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Zarrin [17]
3 years ago
14

4. State Newton's second law of motion

Physics
1 answer:
AleksAgata [21]3 years ago
8 0
  1. It states that the net external force acting on a body is directly proportional to the rate of charge in its momentum.
  2. Interia is the tendency of an object to resist a change in motion.
  3. The electric fan continues to rotate for some time after the current is switched off due to the rotational interia force.
  4. An object interia tends to resist any change in its state of rest or motion.If a carpet is beaten with a stick then the carpet begins to move.But the particles of dust are trying yo resist their state of rest.
  5. A person pushes the water with his hands in the backward direction while swimming .In turn water in the pool pushes him forward due to the reaction force .When a carpet with a stick dust comes out.
  6. As it is in earth's atmosphere the ball experiences an external force from the wind and keeps on slowing down.The idea to lower the hands while catching the ball is to save it from pain or injury.The more the ball remains in air the lesser force it puts as it falls in the hands of the fielder.
  7. acceleration=change in velocity/time acceleration=final velocity-inital velocity/time. =acceleration=v-u/t a=v-u/t at=v-u at+u=v v=at+u
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Calculate the number of free electrons and holes (in m-3) in an intrinsic semiconductor that has electron and hole mobilities of
allochka39001 [22]

Answer:

ni = 2.04e19

Explanation:

we know that in semiconductor like intrinsic, when electron leave the band, it leave a hole in valence band so we have

n = p = ni

from intrinsic carrier concentration

\sigma = n\left | e \right | \mu_e + n\left | e \right | \mu_h

\sigma = ni\left | e \right | \mu_e  + ni\left | e \right | \mu_h

\sigma = ni \left | e \right | ( \mu_e + \mu_h)

1.7 = ni * 1.6*10^{-19} * (.35 + .17)

ni = 2.014 *10^{19} m^{-3}

ni = 2.04e19

5 0
2 years ago
In this diagram, medium A represents water and medium B represents air. The phenomenon in this diagram, moving from A to B, is c
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3 years ago
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If you wanted to find the area of the hot filament in a light bulb, you would have to know the temperature (determinable from th
Sladkaya [172]

Answer:

To find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

Explanation:

The emissive power of a light bulb can be given by the following formula:

E = σεAT⁴

where,

E = Power Input or Emissive Power

σ = Stefan-Boltzmann constant

ε = Emissivity

A = Area

T = Absolute Temperature

Therefore,

A = E/σεT⁴

So, to find out the area of the hot filament of a light bulb, you would need to know the temperature, the power input, the Stefan-Boltzmann constant and <u>Emissivity of the Filament</u>.

3 0
3 years ago
A centrifugal pump is operating at a flow rate of 1 m3/s and a head of 20 m. If the specific weight of water is 9800 N/m3 and th
tamaranim1 [39]

Answer:

<em>The power required by the pump is nearly 230.588 kW</em>

Explanation:

Flow rate of the pump Q = 1 m^3/s

the head flow H = 20 m

specific weight of water γ = 9800 N/m^3

efficiency of the pump η = 85%

First note that specific gravity of water is the product of the density of water and acceleration due to gravity.

γ = ρg

where ρ is density. For water its value is 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

The power to lift this water at this rate will be gotten from the equation

P = ρgQH

but ρg = γ

therefore,

P = γQH

imputing values, we'll have

P = 9800 x 1 x 20 = 196000 W

But the centrifugal pump that will be used will only be able to lift this amount of water after the efficiency factor has been considered. The power of pump needed must be greater than this power.

we can say that

196000 W is 85% of the power of the pump power needed, therefore

196000 = 85% of P_{p}

where P_{p} is the power of the pump needed

85% = 0.85

196000 = 0.85P_{p}

P_{p} = 196000/0.85 = 230588.24 W

<em>Pump power = 230.588 kW</em>

3 0
3 years ago
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