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Over [174]
3 years ago
6

What is the current in a 160V circuit if the resistance is 5Ω?

Chemistry
2 answers:
oee [108]3 years ago
5 0

Answer:

use v=IR

Explanation:

so, 160=I*5

I = 160/5

=32 ampere

Wittaler [7]3 years ago
3 0

The current = 32 A

<h3>Further explanation</h3>

Electric current is a ratio between the voltage and the amount of resistance at the conductor.  

Can be formulated  

\large{\boxed{\bold{I=\frac{V}{R}}}

V = Voltage (volts)  

R = resistance (ohms)  

This formula is also known as Ohm's Law, which states that  

<em>the amount of current flowing is directly proportional to the voltage and inversely proportional to the resistance.</em><em>  </em>

<em></em>

V=160 V

R=5Ω

\tt I=\dfrac{160}{5}=32~A

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The change in internal energy for the combustion of 1.0 molmol of octane at a pressure of 1.0 atmatm is -5084.1 kJkJ . Part A If
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Answer:

W = -10.3 kJ

Explanation:

During combustion, the system performs work and releases heat. Therefore, the change in internal energy is negative, and the change in enthalpy, which is equal to heat at constant pressure, is also negative. Work is then calculated by rearranging the equation for the change in internal energy:

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2 years ago
Does mass change when you add alka seltzer to water??
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Answer & Explanation:

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5 0
3 years ago
Read 2 more answers
The internal energy of 10 moles of helium (a monatomic gas) is 15 kJ. What is the rms speed of the molecules? (The molar mass of
Radda [10]

Answer:

v_{rms}=866.32m/s

Explanation:

Hello,

In this case, since the rms speed of the molecules is computed by:

v_{rms}=\sqrt{\frac{3RT}{M} }

Whereas the absolute temperature is computed from the internal energy (by using the Cp of helium (3.1156 J/g*K) as shown below:

U=nCvT\\\\T=\frac{U}{nCv}=\frac{15kJ*\frac{1000J}{1kJ} }{10mol*\frac{4.00g}{1mol} *3.1156\frac{J}{g*K} }  \\\\T=120.36K

Thereby, the rms speed results:

v_{rms}=\sqrt{\frac{3*8.314\frac{kg*m^2}{s^2*mol*K}*120.36K}{4.00\frac{g}{mol}*\frac{1kg}{1000} } } \\\\v_{rms}=866.32m/s

Regards.

6 0
3 years ago
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