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arlik [135]
2 years ago
14

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the in

cline, how much work is done by that force?

Physics
1 answer:
wel2 years ago
3 0

The question is incomplete! The complete question along with answer and explanation is provided below.

Question:

A 0.5 kg mass moves 40 centimeters up the incline shown in the figure below. The vertical height of the incline is 7 centimeters.

What is the change in the potential energy (in Joules) of the mass as it goes up the incline?  

If a force of 1.0 N pulled up and parallel to the surface of the incline is required to raise the mass back to the top of the incline, how much work is done by that force?

Given Information:  

Mass = m = 0.5 kg

Horizontal distance = d = 40 cm = 0.4 m

Vertical distance = h = 7 cm = 0.07 m

Normal force = Fn = 1 N

Required Information:  

Potential energy = PE = ?

Work done = W = ?

Answer:

Potential energy = 0.343 Joules

Work done = 0.39 N.m

Explanation:

The potential energy is given by

PE = mgh

where m is the mass of the object, h is the vertical distance and g is the gravitational acceleration.

PE = 0.5*9.8*0.07

PE = 0.343 Joules

As you can see in the attached image

sinθ = opposite/hypotenuse

sinθ = 0.07/0.4

θ = sin⁻¹(0.07/0.4)

θ = 10.078°

The horizontal component of the normal force is given by

Fx = Fncos(θ)

Fx = 1*cos(10.078)

Fx = 0.984 N

Work done is given by

W = Fxd

where d is the horizontal distance

W = 0.984*0.4

W = 0.39 N.m

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Answer:

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Explanation:

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3 years ago
In a double slit experiment, the distance between the slits is 0.2 mm and the distance to the screen is 100 cm. What is the phas
Anni [7]

Answer:

The phase difference is  \Delta \phi  = 180^o

Explanation:

From the question we are told that

     The distance between the slits is d = 0.2 \ mm = \frac{0.2}{1000}  = 0.2 *10^{-3} \ m

     The distance to the screen is D = 100 cm = \frac{100}{100} = 1 \ m

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 The distance of the wave from the  central maximum is L =  5mm = 5*10^{-3} m

   

Generally the path difference of this  waves is mathematically represented as

              y = d sin \theta

Here \theta is the angle between the the line connecting the mid-point of the slits with  the screen and the line  connecting the mid-point of the slits to the central maximum

  This implies that

              tan \theta  = \frac{L}{D}

     =>     \theta = tan ^{-1} \frac{L}{D}

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Substituting values into the formula for path difference

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The phase difference is mathematically represented as

          \Delta \phi = \frac{2 \pi }{\lambda }  * y

Substituting values        

         \Delta \phi = \frac{2 \pi }{400 *10^{-9} }  \ * 9.997*10^{-7}

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Converting to degree

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the solution is subtracted by 360° in order to get the actual angle

 

             

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3 years ago
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tatyana61 [14]

Answer:

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As r1 tends to infinity, 1/r1 = 0

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Plugging in the relevant values;

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Answer:

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