What is a voltage source

Power Rating of a Resistor. The power rating of a resistor is the maximum power the resistor can safely dissipate without too gr

IgorLugansk [536]

(a) 273.9 V

The power rating of the resistor is given by

$P=\frac{V^2}{R}$

where

P is the power rating

V is the potential difference across the resistor

R is the resistance

If the maximum power rating is $P=5.0 W$ , and the resistance of the resistor is $R=15 k\Omega = 15000 \Omega$ , then we can find the maximum potential difference across the resistor by re-arranging the previous equation for V:

$V=\sqrt{PR}=\sqrt{(5.0 W)(15000 \Omega)}=273.9 V$

(b) 1.6 W

In this case, we have:

$R=9.0 k\Omega = 9000 \Omega$ is the resistance of the resistor

$V=120 V$ is the potential difference across the resistor

So we can find the power rating by using the same formula of part (a):

$P=\frac{V^2}{R}=\frac{(120 V)^2}{9000 \Omega}=1.6 W$

(c) Maximum voltage: 14.1 V; Rate of heat: 2.00 W and 3.00 W

Here we have two resistors of

$R_1 = 100 \Omega\\R_2 = 150 \Omega$

and each resistor has a power rating of

P = 2.00 W

So the greatest potential difference allowed in the first resistor is

$V=\sqrt{PR_1}=\sqrt{(2.00 W)(100 \Omega)}=14.1 V$

While the greatest potential difference allowed in the second resistor is

$V=\sqrt{PR_2}=\sqrt{(2.00 W)(150 \Omega)}=17.3 V$

So the greatest potential difference allowed not to overheat either of the resistor is 14.1 V.

In this condition, the power dissipated on the first resistor is 2.00 W, while the power dissipated on the second resistor is

$P_2 = \frac{V^2}{R_2}=\frac{(14.1 V)^2}{150 \Omega}=1.33 W$

And this corresponds to the rate of heat generated in the first resistor (2.00 W) and in the second resistor (1.33 W).

4 0

2 years ago

A string along which waves can travel is 4.36 m long and has a mass of 222 g. The tension in the string is 60.0 N. What must be

lora16 [44]

Answer:

frequency is 195.467 Hz

Explanation:

given data

length L = 4.36 m

mass m = 222 g = 0.222 kg

tension T = 60 N

amplitude A = 6.43 mm = 6.43 × $10^{-3}$ m

power P = 54 W

to find out

frequency f

solution

first we find here density of string that is

density ( μ )= m/L ................1

μ = 0.222 / 4.36

density μ is 0.050 kg/m

and speed of travelling wave

speed v = √(T/μ) ...............2

speed v = √(60/0.050)

speed v = 34.64 m/s

and we find wavelength by power that is

power = μ×A²×ω²×v / 2 ....................3

here ω is wavelength put value

54 = ( 0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 ) / 2

0.050 ×(6.43 × $10^{-3}$ )²×ω²× 34.64 = 108

ω² = 108 / 7.160 × $10^{-5}$

ω = 1228.16 rad/s

so frequency will be

frequency = ω / 2π

frequency = 1228.16 / 2π

frequency is 195.467 Hz

7 0

3 years ago

I will fan and give 2 medals!!!!!!

Maksim231197 [3]

Corona

I. Rare gaseous envelope

II. Irregular pearly glow

III. Total solar eclipse

IV. Surrounds darkened moon

V. Caused by diffraction

Solar Prominence

I. loop shape

II. large bright gaseous

III. luminous hydrogen gas

IV. rises above chromospheres

V. anchored to surface (sun’s surface)

Solar Flare

I. Intense High-energy

II. Causes electromagnetic disruptions

III. In solar atmosphere

IV. Varies in brightness

V. Emits radiation

Sunspots

I. Surface dark spots

II. Strong magnetic fields

III. Solar magnetic storms

IV. Visible through telescope

V. Cooler than photospheres

I did this in class I got 100%

7 0

3 years ago

What is one way you are using nonrenewable resources?

lutik1710 [3]

Putting gas in your car is one way you are using non-renewable resources.

You could Reduce, Recycle and Re-use items. You could manage what you do with you're water. Reducing and reusing products cuts down on manufacturing pollution, just as the use of recycled instead of virgin materials prevents pollution in industrial processes. You could add food coloring to your toilet tank. You could also do this: use sprinkler heads that sprinkle mist out instead of droplets of water. to avoid losing water to evaporation.

6 0

3 years ago

Two rigid rods are oriented parallel to each other and to the ground. The rods carry the same current in the same direction. The

Galina-37 [17]

Answer:

220 A

Explanation:

The magnetic force on the floating rod due to the rod held close to the ground is F = BI₁L where B = magnetic field due to rod held close the ground = μ₀I₂/2πd where μ₀ = permeability of free space = 4π × 10⁻⁷ H/m, I₂ = current in rod close to ground and d = distance between both rods = 11 mm = 0.011 m. Also, I₁ = current in floating rod and L = length of rod = 1.1 m.

So, F = BI₁L

F = (μ₀I₂/2πd)I₁L

F = μ₀I₁I₂L/2πd

Given that the current in the rods are the same, I₁ = I₂ = I

So,

F = μ₀I²L/2πd

Now, the magnetic force on the floating rod equals its weight , W = mg where m = mass of rod = 0.10kg and g = acceleration due to gravity = 9.8 m/s²

So, F = W

μ₀I²L/2πd = mg

making I subject of the formula, we have

I² = 2πdmg/μ₀L

I = √(2πdmg/μ₀L)

substituting the values of the variables into the equation, we have

I = √(2π × 0.011 m × 0.1 kg × 9.8 m/s²/[4π × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2 × 10⁻⁷ H/m × 1.1 m])

I = √(0.01078 kgm²/s²/[2.2 × 10⁻⁷ H])

I = √(0.0049 × 10⁷kgm²/s²H)

I = √(0.049 × 10⁶kgm²/s²H)

I = 0.22 × 10³ A

I = 220 A

7 0

3 years ago