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disa [49]
2 years ago
12

While an emf source supplies energy to a circuit, that energy is dissipated when the current passes through resistance. In deali

ng with circuits though, it is often more useful to know the rate at which energy is changing. As you may recall, this quantity is also known as power.Consider a resistor with resistance R. The current through the resistor is I, and the voltage across the resistor is V. Which of these formulae can be used to find the power drawn by the conductor?
Physics
1 answer:
fredd [130]2 years ago
7 0

The question is incomplete, the options are;

RI^2

I^2/R

R/I^2

R/V^2

RV^2

V^2/R

VI

VIR

Select all that apply

Answer:

P=RI^2

P=V^2/R

P=VI

Explanation:

Power is the rate at which energy is changing in a circuit. It is shown by the formulas outlined above from the group of answer choices. Since the current (I), voltage (V), and resistance (R) were mentioned in the question, any of three three formulas could be used to obtain the power drawn by the conductor.

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A car slows down at -5.00 m/s² until it comes to a stop after travelling 15.0 m. What was the initial speed of the car?
lapo4ka [179]
<h2>Answer: 12.24m/s</h2>

According to <u>kinematics</u> this situation is described as a uniformly accelerated rectilinear motion. This means the acceleration while the car is in motion is constant.

Now, among the equations related to this type of motion we have the following that relates the velocity with the acceleration and the distance traveled:

V_{f}^{2}-V_{o}^{2}=2.a.d   (1)

Where:

V_{f} is the Final Velocity of the car. We are told "the car comes to a stop after travelling", this means it is 0.

V_{o} is the Initial Velocity, the value we want to find

a is the constant acceleration of the car (the negative sign means the car is decelerating)

d is the distance traveled by the car

Now, let's substitute the known values in equation (1) and find V_{o}:

0-V_{o}^{2}=2(-5m/{s}^{2})(15m)    (2)

-V_{o}^{2}=-150{m}^{2}/{s}^{2}    (3)

Multiplying by -1 on both sides of the equation:

V_{o}^{2}=150{m}^{2}/{s}^{2}    (4)

V_{o}=\sqrt{150{m}^{2}/{s}^{2}}    (5)

Finally:

V_{o}=12.24m/s >>>This is the Initial velocity of the car

5 0
2 years ago
If Earth were twice as far from the sun, the force of gravity attracting the Earth to the Sun would be
posledela
One quater as your moving away more!
3 0
3 years ago
Please help!! giving a lot of points
Readme [11.4K]

Question 1.

  • mass = 4500 kg
  • potential energy (p.e) = 67500 J

now, we know :

=》

p.e =  mgh

=》

67500 = 4500 \times 10 \times h

=》

67500 = 45000 \times h

=》

h =  \dfrac{67500}{45000}

=》

h = 1.5 \: m

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

  • mass = 4500 kg
  • kinetic energy = 63000 j

we know,

=》

k.e =  \dfrac{1}{2} mv {}^{2}

=》

63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

=》

{v}^{2}  =  \dfrac{63000 \times 2}{4500}

=》

{v}^{2}  = 28

=》

v =  \sqrt{28}

=》

v = 2 \sqrt{7} \:  \:  ms {}^{ - 1}

or

=》

5.29 \:  \: ms {}^{ - 1}

7 0
2 years ago
How is electron movement related to the bonding in sodium chloride?
irina1246 [14]
I think its c sorry if I’m wrong
8 0
3 years ago
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If the current in a wire increases from 5 A to 10 A, what happens to its magnetic field? If the distance of a charged particle f
dsp73

1. The magnitude of the magnetic field doubles

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

I=\frac{\mu_0 I}{2 \pi r}

where \mu_0 is the vacuum permeability, I is the current in the wire, r is the distance from the wire.

As we see from the formula, the intensity of the magnetic field is directly proportional to the current: if the current increases from 5 A to 10 A, it means it doubles, so the magnetic field doubles as well.

2. The magnitude of the magnetic field halves

Explanation: the intensity of the magnetic field produced by a current-carrying wire is given by:

I=\frac{\mu_0 I}{2 \pi r}

We see that the magnitude of the magnetic field is inversely proportional to the distance from the wire (r). In this case, the distance of the particle is changed from 10 cm to 20 cm, so it is doubled: therefore, the magnitude of the field will become half of the initial value.

3. The force reverses direction

Explanation: the force exerted on a charged particle in a magnetic field is:

F=qvB sin \theta

where q is the charge, v is the speed of the particle, B is the magnetic field intensity and \theta the angle between the direction of v and B. If the charge of the particle is switched from 2 µC to –2µC, the magnitude of the force does not change (because the absolute value of q does not change), however the charge q gets a negative sign (-), so the sign of the force changes and gets a negative sign too, so the force reverses direction.

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3 years ago
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