you are given two circuits with two batteries of emf e and internal resistance r1 each. circuit a has the batteries connected in
1 answer:
Answer:
In which direction does the current in circuit A flow?
counterclockwise
<h2>What is the power dissipated by the resistor of resistance R2 for circuit A, given that E=10 V, R1=300ohms, and R2=5000ohms? </h2><h2>Calculate the power to two significant figures.</h2><h2>0.064W</h2><h2 /><h2>For what ratio of R1 and R2 would power dissipated by the resistor of resistance R2 be the same for circuit A and circuit B?</h2><h2>R1/R2 = 1 </h2><h2 /><h2>Under which of the following conditions would power dissipated by the resistance R2 in circuit A be bigger than that of circuit B? </h2><h2>Some answer choices overlap; choose the most restrictive answer.</h2><h2>R2>R1</h2><h2> </h2>Explanation:
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Answer: magnitude of applied force is FA = mg + F
Where F is the resultant force downward that the rope moves with
Explanation:
Force downwards F is,
F = FA - T
T is the upwards tension force on the rope
FA is the actual applied force in pulling the rope down.
Therefore, T = FA - F .....equ. (1)
For the box to move up with force ma ( it's mass times its acceleration upwards) upwards tension on the roap must exceed its own weight mg ( it's mass times acceleration due to gravity 9.8m/s^2)
Therefore, ma = T - mg
T = ma + mg ..... equ. (2)
Equating equ. 1 and 2
T = FA - F = ma + mg
Therefore FA = ma + mg + F
But at constant velocity a = 0
Magnitude of applied force becomes
FA = mg + F
See image below
Answer:
18 radians
Explanation:
The computation is shown below:
As we know that
Torque = Force × Moment arm
= 1N × 1M
= 1N-M
Torque =
Now
Here t = 1 minutes = 60 seconds