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2 years ago

According to ohms law, as the voltage increases across a 40 ohm resistor what happens to the current, resistors, and resistance

Physics

1 answer:

irakobra [83]2 years ago

7 0

Answer:

As the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

Explanation:

Ohm's law states that the current flowing through a circuit is directly proportional to the applied voltage.

V = IR

where;

I is the current

R is the resistance

V is the applied voltage

Based on this law (Ohm's law), as the voltage increases, the current flowing through the circuit increases while the resistance of the resistor remains constant.

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Two neutron stars are separated by a distance of 1.0 x 1012 m. They each have a mass of 1.0 x 1028 kg and a radius of 1.0 x 103

son4ous [18]

To develop this problem it is necessary to apply the concepts related to Gravitational Potential Energy.

Gravitational potential energy can be defined as

$PE = -\frac{GMm}{R}$

As M=m, then

$PE = -\frac{Gm^2}{R}$

Where,

m = Mass

G =Gravitational Universal Constant

R = Distance /Radius

PART A) As half its initial value is u'=2u, then

$U = -\frac{2Gm^2}{R}$

$dU = -\frac{2Gm^2}{R}$

$dKE = -dU$

Therefore replacing we have that,

$\frac{1}{2}mv^2 =\frac{Gm^2}{2R}$

Re-arrange to find v,

$v= \sqrt{\frac{Gm}{R}}$

$v = \sqrt{\frac{6.67*10^{-11}*1*10^{28}}{1*10^{12}}}$

$v = 816.7m/s$

Therefore the velocity when the separation has decreased to one-half its initial value is 816m/s

PART B) With a final separation distance of 2r, we have that

$2r = 2*10^3m$

Therefore

$dU = Gm^2(\frac{1}{R}-\frac{1}{2r})$

$v = \sqrt{Gm(\frac{1}{2r}-\frac{1}{R})}$

$v = \sqrt{6.67*10^{-11}*10^{28}(\frac{1}{2*10^3}-\frac{1}{10^{12}})}$

$v = 1.83*10^7m/s$

Therefore the velocity when they are about to collide is $1.83*10^7m/s$

7 0

3 years ago

Irina rode her bike to work at an average speed of 16 miles per hour. It started to rain, so she got a ride home along the same

KengaRu [80]

Answer:

0.7 hours

Explanation:

From the way back, we can calculate the distance between Irina's work and Irina's home. In fact, we know that the car takes 0.4 hourse traveling at 27 mph, so the distance covered should be

$d=vt=(27 mph)(0.4 h)=10.8 mi$

When Irina rides to work with her bike, she travels at a speed of 16 mph. So we can find the time she takes by dividing the total distance (10.8 miles) by her speed:

$t=\frac{d}{v}=\frac{10.8 mi}{16 mph}=0.7 h$

6 0

3 years ago

Read 2 more answers

suggest an experiment to prove that the rate of evaporation of a liquid depends on its surface area vapour already present in su

gulaghasi [49]

That's two different things it depends on:

-- surface area exposed to the air
AND
-- vapor already present in the surrounding air.

Here's what I have in mind for an experiment to show those two dependencies:

-- a closed box with a wall down the middle, separating it into two closed sections;

-- a little round hole in the east outer wall, another one in the west outer wall,
and another one in the wall between the sections;
So that if you wanted to, you could carefully stick a soda straw straight into one side,
through one section, through the wall, through the other section, and out the other wall.

-- a tiny fan that blows air through a tube into the hole in one outer wall.

Experiment A:

-- Pour 1 ounce of water into a narrow dish, with a small surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
-- Pour 1 ounce of water into a wide dish, with a large surface area.
-- Set the dish in the second section of the box ... the one the air passes through
just before it leaves the box.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
=============================
Show that the 1 ounce of water evaporated faster 
when it had more surface area.
============================================
============================================

Experiment B:

-- Again, pour 1 ounce of water into the wide dish with the large surface area.
-- Again, set the dish in the second half of the box ... the one the air passes
through just before it leaves the box.
-- This time, place another wide dish full of water in the first section of the box,
so that the air has to pass over it before it gets through the wall to the wide dish
in the second section. Now, the air that's evaporating water from the dish in the
second section already has vapor in it before it does the job.
-- Start the fan.
-- Count the amount of time it takes for the 1 ounce of water to completely evaporate.
==========================================
Show that it took longer to evaporate when the air 
blowing over it was already loaded with vapor.
==========================================

6 0

3 years ago

Describe how cool and warm air move in the atmosphere.

Anna [14]

Answer:

it is cool and warm

Explanation:

5 0

3 years ago

What is the new volume of the gas if the pressure on 350 L of oxygen

Iteru [2.4K]

Answer:

420 L

Explanation:

Applying Boyle's Law,

PV = P'V'.................... Equation 1

Where P = Initial pressure, P' = Final pressure, V = Initial volume, V' = Final volume.

make V' the subject of the equation

V' = PV/P'.................... Equation 2

From the question,

Given: P = 720 mmHg, V = 350 L, P' = 600 mmHg

Substitute these values into equation 2

V' = (720×350)/600

V' = 252000/600

V' = 420 L

7 0

3 years ago