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3 years ago

how is it that an electric bulb lights up as soon as we turn the switch on if the drift speed is 1mm /s^2

1 answer:

5 0

Two analogies:
1. The wires are like hoses full of water. As soon as you turn on the water, water is pushed out the end of the hose.
2. Although the electrons only move at 1mm/s, the electomagnetic pulse travels through the circuit at near speed of light (varies btw .1c to .9c). It is this pulse that provides the pressure to push the electrons out the end of the wire into the light

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What is the formula for amplitude?

Vesna [10]

Explanation:

In periodic motion, amplitude is half the distance between the minimum and the maximum.

A = (max - min) / 2

8 0

3 years ago

A block with mass m =6.4 kg is hung from a vertical spring. When the mass hangs in equilibrium, the spring stretches x = 0.28 m.

Zanzabum

Answer

given,

mass of block (m)= 6.4 Kg

spring is stretched to distance, x = 0.28 m

initial velocity = 5.1 m/s

a) computing weight of spring

k x = m g

$k = \dfrac{mg}{x}$

$k = \dfrac{6.4 \times 9.8}{0.28}$

k = 224 N/m

b) $f = \dfrac{\omega}{2\pi}$

$\omega = \sqrt{\dfrac{k}{m}}= \sqrt{\dfrac{224}{6.4}} = 5.92 \ rad/s$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{k}{m}}$

$f = \dfrac{1}{2\pi}\sqrt{\dfrac{224}{6.4}}$

$f =0.94\ Hz$

c) $v_b = -v cos \omega t$

$v_b = -5.1 \times cos (5.92 \times 0.42)$

$v_b = 4.04\ m/s$

d) $a_{max} = v \omega$

$a_{max} = 4.04 \times 5.92$

$a_{max} =23.94\ m/s^2$

e) $Y =- A sin (\omega t)$

$A = \dfrac{v}{\omega}$

$A = \dfrac{4.04}{5.92}$

A = 0.682 m

$Y =- 0.682 \times sin (5.92 \times 0.42)$

$Y =- 0.42$

Force = $m \omega^2 |Y|$

= $6.4 \times 5.92^2\times 0.42$

F = 94.20 N

4 0

3 years ago

Explanation A 5000 kg rocket is at rest in deep space. The rocket burns fuel pushing 10kg of exhaust gases rearward at 4000 m/s.

katovenus [111]

Answer:

F = 4000 N

Explanation:

given,

mass of rocket (M)= 5000 Kg

10 Kg gas burns at speed (m)= 4000 m/s

time = 10 s

average force = ?

at the end the rocket is at rest

by conservation of momentum

M v + m v' = 0

5000 x v - 10 x 4000 = 0

5000 v = 40000

v = 8 m/s

speed of rocket = 8 m/s

now,

we know

change in momentum = F x Δ t

$F = \dfrac{m(v_i-v_f)}{\Delta t}$

$F = \dfrac{5000(8-0)}{10}$

F = 4000 N

Hence, the average force applied to the rocket is equal to F = 4000 N

4 0

3 years ago

Read 2 more answers

Consider three identical electric bulbs of power P. Two of bulbs are connected in series and the third one is connected in paral

Tamiku [17]

Answer: 3P/2

Explanation: Let the resistance of the bulbs be R.

now lets consider a Voltage V is supplied to the parallel circuit such that

$P=VI=V^2/R$

V=IR

both single bulb( bulb 3) and the two bulbs ( bulb 1 and bulb 2) are provided the same Voltage

( as the voltage remains same in parallel circuit)

we can calculate the Current across both circuits

At Bulb 3

Current 1=V/R

Power1=Voltage * Current1

Power1=V*V/R

Power1=P

At Bulb 1 and Bulb 2

Total Resistance= R+R=2R

$Current2=\frac{V}{2R}$

Power2=Voltage * Current2

$Power2=V*\frac{V}{2R} \\Power2=\frac{V^2}{2R} \\Power2=P/2$

$TotalPower=Power1+Power2\\TotalPower=P+P/2\\TotalPower=\frac{3P}{2}$

6 0

2 years ago

Please answer the question below

scoundrel [369]

The correct answer is B.
In a velocity vs time graph, a line going up means an increase in velocity, or speed, which correlates to positive acceleration. Negative acceleration means slowing down.

6 0

3 years ago