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Valentin [98]
4 years ago
10

Maximum current problem. If the current on your power supply exceeds 500 mA it can damage the supply. Suppose the supply is set

for 37 V. What is the smallest resistance you can measure?
Physics
1 answer:
NikAS [45]4 years ago
8 0

Answer:

Smallest resistance will be 74 ohm

Explanation:

We have given that if current is exceeds from 500 mA then it can damage the supply, so current i = 500 mA = 0.5 A

Supply is connected to a voltage source of 37 colt

So V = 37 volt

We have to find the smallest resistance

According to ohm's law resistance will be the ratio of voltage and current and here we have given maximum current which will given smallest resistance

So resistance R=\frac{V}{i}=\frac{37}{0.5}=74ohm

So smallest resistance will be 74 ohm

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It returned to where it had started at the end of the trip

Explanation:

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Answer:

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Explanation:

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3 years ago
A gas undergoes two processes. In the first, the volume remains constant at 0.170 m3 and the pressure increases from 1.50×105 Pa
Mnenie [13.5K]

Answer:

W_{T} = - 24 kJ

Explanation:

The work (W) done by the gas can be calculated using the following equation:

W = p*\Delta V = p*(V_{f} - V_{i})

<u>Where:</u>

p: is the pressure

[tex}V_{f}[/tex]: is the final volume

[tex}V_{i}[/tex]: is the initial volume

In the first process, the work done by the gas is:

W_{1} = p*\Delta V = p*0 = 0

Since the volume remains constant, the total work done by the gas is equal to zero.

In the second process, the work done by the gas is:

W_{2} = p*(V_{f} - V_{i}) = 6.00 \cdot 10^{5} Pa*(0.130 m^{3} - 0.170 m^{3}) = -24 kJ

Now, the total work done by the gas during both processes is:

W_{T} = W_{1} + W_{2} = 0 + (-24 kJ) = - 24 kJ

Therefore, the total work done by the gas during both processes is - 24 kJ.

I hope it helps you!

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