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Lemur [1.5K]
3 years ago
7

If the current through a 20-ω resistor is 8.0 a , how much energy is dissipated by the resistor in 1.0 h ?

Physics
1 answer:
fiasKO [112]3 years ago
4 0
1 hour = 3600 seconds.
Energy dissipated = I²Rt = 8²×20×3600 = 4608000 J
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What people group was not targeted for estermination by the germans?​
madam [21]

Native Americans, South Sea Islanders, Australian aborigines, Uighurs, Inuits, Yupiks, Aleuts, Incas, Mayans, Canaanites, Hittites, Hutu and Tutsi who did not live in Germany, and blond, blue-eyed Aryans who did.  Just about everyone else was on the list distributed to the Orkin men, beginning in the late 1930s.  Especially Jews.

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An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

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A girl pulls on a 10-kg wagon with a constant horizontal force of 30 N. If there are no other horizontal forces, what is the wag
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