Answer:
19.4 ms
Step-by-step explanation:
The electric charge in a capacitor is given as;
Q= CV
Initial volts V = Q/C = 20µC/10µF = 2volts
final volts V = Q/C= 10µC/10µF = 1volts
Voltage on a capacitor undergoing discharge is given as;
v = v₀e^(–t/τ)
Where;
v₀ = initial voltage on the capacitor
v = voltage after time t
R = resistance in ohms,
C = capacitance in farads
t = time in seconds
RC = τ = time constant
Therefore;
τ = 10µF x 2.8kΩ = 28 ms
v = v₀e^(–t/τ)
1 = 2e^(–t/28ms)
e^(–t/28ms) = 0.5
–t/28ms = -0.693
t = (28ms)(0.693) = 19.4 ms
Hey there! :)
We are given the equation:
First, we have to isolate y-term; add -2 both sides.
Then we simplify the expression; distribute -5 in the x+6.
Evaluate:
Therefore, y = -5x-28 is equivalent to y-2=-5(x+6)
<span><span>
cents
( Betty's pay ) - ( what she paid Bobby )
Betty gets to keep $11.70 for the day
( notice that I did all calculations in cents until
the end when I converted to dollars )
</span><span>hope this helps:))</span></span>
50 calories are consumed per milefor 15 miles per hour, calories consumed
= 50*15
=750 calories per hour,
this is equal to 750/3600
= 0.2083 calories/second, given that,
1 calorie per sec
= 4.19 watt, the average power output
= 0.2083*4.19
= 0.873 Watts