Answer:
The value is 
Explanation:
From the question we are told that
The focal length of the objective is 
The focal length of the eyepiece is 
The tube length is 
Generally the magnitude of the overall magnification is mathematically represented as
Where 
is the objective magnification which is mathematically represented as
=> 
=> 
is the eyepiece magnification which is mathematically evaluated as
So
I got you b, V(final)^2=V(initial+2acceleration*displacement
So this turns to (0m/s)^2=(50m/s)^2+2(9.8)(d) so just flip it all around to isolate d so you get
-(50m/s)^2/2(9.8) = d so you get roughly 12.7555 meters up
Answer:
F = - k (x-xo) a graph of the weight or applied force against the elongation obtaining a line already proves Hooke's law.
Explanation:
The student wants to prove hooke's law which has the form
F = - k (x-xo)
To do this we hang the spring in a vertical position and mark the equilibrium position on a tape measure, to simplify the calculations we can make this point zero by placing our reference system in this position.
Now for a series of known masses let's get them one by one and measure the spring elongation, building a table of weight vs elongation,
we must be careful when hanging the weights so as not to create oscillations in the spring
we look for the mass of each weight
W = mg
m = W / g
and we write them in a new column, we make a graph of the weight or applied force against the elongation and it should give a straight line; the slope of this line is sought, which is the spring constant.
The fact of obtaining a line already proves Hooke's law.
Answer:
1.503 J
Explanation:
Work done in stretching a spring = 1/2ke²
W = 1/2ke²........................... Equation 1
Where W = work done, k = spring constant, e = extension.
Given: k = 26 N/m, e = (0.22+0.12), = 0.34 m.
Substitute into equation 1
W = 1/2(26)(0.34²)
W = 13(0.1156)
W = 1.503 J.
Hence the work done to stretch it an additional 0.12 m = 1.503 J
V = I · R
Voltage = (current) · (Resistance)
Voltage = (250 A) · (2.09 x 10⁴)
Voltage = 5,225,000 volts .
I may be out of line here, but I'm pretty sure
that the resistance is 2.09 x 10⁻⁴ .
Then
Voltage = 0.05225 volt (not 5 million and something)
