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Ulleksa [173]
2 years ago
12

Which one of the statements below is true about mechanical waves?

Physics
1 answer:
TEA [102]2 years ago
5 0
They require a medium to travel through
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Which of the following is NOT usually published with a scientific report?
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a parallel circuit is sometimes called a because the current splits up among all the resistors in the parallel circuit
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A parallel circuit is sometimes called a current divider because current splits up among all the resistors in the parallel circuit. In addition, the current through the branches is inversely proportional to the resistance of the branch. If the resistance in each branch is kept constant but the voltage is decreased, the current will decrease.
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16. The energy absorbed in 10minutes by an electrical heater is 1.5 MJ. The supply voltage is 240 V. calculate: a) The current d
dsp73

Answer:

If by 1.5 MJ you mean 1.5E6 Joules then

W = P t    = power X time

W / t = P   power

P = 1.5E6 J / 600 sec = 2500 J / s

P = I V

a) I = 2500 J/s / (240 J/c) = 10.4 C / sec  = 10.4 amps

b) Q = I t = 10.4 C / sec * 300 sec = 3120 Coulombs

c)  E = P * t = 2500 J / sec * 100 hr * 3600 sec / hr = 9.0E8 Joules

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Humid tropical climates are climates that have no winters.
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Through what potential difference would you need to accelerate an alpha particle, starting from rest, so that it will just reach
Svetlanka [38]

Answer:

\Delta V    = 1.8 \times 10^7 V

Explanation:

GIVEN

diameter = 15 fm  =15 \times 10^{-15}m

we use here energy conservation

K_{i}+U_{i} =K_{f}+U_{f}

there will be some initial kinetic  energy but after collision kinetic energy will zero

K_{i} + 0 = 0 + \frac{1}{4 \pi \epsilon _{0}} \frac{(2e)(92e)}{7.5 \times 10^{-15}}

on solving these equations we get kinetic energy initial

KE_{i} = 5.65\times 10 ^{-12} \times \frac  {1 eV}{1.6 \times 10^{-19}}

KE_{i} = 35.33 J ..............(i)

That is, the alpha particle must be fired with 35.33 MeV of kinetic energy. An alpha particle with charge q = 2  e

and gains kinetic energy K  =e∆V  ..........(ii)

 by accelerating through a potential difference ∆V

Thus the alpha particle will

just reach the {238}_U nucleus after being accelerated through a potential difference  ∆V

equating (i) and second equation we get

e∆V  = 35.33 Me V

\Delta V = \frac{35.33}{2}  MV\\\Delta V    = 1.8 \times 10^7 V

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3 years ago
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