The answer is 0.025J.
W=1/2*k*x^2
W=1/2*20*0.050^2
W=0.025J
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:
Explanation:
Given that,
Frequency of radio signal is
f = 800kHz = 800,000 Hz.
Distance from transmitter
d = 8.5km = 8500m
Electric field amplitude
E = 0.9 V/m
The average energy density can be calculated using
U_E = ½•ϵo•E²
Where ϵo = 8.85 × 10^-12 F/m
Then,
U_E = ½ × 8.85 × 10^-12 × 0.9²
U_E = 3.58 × 10^-12 J/m²
The average electromagnetic energy density is 3.58 × 10^-12 J/m²
<span>Energy is neither lost nor gained as it transforms from chemical, to heat, to mechanical energy.</span>
