Answer:
a) 24.4 Ω
b) 4.92 A
c) 495.9 W
d)
c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.
Explanation:
b)
The formula for power is:
P = IV
where,
P = Power of heater = 590 W
V = Voltage it takes = 120 V
I = Current Drawn = ?
Therefore,
590 W = (I)(120 V)
I = 590 W/120 V
<u>I = 4.92 A</u>
<u></u>
a)
From Ohm's Law:
V = IR
R = V/I
Therefore,
R = 120 V/4.92 A
<u>R = 24.4 Ω</u>
<u></u>
c)
For constant resistance and 110 V the power becomes:
P = V²/R
Therefore,
P = (110 V)²/24.4 Ω
<u>P = 495.9 W</u>
<u></u>
d)
If the resistance decreases, it will increase the current according to Ohm's Law. As a result of increase in current the power shall increase according to formula (P = VI). Therefore, correct option is:
<u>c. It will be larger. The resistance will be smaller so the current drawn will increase, increasing the power.</u>
Answer:
0.20kg-m^2
Explanation:
Let the linear velocity of the rope(=of pulley) is v m/s
Using kinematic equation
=> v = u + at
=>v = 0 + 4.9a
=>v = 4.9a ------------ eq1
By v^2 = u^2 + 2as
=>v^2 = 0 + 2 x v/4.9 x 1.2
=>4.9v^2 - 2.4v = 0
=>v(4.9v - 2.4) = 0
=>v = 2.4/4.9 = 0.49 m/s
Thus by v = r x omega
=>omega = v/r = 0.49/0.02 = 24.49 rad/sec
BY W = F x s = 50 x 1.2 = 60 J
=>KE(rotational) = W = 1/2 x I x omega^2
=>60 = 1/2 x I x (24.49)^2
=>I = 0.20 kg-m^2
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