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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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Suppose that you are swimming in a river while a friend watches from the shore. In calm water, you swim at a speed of 1.25 m/s .
aliya0001 [1]

Answer: The observing friend will the swimmer moving at a speed of 0.25 m/s.

Explanation:

  • Let <em>S</em> be the speed of the swimmer, given as 1.25 m/s
  • Let S_{0} be the speed of the river's current given as 1.00 m/s.

  • Note that this speed is the magnitude of the velocity which is a vector quantity.
  • The direction of the swimmer is upstream.

Hence the resultant velocity is given as, S_{R} = S — S 0S_{0}

S_{R} = 1.25 — 1

S_{R} = 0.25 m/s.

Therefore, the observing friend will see the swimmer moving at a speed of 0.25 m/s due to resistance produced by the current of the river.

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3 years ago
What is the the steadiest firing position?
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3 years ago
Help Meeee Please!!!!! This was due so long ago... DO NOT REPORT, THERE IS NO REASON TO REPORT! Assignment: Weathering in Action
zaharov [31]

Answer: Chemical

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22) The sensory cells associated with the deep layers of the epidermis are
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When a man returns to his well-sealed house on a summer day, he finds that the house is at 35°C. He turns on the air conditioner
Paul [167]

Answer:

1353.38 Watt

Explanation:

T₁ = Initial temperature of the house = 35°C

T₂ = Final temperature of the house = 20°C

Δt = Time taken to cool the house = 38 min = 38×60 = 2280 s

m = mass of air in the house = 800 kg

Cv = Specific heat at constant volume = 0.72 kJ/kgK

Cp = Specific heat at constant pressure = 1.0 kJ/kgK

Heat removed

q = mCvΔT

⇒q = 800×720×(35-20)

⇒q = 8640000 J

Average rate of hear removal

Q=\frac{q}{\Delta t}\\\Rightarrow Q=\frac{8640000}{2280}\\\Rightarrow Q=3789.47\ W

COP=\frac{Q}{W}\\\Rightarrow W=\frac{Q}{COP}\\\Rightarrow W=\frac{3789.47}{2.8}\\\Rightarrow W=1353.38\ W

∴ Power drawn by the air conditioner is 1353.38 Watt

6 0
3 years ago
Read 2 more answers
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