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DENIUS [597]
2 years ago
6

What is the resistance of al lamp that allows a current of 10 amps with 120 volts

Physics
1 answer:
BartSMP [9]2 years ago
5 0
V=ir
I=10
v=120
r=?
r=v/i
r=120/10
r=12 ohm
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Because it can easily resist air resistance.

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20 cubic inches of a gas with an absolute pressure of 5 psi is compressed until its pressure reaches 10 psi. What's the new volu
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B. V_{f}= 10\,cubic\,inches

Explanation:

Assuming we are dealing with a perfect gas, we should use the perfect gas equation:

PV=nRT

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P_{i}V_{i}=n_{i}RT_{i} (1)

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V_{f} =\frac{P_{i}V_{i}}{P_{f}} =\frac{(5)(20)}{10}

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6 0
3 years ago
Why is sun called a source of light??​
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Suppose that Hubble's constant were H0 = 51 km/s/Mly (which is not its actual value). What would the approximate age of the univ
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Given the data in the question;

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Age\ of\ Universe; t = \frac{1}{H_0}

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1\ light\ years = 9.46*10^{15}m

so

1\ Million\ light\ years = [9.46 * 10^{15}m] * 10^6 = 9.46 * 10^{21}m

Therefore;

H_0 = 51\frac{km}{\frac{s}{Mly} } = 51000\frac{m}{s\ *\ Mly}  \\\\H_0 = 51000\frac{m}{s\ *\ (9.46*10^{21}m)} \\\\H_0 =  5.39 *10^{-18}s^{-1}\\

Now, we input this Hubble's constant value into our equation;

Age\ of\ Universe; t = \frac{1}{H_0}\\\\t = \frac{1}{ 5.39 *10^{-18}s^{-1}} \\\\t = 1.855 * 10^{17}s\\\\We\ convert\ to\ years\\\\t =  \frac{ 1.855 * 10^{17}}{60*60*24*365}yrs \\\\t = \frac{ 1.855 * 10^{17}}{31536000}yrs\\\\t = 5.88 *10^9 years

Therefore, given the Hubble's constant, the approximate age of the universe is 5.88 × 10⁹ Years.

Learn more: brainly.com/question/14019680

6 0
2 years ago
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4 0
2 years ago
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