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shutvik [7]
3 years ago
8

Consider a 24.0 V storage battery that can transfer, over the course of its useful lifetime, a total charge equivalent to 340000

C. What is the total amount of work that can be done by the battery's chemicals over the useful lifetime of the battery? Give your answer in joules.
Physics
1 answer:
xeze [42]3 years ago
5 0

Answer:

Explanation:

Given that,

Potential difference of battery is

V = 24V

Total charge battery can transferred

q = 340,000C

Work done by the battery?

Work done is given as

W = qV

Where q is charge in Columbs

V is potential difference in Volts

Then, W = qV

W = 340,000×24

W = 8,160,000 J

Work done by the battery over it's useful life time is 8,160,000J

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Answer:

42

Explanation:

range of ball = u×t

= 28×1.5

=42

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TRUE OR FALSE- Doing sports activities as your Cardio Training can be just as effective or more effective than running on a trea
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Suppose Car 1 collides with Car 2; Car 2 was not moving. In an ideal situation with no friction, according to the Law of Conserv
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A spring is mounted horizontally, with its left end fixed. A spring balance attached to the free end and pulled toward the right
malfutka [58]

Answer:

200 N/m

20 rad/s

0.31415 seconds

3.18309 Hz

Explanation:

m = Mass of glider = 0.5 kg

x = Displacement of spring

F = Force on spring = 6 N

From Hooke's law we have relation

F=kx\\\Rightarrow k=\frac{F}{x}\\\Rightarrow k=\frac{6}{0.03}\\\Rightarrow k=200\ N/m

The spring constant is 200 N/m

Angular frequency is given by

\omega=\sqrt{\frac{k}{m}}\\\Rightarrow \omega=\sqrt{\frac{200}{0.5}}\\\Rightarrow \omega=20\ rad/s

The angular frequency is 20 rad/s

Frequency is given by

f=\frac{\omega}{2\pi}\\\Rightarrow f=\frac{20}{2\pi}\\\Rightarrow f=3.18309\ Hz

The frequency is 3.18309 Hz

Time period is given by

T=\frac{1}{f}\\\Rightarrow T=\frac{1}{3.18309}\\\Rightarrow T=0.31415\ s

The time period is 0.31415 seconds

6 0
3 years ago
A blacksmith heats a 1.5 kg iron horseshoe to 500 C, containing 20 kg of water at 18 C then plunges it into a bucket What is the
Tema [17]

Answer:

21.85 C

Explanation:

mass of iron = 1.5 kg, initial temperature of iron, T1 = 500 C

mass of water = 20 kg, initial temperature of water, T2 = 18 C

let T be the equilibrium temperature.

Specific heat of iron = 449 J/kg C

specific heat of water = 4186 J/kg C

Use the principle of caloriemetry

heat lost by the hot body = heat gained by the cold body

mass of iron x specific heat of iron x decrease in temperature = mass of water  x specific heat of water x increase in temperature

1.5 x 449 x (500 - T) = 20 x 4186 x (T - 18)

336750 - 673.5 T = 83720 T - 1506960

1843710 = 84393.5 T

T = 21.85 C

8 0
3 years ago
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