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3 years ago

Compare the resistance of a 1.5-Amp interior light bulb of a car (operating off a 12-V battery) to the

Physics

1 answer:

OLga [1]3 years ago

8 0

Answer:

The resistance of the internal light bulb of the car is higher than that of the household circuitry.

Explanation:

Given the following data;

I. Current = 1.5 A

Voltage = 12 V

To find the resistance;

Voltage = current * resistance

12 = 1.5 * resistance

Resistance = 12/1.5

Resistance = 8 Ohms

II. Power = 100 Watt

Voltage = 110 V

To find the resistance;

Power = voltage/resistance²

100 = 110/resistance²

Cross-multiplying, we have;

100resistance² = 110

R² = 110/100

R² = 1.1

Taking the square root of both sides, we have;

R = √1.1

R = 1.05 Ohms

Therefore, the resistance of the internal light bulb of the car is higher than that of the household circuitry.

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A point charge q1 is at the center of a sphere of radius 20 cm. Another point charge q2 = 10 nC is located at a distance r = 10

jok3333 [9.3K]

Answer:

q₁ = -2.92 nC

Explanation:

Given;

first point charge, q₁ = ?

second point charge, q₂ = 10 nC

net flux through the surface of the sphere, Φ = 800 N.m²/C

According to Gauss’s law, the flux through any closed surface (Gaussian surface), is equal to the net charge enclosed divided by the permittivity of free space.

$\phi = \frac{q_{enc.}}{\epsilon_o}$

where;

Φ is net flux

$q_{enc.}$ net charge enclosed

ε₀ is permittivity of free space.

$q_{enc.}$ = Φε₀

= 800 x 8.85 x 10⁻¹²

= 7.08 x 10⁻⁹ C

$q_{enc.}$ = 7.08 nC

q₁ + q₂ = $q_{enc.}$

q₁ = $q_{enc.}$ - q₂

q₁ = 7.08nC - 10 nC

q₁ = -2.92 nC

4 0

3 years ago

a person takes a trip, driving with a constant speed of 89.5 km/h except for a 22.0 min rest stop. if the person’s average speed

fredd [130]

TheThe distance they have covered for trip is 219Km.

<h3>What do you mean by uniformly accelereted motion?</h3>

When an object is traveling in a straight line with an increase in velocity at equal intervals of time.

The total time for the trip is

T→ t1+ 22 min = t1+ 0.367 h ,where t1 is the time spent traveling at

V1= 89.5 km/ h .

the distance traveled is ∆x = V1t1=Vavg T

after applying value and calculating it we get

t1= 2.44h for a total time of

t total we get 2.81 h .

∆x =V1T1 = VavgTtotal

∆x = 77×2.81= 219 Km

to learn more about Uniform accelereted motion click here brainly.com/question/12920060

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3 0

2 years ago

Is anyone online??just asking

lianna [129]

Answer:

me...:(

Explanation:

3 0

3 years ago

Read 2 more answers

An isotropic point source emits light at wavelength 510 nm, at the rate of 170 W. A light detector is positioned 410 m from the

Wewaii [24]

Answer:

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

Explanation:

As we know that the power emitted by the source is given as

$P = 170 W$

now we know that

$P = \frac{N}{t} (\frac{hc}{\lambda})$

now we know that energy density is given as

$u = \frac{B^2}{2\mu_0} + \frac{\epsilon_0 E^2}{2}$

now we have

$E = B c$

$u = \frac{B^2}{2\mu_0}$

intensity is defined as

$I = \frac{P}{A}$

now we have

$\frac{I}{c} = u = \frac{B^2}{2\mu_0}$ [/tex]

now we have

$\frac{dB}{dt} = \omega B$

$\frac{dB}{dt} = \frac{2\pi c B}{\lambda}$

$\frac{dB}{dt} = \frac{2\pi c \sqrt{2\mu_0 I}}{\lambda\sqrt c}$

here we have

$I = \frac{P}{4\pi r^2}$

$I = \frac{170}{4\pi (410)^2}$

$I = 8.05 \times 10^{-5}$

now we have

$\frac{dB}{dt} = \frac{2\pi\sqrt{2\mu_0 c (8.05 \times 10^{-5})}}{(510 nm)}$

$\frac{dB}{dt} = 3.03 \times 10^6 T/s$

4 0

3 years ago

A crate of mass 190 kg sits on a horizontal floor. The coefficient of static friction between the crate and the floor is 0.4, an

Oxana [17]

Answer:

Explanation:

Mass of 190kg

Coefficient of static friction is 0.4

Coefficient of kinetic friction 0.36

Horizontal force= 500N

Taking g=9.81m/s^2.

The weight of the body my

W=190×9.81=1863.91N

There is a normal acting on the body which is equal to the weight

N=W=1863.91N

Frictional force(fr) is acting on the body and it is opposite the horizontal force.

The minimum force to be overcome before the object can start to move is Fr = μsN

Fr= μsN. μs=0.4

Fr= 0.4×1863.91

Fr=745.56N.

Since the horizontal force (500N) is not up to the minimum force to make the object move, then the force of 500N the body is still at rest.

Then the frictional force at that time is equal to the horizontal force

Therefore

Functional force = 500N

b. Mass of asteroid is

M=2000kg

Asteroid velocity at a particular instant is,

U=(-1.30x10^4, 4.20x10^4, 0)m/s

Magnitude of U is

U=√(-1.30×10^4)^2 +(4.2×10^4)^2+0

U=√1.933E9

U=4.39×10^4m/s

Position of the asteroid from the centre of the earth is,

R= (6.00x10^6, 10.00x10^6, 0)m.

The magnitude of the radius is

R = √(6.00x10^6)^2+ (10.00x10^6)^2+ 0^2

R=√3.6E13+10E13+0

R=√13.6E13

R=1.17E7m

R^2=13.6E13m

The mass of the earth is

Me=5.97x10^24 kg

The momentum of the asteroid after time, t=1.5×10^3s

Given that G=6.67x10^-11Nm^2/kg^2

Momentum is

Mv-Mu=Ft

There the new momentum will be

Mv=Ft+Mu

Now we the to find the force the earth exert on the asteroid by using

F=GMMe/R^2

F=6.67E-11 ×2000× 5.97E24 /13.6E13

F=7.964E17/13.6E13

F=5855.88N

The new momentum

Mv= Mu+Ft

Mv= 2000(4.39E4)+5855.88(1.5E3)

Mv=9.66E7kgm/s

The new momentum is 9.66×10^7 Kgm/s

8 0

3 years ago