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2 years ago

In a circuit, what is responsible for lighting a bulb?

Physics

2 answers:

Yakvenalex [24]2 years ago

6 0

Answer:

3) flow of electrons, if im correct

icang [17]2 years ago

6 0

Answer:

Explanation:

The flow of electrons. A current means the flow of electrons and without a current there would be no electricity therefore the bulb would not light up.

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A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr

Musya8 [376]

Answer:

162500000.

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

$I=\dfrac{q}{t}$

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

$n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}$

n=162500000 t

$\dfrac{n}{t}=16250000$

The number of electron passe per second will be 162500000.

4 0

3 years ago

An electron is initially at rest in a uniform electric field having a strength of 1.85 × 106 V/m. It is then released and accele

kirza4 [7]

Answer:

$W = 462.5 keV$

Explanation:

As we know that when electron moved in electric field then work done by electric field must be equal to the change in kinetic energy of the electron

So here we have to find the work done by electric field on moving electron

So we have

$F = qE$

$F = (1.6 \times 10^{-19})(1.85 \times 10^6)$

$F = 2.96 \times 10^{-13} N$

now the distance moved by the electron is given as

$d = 0.25 m$

so we have

$W = F.d$

$W = (1.6 \times 10^{-19})(1.85 \times 10^6)(0.25)$

$W = 7.4 \times 10^{-14} J$

now we have to convert it into keV units

so we have

$1 keV = 1.6 \times 10^{-16} J$

$W = 462.5 keV$

5 0

3 years ago

The speed of light in a vacuum is approximately 0.3 gm/s. What is the speed of light in meters per second?

klemol [59]

We have to convert Gm/s to m/s.

As $1 \ Gm/s = 10^9 \ m/s$

Therefore the speed of light in vacuum,

$c = 0.3 \ Gm/s = 0.3 \times 10^9 \ m/s \\\\ c= 3 \times 10^8 \ m/s$

Thus, the speed of light in m/s is $3 \times 10^8 \ m/s$

7 0

3 years ago

A. Compute the torque developed by an industrial motor whose ouputis 150kW at an angular speed of 4000 rev/min.

Gnom [1K]

Answer:

A. τ = 358 N.m

B. m = 73kg

C. V = 209.5 m/s

Explanation:

Let's first convert angular speed to rad/s:

ω = 4000 rev/min * 2π / 60 = 419 rad/s

Since Power is P = τ * ω,

τ = P / ω = 358 N.m

For part B: On the hanging weight:

T - m*g = 0

T = m*g

On the drum:

τ = T*R Replacing the expression for Tension of the rope:

τ = m*g*R

Solving for m:

m = 73 kg

For part C:

V = ω * R = 209.5 m/s

5 0

3 years ago

A wave is traveling down a string at 15 m/s. A snapshot of the wave is shown in the figure below.

max2010maxim [7]

(a) Amplitude=1.5 m

Amplitude is the maximum distance from the mean (zero disturbance) position.Here amplitude= 1.5 m

(b) wavelength= 2 m

Wavelength is the length of one complete wave or distance traveled by a wave in a time equal to Time Period. Here wavelength= 2 m

using equation for wave velocity V= f λ

15= f (2)

f=7.5 Hz

so frequency=7.5 Hz

(d) Time period= 0.133 s

Time period= T= 1/f

T= 1/7.5

T=0.133 s

so time period=0.133 s

5 0

3 years ago