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andriy [413]
3 years ago
10

Ali pushes a toy car initially at rest towards javed by exerting a constant horizontal force F of magnitude 10N through a distan

ce of 2 meters. How much work is done onthe toy car? a 5j b) 10j c) 15j d) 20j e) 25j
Physics
1 answer:
mylen [45]3 years ago
8 0

Answer: d) 20J

Explanation:

Work W done by a Force F refers to the release of potential energy from a body that is moved by the application of that force to overcome a resistance along a path.  

It should be noted that it is <u>a scalar magnitude</u>, and its unit in the International System of Units is the Joule (like energy). Therefore, 1 Joule is the work done by a force of 1 Newton when moving an object, in the direction of the force, along 1 meter:  

1J=(1N)(1m)=Nm  

Now, <u>when the applied force is constant and the direction of the force and the direction of the movement are parallel</u>, the equation to calculate it is:

 

W=(F)(d) (1)

When they are not parallel, both directions form an angle, let's call it \alpha. In that case the expression to calculate the Work is:  

W=Fdcos{\alpha} (2)

In this case both (the force and the distance in the path) are parallel, so the work W performed is the product of the force F by the distance traveled d. as shown in equation (1).

Hence:

W=(10N)(2m)  

W=20Nm=20J >>>>This is the work

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Glaciers begin with snowfall building up and __________________ the ice. (Choose the best answer)
Allisa [31]

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compacting

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4 0
3 years ago
A thin spherical shell of radius R has a total charge +Q uniformly distributed over its surface. Of the following distance r fro
grigory [225]

Answer:

The correct answer is B

Explanation:

Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity

         Φ._{E} = ∫ E. dA = q_{int} / ε₀

For this case we create a Gaussian surface that is a sphere.  We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product

        ∫ E dA = q_{int} / ε₀

The area of ​​a sphere is

     A = 4π r²

   

    E 4π r² =q_{int} / ε₀

    E = (1 /4πε₀ )  q / r²

Having the solution of the problem let's analyze the points:

A   ) r = 3R / 4  = 0.75 R.

  In this case there is no charge inside the Gaussian surface therefore the electric field is zero

        E = 0

B) r = 5R / 4 = 1.25R

In this case the entire charge is inside the Gaussian surface, the field is

    E = (1 /4πε₀ )  Q / (1.25R)²

    E = (1 /4πε₀ )  Q / R2 1 / 1.56²

    E₀ = (1 /4π ε₀ )  Q / R²

   E_{B} =  Eo /1.56 ²

  E_{B}  = 0.41 Eo

C) r = 2R

All charge inside is inside the Gaussian surface

    E_{B} =(1 /4π ε₀ ) Q    1/(2R)²

    E_{B} = (1 /4π ε₀ ) q/R²   1/4

    E_{B} = Eo  1/4

    E_{B} = 0.25 Eo

D) False the field changes with distance

The correct answer is B

4 0
3 years ago
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