A conductor of radius r, length and resistivity ρ has resistance r. what is the new resistance if it is stretched to 4 times its
1 answer:
The resistance of a conductor is given by
where L is the length of the wire,
the resistivity of the material and A the cross-sectional area.
We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length:
, then the new resistance is also 4 times the original value:
where L is the length of the wire,
We can see that if all the other quantities do not change, if the new length of the conductor is 4 times the original length:
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Answer:
new atmospheric pressure is 0.9838 × Pa
Explanation:
given data
height = 21.6 mm = 0.0216 m
Normal atmospheric pressure = 1.013 ✕ 10^5 Pa
density of mercury = 13.6 g/cm³
to find out
atmospheric pressure
solution
we find first height of mercury when normal pressure that is
pressure p = ρ×g×h
put here value
1.013 × = 13.6 × 10³ × 9.81 × h
h = 0.759 m
so change in height Δh = 0.759 - 0.0216
new height H = 0.7374 m
so new pressure = ρ×g×H
put here value
new pressure = 13.6 × 10³ × 9.81 × 0.7374
atmospheric pressure = 98380.9584
so new atmospheric pressure is 0.9838 × Pa
Answer:
Approximately .
Explanation:
.
.
Notice that sine waves and
share the same frequency and wavelength. The only distinction between these two waves is the
in
.
Therefore, the sum would still be a sine wave. The amplitude of
could be found without using calculus.
Consider the sum-of-angle identity for sine:
.
Compare the expression to
. Let
and
. Apply the sum-of-angle identity of sine to rewrite
.
.
Therefore, the sum would become:
.
Consider: would it be possible to find and
that satisfy the following hypothetical equation?
.
Simplify this hypothetical equation:
.
Apply the sum-of-angle identity of sine to rewrite the left-hand side:
.
Compare this expression with the right-hand side. For this hypothetical equation to hold for all real and
, the following should be satisfied:
, and
.
Consider the Pythagorean identity. For any real number :
.
Make use of the Pythagorean identity to solve this system of equations for . Square both sides of both equations:
.
.
Take the sum of these two equations.
Left-hand side:
.
Right-hand side:
.
Therefore:
.
.
Substitute back to the system to find
. However, notice that the exact value of
isn't required for finding the amplitude of
.
(Side note: one possible value of is
radians.)
As long as is a real number, the amplitude of
would be equal to the absolute value of
.
Therefore, the amplitude of would be:
.