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Lady bird [3.3K]
2 years ago
12

A child pulls a sled by a rope across at a constant speed

Physics
1 answer:
andreyandreev [35.5K]2 years ago
8 0

Answer:

Normal, Gravitational, Applied, Friction, and Tension

Explanation:

Because the sled is on the ground, or rather because it is on the earth, it will have a force of gravity, or its weight, pulling it down.

There is a normal force because the sled is not moving into the ground, so there needs to be a force pushing on the sled from the ground. Additionally, Newton's 3rd law states that for every action there is an equal but opposite reaction, if it's in equilibrium or not moving.

There will be an applied force since the child is pulling the sled and is applying a force to do so.

There will be a friction force that is pulling the sled opposite the direction it is moving, since it is on the ground that is presumably not fricitonless.

There will be a tension force pulling on the sled by the rope. However there is also a tension force pulling on the child's hand by the rope, meaning that these are internal forces and essentially balance out, but they are still there.

There is <em>no </em>air resistance force, or very little of it, since the object is not falling.

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Red light of wavelength 633 nm from a helium-neon laser passes through a slit 0.400mm wide. The diffraction pattern is observed
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Answer:

a)y_{first}=5.3mm

b)y_{second}=10.6-5.3 =5.3 mm  

Explanation:

a)

The width of the central bright in this diffraction pattern is given by:

y=\frac{m\lambda D}{a} when m is a natural number.

here:

  • m is 1 (to find the central bright fringe)                
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  • λ is the wavelength

So we have:

y_{first}=\frac{633*10^{9}*3.35}{0.0004}

y_{first}=5.3mm

b)

Now, if we do m=2 we can find the distance to the second minima.

y_{2}=\frac{2*633*10^{9}*3.35}{0.0004}

y_{2}=10.6 mm

Now we need to subtract these distance, to get the width of the first bright fringe :

y_{second}=10.6-5.3 =5.3 mm    

I hope it heps you!

     

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