If a wire of resistance R is stretched uniformly to 2.8 times its initial length, by what factor does the power dissipated in the wire change, assuming it remains hooked up to the same voltage source? Assume the wire's volume and density remain constant. (The answer is not 7.84)
"the wire's volume and density remain constant"
... so the cross-sectional area is decreased
... two cylinders with the same volume , but one is 2.8 times as long
the new resistance is related to the lower cross-section
... as well as the longer length
the new resistance should be ... R * 2.8^2 ... or ... R * 7.84
power = (voltage)^2 / resistance
the power changes by a factor of ... 1 / 7.84
(power new) / (power old) = 1 / 7.84
Well, you're right. The answer is not 7.84. It's actually banana! Just kidding. Let me calculate the right answer for you.
To find out the change in power dissipated in the wire, we need to consider the relationship between resistance and length.
The resistance of a wire is directly proportional to its length. So if the length of the wire is stretched uniformly by a factor of 2.8, the resistance of the wire also increases by the same factor.
Now, power is given by the equation:
Power = (Voltage)^2 / Resistance
Since the voltage source remains the same, the change in power can be calculated by taking the ratio of the final power to the initial power.
Let's assume the initial power is P.
Initial power = (Voltage)^2 / Initial Resistance
Final power = (Voltage)^2 / Final Resistance = (Voltage)^2 / (2.8 * Initial Resistance)
Now, to find the factor by which the power changes, we can divide the final power by the initial power:
Change in power = Final power / Initial power
= [(Voltage)^2 / (2.8 * Initial Resistance)] / [(Voltage)^2 / Initial Resistance]
= 1 / 2.8
Therefore, the power dissipated in the wire changes by a factor of 1/2.8 or approximately 0.357.
To find the factor by which the power dissipated in the wire changes, we can use the formula:
Power (P) = (Voltage^2) / Resistance
Initially, let's assume the original length of the wire is L and its resistance is R. Therefore, the initial resistance is R.
When the wire is stretched uniformly to 2.8 times its initial length, the new length of the wire becomes 2.8L.
According to the given information, the wire's volume and density remain constant. This implies that the wire's cross-sectional area remains the same. Let's represent the original cross-sectional area of the wire as A.
Since the wire's volume remains constant, we have:
Volume = Length × Cross-sectional area
Initially, the volume of the wire is L × A, and when it is stretched, the new volume remains the same:
(L × A) = (2.8L) × (A')
Simplifying the equation, we find:
2.8 × A' = A
Now, we can find the new cross-sectional area, A':
A' = A / 2.8
As the wire's volume and density remain constant, the resistance is inversely proportional to the cross-sectional area. This means that the new resistance, R', is related to the original resistance, R, in the following way:
R' = R × (A / A')
Substituting the values of A and A', we have:
R' = R × (2.8 / 1)
R' = R × 2.8
So, the new resistance, R', is 2.8 times the original resistance, R.
Now, let's calculate the factor by which the power changes:
Original Power, P = (Voltage^2) / R
New Power, P' = (Voltage^2) / R'
The factor by which the power changes can be found by dividing the new power by the original power:
Factor = P' / P
Using the given information that the wire remains hooked up to the same voltage source, we can simplify the formula as follows:
Factor = (Voltage^2) / R' × R / (Voltage^2)
= R / R'
= R / (2.8R)
= 1 / 2.8
≈ 0.357
Therefore, the factor by which the power dissipated in the wire changes is approximately 0.357.