In order to read the publications of his peers, or read his own notes of the work
that he did on the previous day, or find his coffee mug on his desk in the lab, the
research scientist must arrange to have each of them illuminated with visible
wavelengths of light, and then he must catch the light reflected from each of them
with his eyes.
Answer:
Length = 2.32 m
Explanation:
Let the length required be 'L'.
Given:
Resistance of the resistor (R) = 3.7 Ω
Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]
Resistivity of the material of rod (ρ) = 
First, let us find the area of the circular rod.
Area is given as:
Now, the resistance of the material is given by the formula:
Express this in terms of 'L'. This gives,
Now, plug in the given values and solve for length 'L'. This gives,
Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.
Answer:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
Explanation:
To calculate the predicted surface elevation of a 50km thick crust above a surface of 2.5km we are given a density of 3 gram per centimeter cube.
The displacement of the material will be calculated by subtracting the surface elevation of 2.5 km from the 50 km thick crust. Therefore 50-25= 47.5 km.
Thus let the density of the material be Pm
50*3= 47.5*Pm
Therefore: Pm= (50*3)/47.5= 3.16gram per centimeter cube
Thus with an average density of 2.8gram per centimeter cube
50*2.8= (50-x)*3.16
(50-x)= (50*2.8)/3.16
50-x=44.3
x=50-44.3= 5.7
I believe this is what you have to do:
The force between a mass M and a point mass m is represented by
So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁
So F₁ = G(Mm/r^2)
Now the distance has doubled so lets account for this in F₂:
F₂ = G(Mm/(2r)^2)
Now square the 2 that gives you four and we can pull that out in front to give
F₂ = 
G(Mm/r^2)
Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations
now we see that:
F₂ = F₁
So the second force will be 0.25 (1/4) x 1600 or 400 N.
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