating the power bill, power companies use kilowatt-hours. This unit is a derived unit of energy equal to 3.6 MJ. If energy is being transmitted or used at a constant rate (power) over a period of time, the total energy in kilowatt-hours is the product of the power in kilowatts and the time.
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A wheel that was initially spinning is accelerated at a constant angular acceleration of 5.0 rad/s^2. After 8.0 s, the wheel is
notka56 [123]

Answer:

a)  Initial angular speed = 30 rad/s

b) Final angular speed = 70 rad/s        

Explanation:

a) We have equation of motion s = ut + 0.5at²

    Here s = 400 radians

              t = 8 s

              a = 5 rad/s²

    Substituting

             400 = u x 8 + 0.5 x 5 x 8²

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b) We have equation of motion v = u + at

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              a = 5 rad/s²  

    Substituting

             v = 30 + 5 x 8 = 70 rad/s    

   Final angular speed = 70 rad/s        

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3 years ago
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