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Paha777 [63]
2 years ago
7

1. write the meaning of the following terms:electrostatic,neutral, positively charged, negatively charged, coulomb,microcoulomb,

nanocoulomb,conservation of charge,quantisation of charge​
Physics
1 answer:
dybincka [34]2 years ago
5 0

ELECTROSTATIC:

relating to stationary electric charges or fields as opposed to electric currents.

NEUTRAL:

nor negative nor positive/having no charge

POSITIVELY CHARGED:

positive charge occurs when the number of protons exceeds the number of electrons

NEGATIVELY CHARGED:

negative charge occurs when the number of electrons exceeds the number of protons.

COULOMB:

SI unit for electric charge. One coulomb is equal to the amount of charge from a current of one ampere flowing for one second.

MICROCOULOMB:

a unit of electrical charge equal to one millionth of a coulomb.

NANOCOULOMB:

Nanocoulombs are a unit of charge 1,000,000,000 times smaller than Coulomb.

CONSERVATION OF CHARGE:

constancy of the total electric charge in the universe or in any specific chemical or nuclear reaction

QUANTISATION OF CHARGE:

Charge quantization is the principle that the charge of any object is an integer multiple of the elementary charge.

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Answer:

a) Check Explanation

b) Check Explanation

c) The rate at which water is being pumped into the tank = 2.631 cm³/min

Explanation:

Let the rate of flow of water into the tank be k cm³/min

a) The image of the conical tank is presented in the attached image

Note, the radius and height of a cone are related through the similar triangles principle.

As shown in the attached image, it is evident that

r/h = 3/10

r = 3h/10 = 0.3 h

b) The quantities given in the problem.

- Shape of the tank, conical tank, Hence volume of the tank = πr²h/3

- total height of the tank, H = 10 cm

- Radius of the tank at the top, R = D/2 = 6/2 = 3 cm

- rate at which water is leaking from the tank = 1.5 cm³/min

- water is being pumped into the tank at constant rate of k cm³/min

- As at height of water, h = 2 cm, the rate of rise in water level = 1 cm/min

c) volume of the tank at any time = πr²h/3

Rate of change in the volume of water in the tank = (rate of flow into the tank) - (Rate of water flow out of the tank)

dV/dt = k - 1.5

V = πr²h/3 and r = 0.3 h, r² = 0.09 h²

V = 0.03πh³

dV/dt = (dV/dh) × (dh/dt)

dV/dh = 0.09π h²

dV/dt = 0.09π h² (dh/dt)

dV/dt = k - 1.5

0.09π h² (dh/dt) = k - 1.5

But at h = 2 cm, (dh/dt) = 1.0 cm/min

0.09π h² (dh/dt) = k - 1.5

0.09π 2² (1) = k - 1.5

k - 1.5 = 1.131

k = 1.5 + 1.131 = 2.631 cm³/min

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3 years ago
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Answer:

m  200 g , T  0.250 s,E 2.00 J

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Explanation:

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V_{f}=V_{o}+a.t  (1)

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V_{f} is the final velocity of the supply bag

V_{o}=0 is the initial velocity of the supply bag (we know it is zero because we are told it was "dropped", this means it goes to ground in free fall)

a=g=-9.8m/s^{2} is the acceleration due gravity (the negtive sign indicates the gravity is downwards, in the direction of the center of the Earth)

t=3.2s is the time

Knowing this, let's solve (1):

V_{f}=0+(-9.8m/s^{2})(3.2s)  (2)

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