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Goryan [66]
3 years ago
14

Power is expressed in ________?

Physics
1 answer:
Anna [14]3 years ago
6 0

Answer:

(B) Joules per second

Explanation:

Power is the amount of work done per unit of time. Since work is measured in Joules, the SI unit of power is Joules per second (J/s), or Watts (W).

Of course, the cooler unit of power is the Horsepower (hp), originally introduced by James Watt himself, using which people nowadays still boast about their favorite car's "strength."

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An aurora occurs when ____
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The short answer to how the aurora happens is that energetic electrically charged particles (mostly electrons) accelerate along the magnetic field lines into the upper atmosphere, where they collide with gas atoms, causing the atoms to give off light.

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2 years ago
What type of simple machine is found in a wheelbarrow?<br><br> Lever<br><br> Wedge
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Answer: Lever

A wheelbarrow consists of a lever to be able to lift the material, and a wheel to be able to move it horizontally. So in a sense, a wheelbarrow is a complex machine consisting of two simple machines.

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3 years ago
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The first four steps in the formation of the sun are listed below.
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(B)Option b is your answer..........
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Explain the difference between displacement and distance
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3 years ago
Early black-and-white television sets used an electron beam to draw a picture on the screen. The electrons in the beam were acce
lutik1710 [3]

Answer:

speed of electrons = 3.25 × 10^{7} m/s

acceleration in term g is 3.9 × 10^{17} g.

radius of circular orbit is 2.76 × 10^{-4} m

Explanation:

given data

voltage = 3 kV

magnetic field = 0.66 T

solution

law of conservation of energy

PE = KE

qV = 0.5 × m × v²

v = \sqrt{\frac{2qV}{m}}

v = \sqrt{\frac{2\times 1.6 \times 10^{-19}\times 3}{9.1\times 10^{-31}}

v = 3.25 × 10^{7} m/s

and

magnetic force on particle movie in magnetic field

F = Bqv

ma = Bqv

a = \frac{Bqv}{m}  

a =  \frac{0.67\times 1.6\times 10^{-19}\times 3.25\times 10^7}{9.1\times 10^{-31}}

a = 3.82 × 10^{18} m/s²

and acceleration in term g

a = \frac{3.82\times 10^{18}}{9.81}  

a = 3.9 × 10^{17} g

acceleration in term g is 3.9 × 10^{17} g.

and

electron moving in circular orbit has centripetal force

F = \frac{mv^2}{r}  

Bqv = \frac{mv^2}{r}  

r = \frac{mv}{Bq}  

r = \frac{9.1\times 10^{-31}\times 3.25\times 10^7}{0.67\times 1.6\times 10^{-19}}  

r = 2.76 × 10^{-4} m

radius of circular orbit is 2.76 × 10^{-4} m

8 0
3 years ago
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