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FinnZ [79.3K]
3 years ago
8

A spring with a mass attached is initially at rest. It is then stretched 2 cm by a physics student. Another physics student with

a similar mass and spring system then stretches their spring to 4 cm. Compare and contrast the potential energies of the springs after they are stretched by the students. *
Physics
1 answer:
Andrew [12]3 years ago
3 0
The student who displaced the spring by 2 cm has less potential energy than the student who displaced the spring by 4 cm, this is because potential energy (elastic) is directly proportionate to extension (displaced amount), so as the amount of displacement of the spring is higher, then the potential energy of the springs is higher and vice versa.
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Synaptic pruning is an essential part of brain development. By getting rid of the synapses that are no longer used, the brain becomes more efficient as you age.

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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

      The wavelength is  \lambda = 400nm

 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}

             \theta = tan ^{-1} [\frac{5*10^{-3}}{1}]

           \theta =0.2865

Substituting values into the formula for path difference

       y = 0.2 *10^{-3} sin(0.2864)  

       y = 9.997*10^{-7} \ m  

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}

         \Delta \phi =5 \pi

Converting to degree

         \Delta \phi =5 \pi radians = 5 (180^o)  =  180^o  

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

 

             

4 0
3 years ago
A 25 kg child runs at a speed of 5.0 m/s and jumps onto a stationary shopping cart and holds on for dear life. The cart has mass
makkiz [27]

Answer:

3.38 m/s

Explanation:

Mass of child = m₁ = 25

Initial speed of child = u₁ = 5 m/s

Initial speed of cart = u₂ = 0 m/s

Mass of cart = m₂ = 12 kg

Velocity of cart with child on top = v

This is a case of perfectly inelastic collision

m_1u_1+m_2u_2=(m_1+m_2)v\\\Rightarrow v=\frac{m_1u_1+m_2u_2}{m_1+m_2}\\\Rightarrow v=\frac{25\times 5+12\times 0}{25+12}\\\Rightarrow v=\frac{125}{37}\\\Rightarrow v=3.38\ m/s

Velocity of cart with child on top is 3.38 m/s

7 0
3 years ago
A man pushes an 250-N crate at constant speed a distance of 30.0 m upward along a rough slope that makes an angle of 60° with th
mel-nik [20]

Answer:

6495.19 Joule

Explanation:

F = Weight of the crate = 250 N

d = Distance the cart is pushed = 30 m

θ = Angle of inclination = 60°

The weight of the crate will be resloved into two components

Fdsinθ and Fdcosθ

Work done by the force of gravity is

W = Fdsinθ

⇒W = 250×30×sin60

⇒W = 6495.19 Joule

∴ The work done by the force of gravity is 6495.19 Joule

8 0
3 years ago
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