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Nimfa-mama [501]
1 year ago
12

When does a lightbulb carry more current, (a) immediately after it is turned on and the glow of the metal filament is increasing

on (b) after it has been on for a few milliseconds and the glow is steady?
Physics
1 answer:
emmasim [6.3K]1 year ago
8 0

A  lightbulb carries more current immediately after it is turned on and the glow of the metal filament is increasing; option A.

<h3>What is current?</h3>

Current refers to the flow of electric charges typically electrons.

Current flowing through a metallic material decreases with increase in temperature of the material.

This is because the resistance of the metal increases with increase in temperature.

Therefore, for a light bulb, the current flow through it will be maximum when it is just turned on because the temperature, and hence the resistance of the  filament is at its lowest.

In conclusion, current flow decreases with increase in resistance.

Learn more about current and resistance at: brainly.com/question/24858512

#SPJ4

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1. A small package is dropped from the Golden Gate Bridge. What is the velocity of the package
Sidana [21]

The velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

From the question,

A small package is dropped from the Golden Gate Bridge.

This means the initial velocity of the package is 0 m/s.

We are to calculate the velocity of the package  after it has fallen for 3.0 s.

From one of the equations of kinematics for objects falling freely,

We have that,

v = u + gt

Where

v is the final velocity

u is the initial velocity

g is the acceleration due to gravity

and t is time

To calculate the velocity of the package  after it has fallen for 3.0 s

That means, we will determine the value of v, at time t = 3.0 s

The parameters are

u = 0 m/s

g = 9.8 m/s²

t = 3.0 s

Putting these values into the equation

v = u + gt

We get

v = 0 + (9.8×3.0)

v = 0 + 29.4

v = 29.4 m/s

Hence, the velocity of the package  after it has fallen for 3.0 s is 29.4 m/s

Learn more here: brainly.com/question/13327816

6 0
2 years ago
A student wearing frictionless in-line skates on a horizontal surface is pushed by a friend with a constant force of 47 N. How f
Novosadov [1.4K]

Answer:

7.53 m

Explanation:

Force, F = 47 N

initial velocity, u = 0

Final kinetic energy, Kf = 354 J

Let the distance traveled by the student is s.

According to the work energy theorem,

Work done by all forces = Change in kinetic energy

Force x distance = final kinetic energy - initial kinetic energy

F x s = kf - ki

47 x s = 354

s = 7.53 m

4 0
3 years ago
Which of the following processes could NOT be used to separate dissolved particles from the liquid in a solution? *
DIA [1.3K]

Answer:

Vaporation

Explanation:

In the vaporization or boiling, the passage of particles from the liquid state to the gaseous state occurs completely

5 0
3 years ago
Read 2 more answers
Point charges q1= - 4.30 nC and q2=+ 4.30 nC are separated by a distance of 3.00 mm, forming an electric dipole.
sammy [17]

Answer:

a) p = 25.8 10⁻¹² C m , b) The direction of the dipole moment is directed from the negative to the positive charge, c)  E = 4.65  10² N/C  

Explanation:

a) The dipole moment is

        p = 2qa

        p = 2 4.30 10⁻⁹ 3.00 10⁻³

        p = 25.8 10⁻¹² C m

b) The direction of the dipole moment is directed from the negative to the positive charge, that is, in the opposite direction to the electric field.

c) The torque is

          τ = p x E

          τ = p E sin  θ

          E = τ / p sin θ

          E = 7.20 10⁻⁹ /(25.8 10⁻¹²  sin 36.9)

          E = 4.65  10² N/C            

We calculate

           τ = 15.49 10⁻¹² 4.7 10²

           τ = 7.28 10⁻⁹ N m

4 0
3 years ago
A soap bubble, when illuminated with light of frequency 5.27 Hz × 1014 Hz, appears to be especially reflective. If it is surroun
denis23 [38]

Answer:

1.07004\times 10^{-7}\ m

Explanation:

n_s = Refractive index of bubble = 1.33

f = Frequency of light = 5.27\times 10^{14}\ Hz

c = Speed of light = 3\times 10^8\ m/s

The wavelength of light is given by

\lambda=\dfrac{2n_st}{m-\dfrac{1}{2}}

Wavelength is also given by

\lambda=\dfrac{c}{f}

m = 1 for minimum thickness

\dfrac{c}{f}=\dfrac{2n_st}{m-\dfrac{1}{2}}\\\Rightarrow t=\dfrac{m-\dfrac{1}{2}c}{2n_sf}\\\Rightarrow t=\dfrac{(1-\dfrac{1}{2})\times 3\times 10^8}{2\times 1.33\times 5.27\times 10^{14}}\\\Rightarrow t=1.07004\times 10^{-7}\ m

The minimum thickness is 1.07004\times 10^{-7}\ m

4 0
3 years ago
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