Answer:
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I believe that the answer to the question provided above is that with increase in resistance provided with constant current, Power dissipated will be lessen since power loss is high. Low power dissipation has low heat production.
Hope my answer would be a great help for you.
Answer:
Explanation:
charge, q = 2e = 2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C
mass, m = 4 u = 4 x 1.661 x 10^-27 kg = 6.644 x 10^-27 kg
Radius, r = 4.5 cm = 0.045 m
Magnetic field, B = 1.20 T
(a) Let the speed is v.


v = 2.6 x 10^6 m/s
(b) Let T be the period of revolution


T = 1.09 x 10^-7 s
(c) The formula for the kinetic energy is


K = 2.25 x 10^-14 J
(d) Let the potential difference is V.
K = qV


V = 70312.5 V
Answer: Is there suppose to be a picture?
Explanation:
Answer:
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Explanation:
= Temperature of the room = 22.0 °C = 22 + 273 = 295 K
= Temperature of the skin = 33.0 °C = 33 + 273 = 306 K
= Surface area = 1.50 m²
= emissivity = 0.97
= Stefan's constant = 5.67 x 10⁻⁸ Wm⁻² K⁻⁴
Rate of heat transfer is given as


W