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3 years ago

Which of the following equations represents Ohm’s Law? Select all that apply.

Physics

2 answers:

Vera_Pavlovna [14]3 years ago

6 0

Answer : The correct options are, V=IR, I=V/R and R=V/I

Explanation :

Ohm's law : It is defined as an electric current is directly proportional to voltage and inversely proportional to resistance.

Formula of Ohm's law :

$V=I\times R$

Or,

$R=\frac{V}{I}$

or,

$I=\frac{V}{R}$

where,

R = resistance of a circuit

V = voltage of circuit

I = current flowing in a circuit

Hence, the correct options are, V=IR, I=V/R and R=V/I

yaroslaw [1]3 years ago

3 0

The first three choices are nonsense.

The last three choices are alternative statements of Ohm's law:

R = V / I

V = I · R

I = V / R

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A 98.0 N grocery cart is pushed 12.0 m along an aisle by a shopper who exerts a constant horizontal force of 40.0 N. If all fric

Digiron [165]

Answer:

9.8 m/s

Explanation:

The work done by the force pushing the cart is equal to the kinetic energy gained by the cart:

$W=K_f -K_i$

where

W is the work done

$K_f$ is the final kinetic energy of the cart

$K_i$ is the initial kinetic energy of the cart, which is zero because the cart starts from rest, so we can write:

$W=K_f$

But the work is equal to the product between the pushing force F and the displacement, so

$W=Fd=(40.0 N)(12.0 m)=480 J$

So, the final kinetic energy of the cart is 480 J. The formula for the kinetic energy is

$K_f=\frac{1}{2}mv^2$ (1)

where m is the mass of the cart and v its final speed.

We can find the mass because we know the weight of the cart, 98.0 N:

$m=\frac{F_g}{g}=\frac{98.0 N}{9.8 m/s^2}=10 kg$

Therefore, we can now re-arrange eq.(1) to find the final speed of the cart:

$v=\sqrt{\frac{2K_f}{m}}=\sqrt{\frac{2(480 J)}{10 kg}}=9.8 m/s$

7 0

3 years ago

Determine the speed

Paraphin [41]

Calculating the average speed is simple using the formula <span>speed = distance/time</span>

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4 years ago

What happens to the speed of the electrons when the resistance is decreased?

3241004551 [841]

Answer:

they slow down

Explanation:

Because number of electrons remain same in the closed circuit or wire, due to conservation of charge in any case. But due to addition of a external resistor current will decreases, by applying ohm's law( V=IR).

7 0

3 years ago

A coil 3.95 cm radius, containing 520 turns, is placed in a uniform magnetic field that varies with time according to B=( 1.20×1

Pie

Answer:

(a) E= 3.36×10−2 V +( 3.30×10−4 V/s3 )t3

(b) $I=0.0085\ A$

Explanation:

Given:

radius if the coil, $r=0.0395\ m$

no. of turns in the coil, $n=520$

variation of the magnetic field in the coil, $B=(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4$

resistor connected to the coil, $R=560\ \Omega$

(a)

we know, according to Faraday's Law:

$emf=n.\frac{d\phi}{dt}$

where:

$d \phi=$ change in associated magnetic flux

$\phi= B.A$

where:

A= area enclosed by the coil

Here

$A=\pi.r^2$

$A=\pi\times 0.0395^2$

$A=0.0049\ m^2$

$\therefore \phi=((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049$

So, emf:

$emf= 520\times \frac{d}{dt} [((1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)\times 0.0049]$

$emf= 520\times 0.0049\times \frac{d}{dt} [(1.2\times 10^{-2})t+(3.45\times 10^{-5})t^4)]$

$emf= 2.548\times [0.012+(13.8\times 10^{-5})t^3)]$

$emf= 0.0306+3.516\times 10^{-4}\ t^3$

(b)

Given:

$t_0=5.25\ s$

Now, emf at given time:

$emf=4.7755\times 10^{-2}\ V$

∴Current

$I=\frac{emf}{R}$

$I=\frac{4.7755\times 10^{-2}}{560}$

$I=8.5\times 10^{-5} A$

6 0

4 years ago

When an ion accelerates through a potential difference of 2160 V its electric potential energy decreases by 3.46×10−16 J . What

Anastaziya [24]

We do as follows:

energy = charge x voltage change .

Let q be the charge then E = q V

<span>and solving for q we have

q = E / V = -1.37e-5J / 2850V = -4.81x10-9 C
</span>
Hope this answers the question. Have a nice day.

5 0

4 years ago