Answer:
(a) = -0.16%
(b) = smaller
Explanation:
given
power = 460 W
potential difference = 120 V
(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?
we know 
.....................(i)
we need to find change in power
..............(ii)
from equations we get
(b)
if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power
and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller
Answer:
The second law of a vibrating string states that for a transverse vibration in a stretched string, the frequency is directly proportional to the square root of the string's tension, when the vibrating string's mass per unit length and the vibrating length are kept constant
The law can be expressed mathematically as follows;
The second law of the vibrating string can be verified directly, however, the third law of the vibrating string states that frequency is inversely proportional to the square root of the mass per unit length cannot be directly verified due to the lack of continuous variation in both the frequency, 'f', and the mass, 'm', simultaneously
Therefore, the law is verified indirectly, by rearranging the above equation as follows;
From which it can be shown that the following relation holds with the limits of error in the experiment
m₁·l₁² = m₂·l₂² = m₃·l₃² = m₄·l₄² = m₅·l₅²
Explanation:
the resistance is 80.8Ω and average power 178.2 W
a) the value of resistance is given as,
R=rms voltage / rms current
and rms current = 2.1/√2
= I°/√2
rms current = 2.1/√2
rms current = 1.48
therefore R = 120/1.48
=80.4Ω
b) average power is product of rms current and rms voltage
P=120× 1.48
=178.2 W
learn more about power and current here:
brainly.com/question/19723198
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If you move south at 30 m/s for 30 minutes, your displacement is
D = (30 mtr/s) x (30 min) x (60 sec/min)
D = (30 x 30 x 60) (mtr-min-sec/sec-min)
D = 54,000 meters south
D = <em>54 km south</em>
