I believe the answer is A. shorter wires
The heat Q transferred to cause a temperature change depends on the magnitude of the temperature change, the mass of the system, and the substance and phase involved.
Explanation:
https://courses.lumenlearning.com/physics/chapter/14-2-temperature-change-and-heat-capacity/
The capacitive reactance is reduced by a factor of 2.
<h3>Calculation:</h3>
We know the capacitive reactance is given as,

where,
= capacitive reactance
f = frequency
C = capacitance
It is given that frequency is doubled, i.e.,
f' = 2f
To find,
=?




Therefore, the capacitive reactance is reduced by a factor of 2.
I understand the question you are looking for is this:
A capacitor is connected across an AC source. Suppose the frequency of the source is doubled. What happens to the capacitive reactant of the inductor?
- The capacitive reactance is doubled.
- The capacitive reactance is traduced by a factor of 4.
- The capacitive reactance remains constant.
- The capacitive reactance is quadrupled.
- The capacitive reactance is reduced by a factor of 2.
Learn more about capacitive reactance here:
brainly.com/question/23427243
#SPJ4
Answer: vl = 2.75 m/s vt = 1.5 m/s
Explanation:
If we assume that no external forces act during the collision, total momentum must be conserved.
If both cars are identical and also the drivers have the same mass, we can write the following:
m (vi1 + vi2) = m (vf1 + vf2) (1)
The sum of the initial speeds must be equal to the sum of the final ones.
If we are told that kinetic energy must be conserved also, simplifying, we can write:
vi1² + vi2² = vf1² + vf2² (2)
The only condition that satisfies (1) and (2) simultaneously is the one in which both masses exchange speeds, so we can write:
vf1 = vi2 and vf2 = vi1
If we call v1 to the speed of the leading car, and v2 to the trailing one, we can finally put the following:
vf1 = 2.75 m/s vf2 = 1.5 m/s