Answer:
he would have to run at a 7 and a half mile to get 8 miles in an hour
Explanation:
(i) The total capacitance for the circuit is 5 μF.
(ii) The total charge stored in the circuit is 1 x 10⁻⁴ C.
(iii) The charge stored in 3μF capacitor is 6 x 10⁻⁶ C.
<h3>Total capacitance of the circuit</h3>
The total capacitance of the circuit is determined by reolving the series capacitors separate and parallel capacitors separate as well.
<h3>C1 and C2 are in series </h3>

<h3>C1 and C2 are parallel to C3</h3>

<h3>C(123) is series to C5 and C6</h3>

<h3>C7 and C8 are in series</h3>

<h3>Total capaciatnce of the circuit</h3>
Ct + C(78) = 2 μF + 3 μF = 5 μF
<h3 /><h3>Total charge stored in the circuit</h3>
The total charge stored in the capacitor is calculated as follows;
Q = CV
Q = (5 x 10⁻⁶) x (20)
Q = 1 x 10⁻⁴ C
<h3>Charge stored in 3μF capacitor</h3>
Q = (3 x 10⁻⁶) x (20)
Q = 6 x 10⁻⁶ C
Learn more about capacitance of capacitor here: brainly.com/question/13578522
You'd get an extra 40/60 of the energy, or 2/3. Multiply 5/3 by the required energy to get the actual consumption.
Answer:
v=12.5 i + 12.5 j m/s
Explanation:
Given that
m₁=m₂ = m
m₃ = 2 m
Given that speed of the two pieces
u₁=- 25 j m/s
u₂ =- 25 i m/s
Lets take the speed of the third mass = v m/s
From linear momentum conservation
Pi= Pf
0 = m₁u₁+m₂u₂ + m₃ v
0 = -25 j m - 25 i m + 2 m v
2 v=25 j + 25 i m/s
v=12.5 i + 12.5 j m/s
Therefore the speed of the third mass will be v=12.5 i + 12.5 j m/s