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liraira [26]
2 years ago
15

A nuclear explosion may release tremendous amounts of energy in the form of noise, heat, visible light, radiation and an atmosph

eric shock wave. An observer watching the explosion from the vacuum of nearby space would be able to experience which of these types of energy?
A) heat and radiation
B) sound and radiation
C) light and radiation
D) heat, light, and the shock wave
Physics
2 answers:
posledela2 years ago
7 0

light and radiation is able to experience by an observer watching the explosion from the vacuum.

<u>Explanation:</u>

  • Basically in the nuclear explosion, there will be an enormous of energy released as noise, heat, visible light, radiation and atmospheric wave.
  • Usually sound and wave propagate through medium but in vacuum, there will be no medium to transfer this type of energy. so there is no chance of sound and wave transfer.  
  • Light and radiation travel in vacuum because they didn't need the medium to transfer .  

katrin [286]2 years ago
4 0

Answer:c light and radiation

Explanation:

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jonny [76]

Answer:

8400m

Explanation:

The engine that falls off would have the same constant horizontal velocity as the airplane's when if falls off if we ignore air resistance. So it would have a horizontal velocity of 280m/s for 30seconds before it hits the ground.

Therefor the horizontal distance the engine travels during its fall is

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6 0
3 years ago
Suppose that a charged particle of diameter 1.00 micrometer moves with constant speed in an electric field of magnitude 1.00×105
Dovator [93]
It's a bit of a trick question, had the same one on my homework. You're given an electric field strength (1*10^5 N/C for mine), a drag force (7.25*10^-11 N) and the critical info is that it's moving with constant velocity(the particle is in equilibrium/not accelerating). 
<span>All you need is F=(K*Q1*Q2)/r^2 </span>
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3 0
3 years ago
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The joule (J) is a unit of energy. Recall that energy may be converted between many different forms such as mechanical energy, t
REY [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The workdone is  W = -177.275J

Explanation:

From the question we are told that

      The initial Volume is  Vi = 0.160 L

      The final volume is  V_f = 0.510L

      The external pressure is  P = 5.00 \ atm

Generally the change in volume is

           \Delta V = V_f - V_i

Substituting values we have

           \Delta V = 0.510 -0.160

                 = 0.350L

Generally workdone is mathematically represented as

           W = -P \Delta V

W is negative because the working is done on the environment by the system which is indicated by volume increase

     Substituting values

                W = - 5* 0.350

                    = -1.75 \ L \ \cdot atm

Now  1 \  L \cdot atm = 101.3J

  Therefore  W = -1.75* 101.3

                          = -177.275J

   

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1. - 12N

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