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Citrus2011 [14]
3 years ago
7

A 9.0-V battery moves 20 mC of charge through a circuit running from its positive terminal to its negative terminal. How much en

ergy was delivered to the circuit?
Physics
1 answer:
dem82 [27]3 years ago
7 0

Answer:

E = 0.18 J

Explanation:

given,

Potential of the battery,V = 9 V

Charge on the circuit, Q = 20 m C

                                        = 20 x 10⁻³ C

energy delivered in the circuit

E = Q V

E = 20 x 10⁻³ x 9

E =  180 x 10⁻³

E = 0.18 J

Energy delivered in the circuit is equal to E = 0.18 J

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Setler79 [48]
The power of the lamp would be calculated with the equation of ohm laws. P = U x I = 122V x 0.1A = 12.2W
4 0
2 years ago
You pull a solid nickel ball with a density of 8.91 g/cm3 and a radius of 1.40 cm upward through a fluid at a constant speed of
Sunny_sXe [5.5K]

Answer:

P = 1.090\,N

Explanation:

The constant speed means that ball is not experimenting acceleration. This elements is modelled by using the following equation of equilibrium:

\Sigma F = P - W + F_{D}

\Sigma F = P - \rho \cdot V \cdot g + c\cdot v = 0

Now, the exerted force is:

P = \rho \cdot V \cdot g - c\cdot v

The volume of a sphere is:

V = \frac{4\cdot \pi}{3}\cdot R^{3}

V = \frac{4\cdot \pi}{3}\cdot (0.014\,m)^{3}

V = 1.149\times 10^{-5}\,m^{3}

Lastly, the force is calculated:

P = (8910\,\frac{kg}{m^{3}} )\cdot (1.149\cdot 10^{-5}\,m^{3})\cdot (9.81\,\frac{m}{s^{2}} )+(0.950\,\frac{kg}{s})\cdot (0.09\,\frac{m}{s} )

P = 1.090\,N

5 0
3 years ago
How do you find the capacitance in this?
Lostsunrise [7]

Answer:

Explanation:

parallel capacitances add directly

Series capacitances add by reciprocal of sum of reciprocals.

Ceq = [ C ] + [1 / (1/C + 1/C)] + [1 / (1/C + 1/C + 1/C)]

Ceq = [ C ] + [C / 2] + [C / 3]

Ceq = [ 6C/6 ] + [3C / 6] + [2C / 6]

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2 years ago
Which of these nebulae is the odd one out?
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Answer: The answer is D!

Explanation:

5 0
2 years ago
Read 2 more answers
A litre of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure,if the pressure is made 75 atm then at which temperat
liberstina [14]

Answer:

45000 K .

Explanation:

Given :

A liter of a gas weigh 2 gram at 300 kelvin temperature and 1 atm pressure

We need to find the temperature in which 1 litre of the same gas weigh 1 gram

in pressure 75 atm.

We know, by ideal gas equation :

PV=nRT

Here , n is no of moles , n=\dfrac{Given \ Weight }{Molecular\ Mass}=\dfrac{w}{M}

Putting initial and final values and dividing them :

\dfrac{P_1V_1}{P_2V_2}=\dfrac{\dfrac{w_1}{M}T_1}{\dfrac{w_2}{M}T_2}

\dfrac{1\times 1}{75\times 1}=\dfrac{\dfrac{2}{M}\times 300}{\dfrac{1}{M}\times T_2}\\ \\T_2=45000\ K.

Hence , this is the required solution.

7 0
3 years ago
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