Answer:
Atoms in a element on the periodic table I dont know which one tho so try google that sorry I am not much of a help
Explanation:
I'm pretty sure what you are trying to ask for is radiative energy, light energy, and electronic energy.
Radiative since the microwave is releasing radiation,
Light since there is light inside the microwave,
Electronic since it is plugged in and uses electricity.
You can also use sound, but I don't think every microwave makes sound.
Answer
The answer for the first one I think is false.
The second one would be true i think. I hope i got it right and have a wonderful day
Answer:
it will take 36.12 ms to reduce the capacitor's charge to 10 μC
Explanation:
Qi= C×V
then:
Vi = Q/C = 30μ/20μ = 1.5 volts
and:
Vf = Q/C = 10μ/20μ = 0.5 volts
then:
v = v₀e^(–t/τ)
v₀ is the initial voltage on the cap
v is the voltage after time t
R is resistance in ohms,
C is capacitance in farads
t is time in seconds
RC = τ = time constant
τ = 20µ x 1.5k = 30 ms
v = v₀e^(t/τ)
0.5 = 1.5e^(t/30ms)
e^(t/30ms) = 10/3
t/30ms = 1.20397
t = (30ms)(1.20397) = 36.12 ms
Therefore, it will take 36.12 ms to reduce the capacitor's charge to 10 μC.
Answer:
Resistance of the second wire is twice the first wire.
Explanation:
Let us first see the formula of resistance;
R = pxL/A
Here L is the lenght of the wire, A the area and p is the resistivity of wire.
As we are given that the length of second wire is double than that of the first wire, hence the resistance of second wire would be double.
Since we have two loop in second case, inducing double voltage but as resistance is doubled so the current would remain same according to ohms law
I = V/R
