Ask question

Ask question

All categories

sertanlavr [38]

3 years ago

9

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential diffe

rence. Part A What is the resistance of a 100 W bulb?

2 answers:

Aleonysh [2.5K]3 years ago

7 0

Complete Question:

The power rating of a light bulb (such as a 100-W bulb) is the power it dissipates when connected across a 120-V potential difference.

Part A: What is the resistance of a 100 W bulb?

part B: Current flowing through a 100 W bulb?

part C: Find the resistance of a 40 W bulb

part D: Current flowing through a 40W bulb

Answer:

a)Resistance of a 100 W bulb, R = 144 ohms

b) Current flowing through a 100 W bulb, I =0.833 A

c)Resistance of a 40W bulb, R = 360 ohms

d)current in a 40W bulb, I = 0.33 A

Explanation:

a) Resistance of the 100 W bulb

Power, P = 100 W

Potential difference, V = 120 V

P = V²/R

R = V²/P

R = 120²/100

R = 144 ohms

b) the current flowing through the bulb

According to Ohm's law, V = IR

I = V/R

I = 120/144

I = 0.833 A

c) Find the resistance of a 40 W bulb

Since the voltage supplied is constant, P = V²/R

P = 40 W

V = 120 V

40 = 120²/R

R = 120²/40

R = 360 ohms

d)current flowing through the 40 W bulb

V = IR

I = V/R

I = 120/360

I = 0.33 A

Leni [432]3 years ago

6 0

Answer:

144 Ω

Explanation:

Power: This can be defined as the rate at which Energy is used up in an electric circuit. The S.I unit of power is Watt(W).

The expression for electric power is given as

P = V²/R............... Equation 1

Where P = power, V = Voltage, R = Resistance.

make R the subject of the equation

R = V²/P............. Equation 2

Given: V = 120 V, P = 100 W

Substitute into equation 2

R = 120²/100

R = 14400/100

R =144 Ω

You might be interested in

A source at rest emits light of wavelength 500 nm. When it is moving at 0.90c toward an observer, the observer detects light of

Vesna [10]

Answer:

The observer detects light of wavelength is 115 nm.

(b) is correct option

Explanation:

Given that,

Wavelength of source = 500 nm

Velocity = 0.90 c

We need to calculate the wavelength of observer

Using Doppler effect

$\lambda_{o}=\sqrt{\dfrac{1-\beta}{1+\beta}}\lambda_{s}$

Where, $\beta=\dfrac{c}{v}$

$\lambda_{o}=\sqrt{\dfrac{c-0.90c}{c+0.90c}}\times500\times10^{-9}$

$\lambda_{o}=115\ nm$

Hence, The observer detects light of wavelength is 115 nm.

8 0

3 years ago

In an Atwood's machine, one block has a mass of 602.0 g, and the other a mass of 717.0 g. The pulley, which is mounted in horizo

Wittaler [7]

Answer:

The acceleration of the both masses is 0.0244 m/s².

Explanation:

Given that,

Mass of one block = 602.0 g

Mass of other block = 717.0 g

Radius = 1.70 cm

Height = 60.6 cm

Time = 7.00 s

Suppose we find the magnitude of the acceleration of the 602.0-g block

We need to calculate the acceleration

Using equation of motion

$s=ut+\dfrac{1}{2}at^2$

Where, s = distance

t = time

a = acceleration

Put the value into the formula

$60.0\times10^{-2}=0+\dfrac{1}{2}\times a\times(7.00)^2$

$a=\dfrac{60.0\times10^{-2}\times2}{(7.00)^2}$

$a=0.0244\ m/s^2$

Hence, The acceleration of the both masses is 0.0244 m/s².

5 0

3 years ago

A planet is discovered orbiting around a star in the galaxy Andromeda at four times the distance from the star as Earth is from

Dmitriy789 [7]

Answer:

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

Explanation:

The orbital period of a planet around a star can be expressed mathematically as;

T = 2π√(r^3)/(Gm)

Where;

r = radius of orbit

G = gravitational constant

m = mass of the star

Given;

Let R represent radius of earth orbit and r the radius of planet orbit,

Let M represent the mass of sun and m the mass of the star.

r = 4R

m = 16M

For earth;

Te = 2π√(R^3)/(GM)

For planet;

Tp = 2π√(r^3)/(Gm)

Substituting the given values;

Tp = 2π√((4R)^3)/(16GM) = 2π√(64R^3)/(16GM)

Tp = 2π√(4R^3)/(GM)

Tp = 2 × 2π√(R^3)/(GM)

So,

Tp/Te = (2 × 2π√(R^3)/(GM))/( 2π√(R^3)/(GM))

Tp/Te = 2

Therefore, the orbital period of the planet is twice that of the earth's orbital period.

5 0

3 years ago

Read 2 more answers

Calculate the charge that flows through the cell in 1 minute. Each filament lamp has a power of 3 W and a resistance of 12 Ω

lapo4ka [179]

Answer:

24 Coulumbs

Explanation:

Given data

time= 1 minute= 6 seconds

P=2 W

R= 12 ohm

We know that

P= I^2R

P/R= I^2

2/12= I^2

I^2= 0.166

I= √0.166

I= 0.4 amps

We know also that

Q= It

substitute

Q= 0.4*60

Q= 24 Columbs

Hence the charge is 24 Coulumbs

5 0

3 years ago

An air hockey game has a puck of mass 30 grams and a diameter of 100 mm. The air film under the puck is 0.1 mm thick. Calculate

OverLord2011 [107]

Answer:

time required after impact for a puck is 2.18 seconds

Explanation:

given data

mass = 30 g = 0.03 kg

diameter = 100 mm = 0.1 m

thick = 0.1 mm = 1 × $10^{-4}$ m

dynamic viscosity = 1.75 × $10^{-5}$ Ns/m²

air temperature = 15°C

to find out

time required after impact for a puck to lose 10%

solution

we know velocity varies here 0 to v

we consider here initial velocity = v

so final velocity = 0.9v

so change in velocity is du = v

and clearance dy = h

and shear stress acting on surface is here express as

= µ $\frac{du}{dy}$

so

= µ $\frac{v}{h}$ ............1

put here value

= 1.75× $10^{-5}$ × $\frac{v}{10^{-4}}$

= 0.175 v

and

area between air and puck is given by

Area = $\frac{\pi }{4} d^{2}$

area = $\frac{\pi }{4} 0.1^{2}$

area = 7.85 × $\frac{v}{10^{-3}}$ m²

so

force on puck is express as

Force = × area

force = 0.175 v × 7.85 × $10^{-3}$

force = 1.374 × $10^{-3}$ v

and now apply newton second law

force = mass × acceleration

- force = $mass \frac{dv}{dt}$

- 1.374 × $10^{-3}$ v = $0.03 \frac{0.9v - v }{t}$

t = $\frac{0.1 v * 0.03}{1.37*10^{-3} v}$

time = 2.18

so time required after impact for a puck is 2.18 seconds

3 0

3 years ago