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3 years ago

What happens when the voltage increases and the resistance stays the same in a electrical circuit?

Physics

1 answer:

Orlov [11]3 years ago

4 0

Answer:

The current in the circuit increases

Explanation:

The ohm's law states that the potential across a circuit is proportional to the current in the circuit.

V ∝ I

Where 'V' is the potential difference across the circuit and 'I' is the current in the circuit.

The proportionality constant present in the equation is the resistance of the circuit. Hence, the equation becomes

V = IR

According to the equation, when V is directly proportional to 'I' where 'R' remains as constant, then the change in 'V is brings change in 'I' to make the equation valid.

So, when there is an increase in the voltage, the current on the circuit increases.

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Question 3 of 15

omeli [17]

Answer:

B) Degrees

Explanation:

The directions of the vectors are often defined in terms of due East, due North, due West and due South. A direction exactly in between of North and East can be described as Northeast, similarly we can describe directions in terms of Northwest, Southeast and South west.

From these, the direction of a vector can be easily expressed in degrees, which is measured counter clockwise about its tail from due East. Considering that we can say that East is at 0° , North is at 90° , West is at 180 and South is at 270° counter clockwise rotation from due East.

So, we know that the direction of a vector lying somewhere between due East i.e 0° and due North i.e 90°, will be measured in degrees, which will have a value between 0°-90°

4 0

2 years ago

Two separate masses on two separate springs undergo simple harmonic motion indefinitely (the surface is frictionless). In CASE 1

Artemon [7]

Answer:

$\frac{F_1}{F_2} =\frac{1}{2}$

Explanation:

Assuming the we have to find ratio maximum forces on the mass in each case

we know that in a spring mass system

F= Kx

K= spring constant

x= spring displacement

Case 1:

$F_1=2k\times d$

case 2:

$F_2=2k\times 2d$

therefore, $\frac{F_1}{F_2} = \frac{2K\times d}{2K\times 2d}$

$\frac{F_1}{F_2} =\frac{1}{2}$

4 0

3 years ago

How long does it take a car traveling at 45km/h to travel 100.0 m?

djyliett [7]

Answer:

8 seconds

Explanation:

Since the carspeed is in km/h, we need equal units, so we will make 100.0m 0.1000km.

Then we need to find how long it takes the car to travel 0.1km

We can use the formula distance=speed * time and get

0.1=45 * time

Therefore we get .002222... hours

Multiplying this by 3600 (to get seconds, 60x60), we get 8 seconds

6 0

2 years ago

Read 2 more answers

Suppose you have a sphere of radius 1.00 m that carries a charge of +1.00 µC at its center. The radius of the sphere is changed

Gekata [30.6K]

Answer:

The flux remains the same and the field increases.

Explanation:

According to the Gauss' law, the surface integral of the electric field over a closed surface, called the Gaussian surface, is equal to $\dfrac{1}{\epsilon_o}$ times the net charge $q$ enclosed by the surface.

$\oint \vec E \cdot d\vec A=\dfrac{q}{\epsilon_o}\ \ \ \ ............\ (1).$

<em>where,</em>

$\epsilon_o$ = electrical permititivity of free space.

The term on the LHS of Gauss law' $\oint \vec E \cdot d\vec A$ is the electric flux $\phi$ through the Gaussian surface.

Therefore,

$\phi=\dfrac{q}{\epsilon_o}\ \ \ .....................\ (2).$

From equation (2), the electric flux through the sphere depends only on the charge enclosed by the sphere and not on the size of the sphere, therefore, when the radius of the sphere is changed to 0.500 m from 1.00 m, the electric flux through the surface will not change.

Consider the Gaussian sphere of radius $r$ same as that of the given sphere and concentric with the given sphere, then,

$\oint \vec E \cdot d\vec A=\oint E\ dA$

This is because the direction of the electric field through a Gaussian spherical surface and its surface area element $d\vec A$ is always normal to it.

Therefore,

$\oint \vec E \cdot d\vec A = E\oint dA=E\ 4\pi r^2$

Using equation (1),

$E\ 4\pi r^2=\dfrac{q}{\epsilon_o}\\E=\dfrac{q}{4\pi r^2 \epsilon_o}\\E\propto \dfrac{q}{r^2}$

The electric field is inversely proportional to the square of the radius of the sphere, therefore, on decreasing the radius from 1.00 m to 0.500 m, the electric field increases.

Thus, the correct answer is "The flux remains the same and the field increases".

7 0

3 years ago

Which of the following is a characteristic of an ionic compound?

Eva8 [605]

It is made of a metal and nonmetal.

7 0

3 years ago

Read 2 more answers