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tia_tia [17]
3 years ago
10

How much voltage is required to run 0.42 A of current through a 150

Physics
2 answers:
IRINA_888 [86]3 years ago
8 0

Answer:

like the other person said, the answer is 6.3

Explanation:

igomit [66]3 years ago
4 0

The voltage in the resistor is 63 V

Explanation:

We can solve the problem by applying Ohm's law, which states the relationship between voltage, current and resistance in a resistor:

V=RI

where

V is the voltage

R is the resistance

I is the current

For the resistor in this problem, we have:

I = 0.42 A is the current

R=150 \Omega is the resistance

Substituting into the equation, we find the voltage needed:

V=(0.42)(150)=63 V

Learn more about voltage and current:

brainly.com/question/4438943

brainly.com/question/10597501

brainly.com/question/12246020

#LearnwithBrainly

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How much heat do you need to raise the temperature of 150 g of oxygen from -30c to -15c?
Naddika [18.5K]
The amount of heat needed to raise the temperature of a substance by \Delta T is given by
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For oxygen, the specific heat capacity is approximately 
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The variation of temperature for the sample in our problem is 
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while the mass is m=150 g, so the amount of heat needed is
Q=m C_s \Delta T=(150 g)(0.92 J/g K)(15 K)=2070 J
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Answer:

As the capacitor is discharging, the current is increasing

Explanation:

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As the capacitor is discharging, the current is increasing

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A lens forms an image of an object. The object is 16.0 cm from the lens. The image is 12.0 cm from the lens on the same side as
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(a) -48.0 cm, diverging

We can use the lens equation:

\frac{1}{f}=\frac{1}{p}+\frac{1}{q}

where

f is the focal length

p = 16.0 cm is the object distance

q = -12.0 cm is the image distance (with a negative sign because the image is on the same side as the object, so it is virtual)

Solving for f, we find the focal length of the lens:

\frac{1}{f}=\frac{1}{16.0 cm}+\frac{1}{-12.0 cm}=-0.021 cm^{-1}

f=\frac{1}{-0.021 cm^{-1}}=-48.0 cm

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(b) 6.38 mm, erect

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\frac{y'}{y}=-\frac{q}{p}

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(c)

See attached picture.

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