If the award weighs 200 newtons and 200 newtons equals 44.96 pounds of force even though it is of such a force if it hits the ground the energy will either discharge into the air doing nothing but creating a loud sound or it will discharge into the ground altering the ground that it landed on.
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Answer:
110 N
Explanation:
When a force is applied on a body and body does not move, it means the body remains at rest.
In this condition, there is a contact force between the body and the floor which is called static friction.
Th static friction force is a self adjusting force and comes into play when the body is at rest.
Here, the applied force is 110 N and the chest is not moving, that means a static friction force is acting between the chest and the floor. This static friction force is the force of contact between the chest and the floor. The static friction force is equal to the applied force when the body does not move.
So, the contact force between the chest and the floor is 100 N.
Answer:
a) d₁ = 247.8 μm
d₂ = 205.3 μm
b) d₂ = 20.53 x 10⁻⁵ m = 205.3 μm
Explanation:
a)
The formula for Michelson Interferometer is derived to be:
d = mλ/2
where,
d = distance moved
m = no. of fringes
λ = wavelength of light
For JAN, we have following data
d = d₁
m = 818
λ = 606 nm = 606 x 10⁻⁹ m
Therefore,
d₁ = (818)(606 x 10⁻⁹ m)/2
<u>d₁ = 24.78 x 10⁻⁵ m = 247.8 μm</u>
For LINDA, we have following data
d = d₂
m = 818
λ = 502 nm = 502 x 10⁻⁹ m
Therefore,
d₂ = (818)(502 x 10⁻⁹ m)/2
<u>d₂ = 20.53 x 10⁻⁵ m = 205.3 μm</u>
b)
The resultant displacement can be found out from the difference between both displacement. And the direction of resultant displacement will be the same as the direction of greater displacement. Therefore,
Resultant Displacement = Δd = d₁ - d₂
Δd = 247.8 μm - 205.3 μm
<u>Δd = 42.5 μm (in the direction of JAN)</u>
