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2 years ago

7

For a fixed resistance, R, give the relation (i.e. cite the symbolic formula) between the electric potential, V, across the resi

stor

1 answer:

docker41 [41]2 years ago

5 0

Ohm's Law is a formula used to calculate the relationship between voltage, current and resistance in an electrical circuit.

what is ohm's law?

Ohm’s law states that the voltage or potential difference between two points is directly proportional to the current or electricity passing through the resistance, and directly proportional to the resistance of the circuit. The formula for Ohm’s law is V=IR.

Mathematically, Ohm’s law can be expressed as,

I ∝V

Introducing the constant of proportionality, the resistance R in the above equation, we get,

I=V/R or V= I×R

Where,

R is the resistance of the conductor in Ohm

I is the current through the conductor in Amperes (A),

V is the voltage or potential difference measured across the conductor in Volts (V).

learn more about ohm's law from here: brainly.com/question/28208985

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An energy plant produces an output potential of 1500 kV and serves a city 143 km away. A high-voltage transmission line carries

jekas [21]

Answer:

2123.55 $/hr

Explanation:

Given parameters are:

$V_{plant} = 1500$ KV

L = 143 km

I = 500 A

$\rho = 2.4$ $\Omega / km$

So, we will find the voltage potential provided for the city as:

$V_{wire} =IR = I\rho L = 1500*2.4*143 = 514.8$ kV

$V_{city} = V_{plant}- V_{wire} = 1500-514.8 = 985.2$ kV

Then, we will find dissipated power because of the resistive loss on the transmission line as:

$P = I^2R = I^2\rho L=500^2*2.4*143 = 8.58*10^7$ W

Since the charge of plant is not given for electric energy, let's assume it randomly as $x = \frac{\dollar 0.081}{kW.hr}$

Then, we will find the price of energy transmitted to the city as:

$Cost = P * x = 8.58*10^7 * 0.081 * 0.001 = 6949.8$ $/hr

To calculate money per hour saved by increasing the electric potential of the power plant:

Finally,

$I_{new} = P/V_{new} = I/1.2\\P_{new} = I_{new}^2R_{wire}\\Cost = P_{new}/1.44=6949.8/1.44 = 4826.25$ $/hr

The amount of money saved per hour = $6949.8 - 6949.8/1.44 = 2123.55$ $/hr

Note: For different value of the price of energy, it just can be substituted in the equations above, and proper result can be found accordingly.

3 0

3 years ago

There are two main types of exercise: (1) exercise, which uses oxygen for energy, and (2) exercise, which does not.

Goryan [66]

Answer:

b

Explanation:

6 0

3 years ago

Read 2 more answers

Tala wants to create a sound echo. She can create the echo using a pane of glass or a carpet square. Which one does she choose?

Alisiya [41]

Carpet asorbs sound, so glass should make it echo.

7 0

3 years ago

A ball is thrown straight up from the edge of the roof of a building. A second ball is dropped from the roof a time of 1.03 s la

Nookie1986 [14]

Answer:

$h=53.09m$ (2)

$v_{min}>5.05m/s$

$v_{max}$

Explanation:

<u>a)Kinematics equation for the first ball:</u>

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=8.9m/s$

The ball reaches the ground, y=0, at t=t1:

$0=h+v_{o}t_{1}-1/2*g*t_{1}^{2}$

$h=1/2*g*t_{1}^{2}-v_{o}t_{1}$ (1)

Kinematics equation for the second ball:

$v(t)=v_{o}-g*t$

$y(t)=y_{o}+v_{o}t-1/2*g*t^{2}$

$y_{o}=h$ initial position is the building height

$v_{o}=0$ the ball is dropped

The ball reaches the ground, y=0, at t=t2:

$0=h-1/2*g*t_{2}^{2}$

$h=1/2*g*t_{2}^{2}$ (2)

the second ball is dropped a time of 1.03s later than the first ball:

t2=t1-1.03 (3)

We solve the equations (1) (2) (3):

$1/2*g*t_{1}^{2}-v_{o}t_{1}=1/2*g*t_{2}^{2}=1/2*g*(t_{1}-1.03)^{2}$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$g*t_{1}^{2}-2v_{o}t_{1}=g*(t_{1}^{2}-2.06*t_{1}+1.06)$

$-2v_{o}t_{1}=g*(-2.06*t_{1}+1.06)$

$2.06*gt_{1}-2v_{o}t_{1}=g*1.06$

$t_{1}=g*1.06/(2.06*g-2v_{o})$

vo=8.9m/s

$t_{1}=9.81*1.06/(2.06*9.81-2*8.9)=4.32s$

t2=t1-1.03 (3)

t2=3.29sg

$h=1/2*g*t_{2}^{2}=1/2*9.81*3.29^{2}=53.09m$ (2)

b) $t_{1}=g*1.06/(2.06*g-2v_{o})$

t1 must : t1>1.03 and t1>0

limit case: t1>1.03:

$1.03>9.81*1.06/(2.06*g-2v_{o})$

$1.03*(2.06*9.81-2v_{o})$

$20.8-2.06v_{o}$

$(20.8-10.4)/2.06$

$v_{min}>5.05m/s$

limit case: t1>0:

$g*1.06/(2.06*g-2v_{o})>0$

$2.06*g-2v_{o}>0$

$v_{o}$

$v_{max}$

8 0

4 years ago

A 1250-kg compact car is moving with velocity v1 =36.2i^+12.7j^m/s. It skids on a frictionless icy patch and collides with a 448

MA_775_DIABLO [31]

Momentum is conserved, so the sum of the separate momenta of the car and wagon is equal to the momentum of the combined system:

(1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s) = ((1250 + 448) kg) <em>v</em>

where <em>v</em> is the velocity of the system. Solve for <em>v</em> :

<em>v</em> = ((1250 kg) ((36.2 <em>i</em> + 12.7 <em>j </em>) m/s) + (448 kg) ((13.8 <em>i</em> + 10.2 <em>j</em> ) m/s)) / (1698 kg)

<em>v</em> ≈ (30.3 <em>i</em> + 12.0 <em>j</em> ) m/s

7 0

3 years ago