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masya89 [10]
3 years ago
15

How does the current in a resistor change if the voltage across the resistor is

Physics
1 answer:
andrey2020 [161]3 years ago
3 0

Answer:

increases by a factor of 4.

Explanation:

The power dissipated through a resistor is I2RI2R

Current=IResistance=RPower=(current)2×(resistance)=I2RIf we double the current, P=(2I)2R=4I2RCurrent=IResistance=RPower=(current)2×(resistance)=I2RIf we double the current, P=(2I)2R=4I2R

Thus, doubling the current increases the power dissipated through a resistor by a factor of four.

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A string exerts a force of 20 N on a box at an angle of 38° from the horizontal. What is the horizontal
Gnoma [55]

Answer:

15.76N

Explanation:

horizontal component =Fcos¢ = 20cos38 = 15.76N

4 0
2 years ago
Read 2 more answers
Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would
Butoxors [25]

Suppose the entire solar nebula had cooled to 50 K before the solar wind cleared the early solar system of its gases. How would the composition and sizes of the planets of the inner solar system be different from what we see today is given below

Explanation:

1.In astronomy or planetary science, the frost line, also known as the snow line or ice line, is the particular distance in the solar nebula from the central protostar where it is cold enough for volatile compounds such as water, ammonia, methane, carbon dioxide, carbon monoxide to condense into solid ice grains.

2.The frost line in the solar nebula lies between Mars and Jupiter. It is the distance where it was cold enough for hydrogen compounds to condense into ices. Frost line: Explain how temperature differences led to the formation of two distinct types of planets.

3. The frost line is the point moving away from the Sun where it is cool enough for hydrogen compounds to freeze. Since the solar nebula was hotter near the center of the disk, hydrogen compounds such as water stayed gaseous in the inner solar system. Outside of the frost line, they froze.

4.The solar nebula flattened into a rotating disk. As gas became dense and hot, then it spins faster and pulled towards the center whereby the sun is formed. Solar nebula they collapse where the protostellar disk rotates. In the center of of nebula, there is a fusion begins and then sun is being formed

5.When it comes to the formation of our Solar System, the most widely accepted view is known as the Nebular Hypothesis. In essence, this theory states that the Sun, the planets, and all other objects in the Solar System formed from nebulous material billions of years ago.

4 0
3 years ago
A man takes 20 seconds to climb 5m up a ladder. He weighs 720N. Calculate the power he must deliver to do this.
Inessa05 [86]

Answer:

180 W

Explanation:

The work done by the man against gravity is equal to its gain in gravitational potential energy:

W=mg\Delta h

where

(mg) = 720 N is the weight of the man

\Delta h= 5 m is the change in height

Substituting,

W=(720)(5)=3600 J

The power he must deliver is given by

P=\frac{W}{t}

where

W = 3600 J

t = 20 s is the time taken

Substituting,

P=\frac{3600}{20}=180 W

3 0
4 years ago
Which technological advance allows scientists to handle these objects enough to feel their properties while still protecting the
muminat
The answer depends heavily on what 'objects' you're talking about.
5 0
4 years ago
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A bowling ball of mass 5 kg rolls down a slick ramp 20 meters long at a 30 degree angle to the horizontal. What is the work done
Elena-2011 [213]

Answer:

The work done by gravity during the roll is 490.6 J

Explanation:

The work (W) is:

W = F*d

<em>Where</em>:

F: is the force

d: is the displacement = 20 m

The force is equal to the weight (W) in the x component:

F = W_{x} = mgsin(\theta)

<em>Where:</em>

m: is the mass of the bowling ball = 5 kg

g: is the gravity = 9.81 m/s²    

θ: is the degree angle to the horizontal = 30°        

F = mgsin(\theta) = 5 kg*9.81 m/s^{2}*sin(30) = 24.53 N    

Now, we can find the work:

W = F*d = 24.53 N*20 m = 490.6 J      

Therefore, the work done by gravity during the roll is 490.6 J.

I hope it helps you!

6 0
3 years ago
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