In series circuit, Req = R₁ + R₂ + R₃ + ···
In parallel circuit, 
<h3>Q7.</h3>
total resistance in the upper branch = R₂ + R₃ = R₂ + 2


R₂ + 2 = 12
R₂ = 10Ω
<h3>Q8.</h3>


Req = 1.7Ω
Hello!

Use the equation F = m · a (Newton's Second Law) to solve. Substitute in the given values:
F = 5 · 20
F = 100N
Answer:
Your answer would be letter <em><u>B</u></em><em><u>.</u></em><em><u> </u></em><em><u>Electrons</u></em><em><u> </u></em><em><u>orbit</u></em><em><u> </u></em><em><u>the</u></em><em><u> </u></em><em><u>nucleus</u></em><em><u> </u></em><em><u>in</u></em><em><u> </u></em><em><u>energy</u></em><em><u> </u></em><em><u>level</u></em><em><u>.</u></em>
Explanation:
Hope it helps..
Just correct me if I'm wrong, okay?
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It’s either C or D. let me know but I would do C!
Answer:
a)30.14 rad/s2
b)43.5 rad/s
c)60633 J
d)42 kW
e)84 kW
Explanation:
If we treat the propeller is a slender rod, then its moments of inertia is

a. The angular acceleration is Torque divided by moments of inertia:

b. 5 revolution would be equals to
rad, or 31.4 rad. Since the engine just got started


c. Work done during the first 5 revolution would be torque times angular displacement:

d. The time it takes to spin the first 5 revolutions is

The average power output is work per unit time
or 42 kW
e.The instantaneous power at the instant of 5 rev would be Torque times angular speed at that time:
or 84 kW