Ask question

Ask question

All categories

3 years ago

15

The difference in electric potential between a thunder cloud and the ground is 1.53 108 V. Electrons move from the ground which

is at a lower potential to the cloud which is at a higher potential. Determine the change in electric potential energy of one of the electrons that move to the cloud.

1 answer:

Kryger [21]3 years ago

8 0

Answer:

=−2.451 330 152 1*10^27J

Explanation:

The electric potential=the Voltage * Charge:

E = VQ

V = 1.53x10^8 V (positive, because the cloud has a higher potential)

Q = -1.60217657 x10^19 C (the charge of an electron)

E = (1.53x10^8 V )* (-1.60217657 x10^19 C)

E=−2.451 330 152 1*10^27J

The negative sign indicates that the potential energy is decreased by the movement of the electron.

You might be interested in

How does a crowbar work?

mylen [45]

It works by you putting leverage on one side makes more force go to the other side so if you put a crowbar in between a door and you push on one side the other will push the opposite side with more force<span />

4 0

3 years ago

Two charges (q1 = 3.8*10-6C, q2 = 3.2*10-6C) are separated by a distance of d = 3.25 m. Consider q1 to be located at the origin.

Sergio039 [100]

Answer:

The distance is 1.69 m.

Explanation:

Given that,

First charge $q_{1}= 3.8\times10^{-6}\ C$

Second charge $q_{2}=3.2\times10^{-6}\ C$

Distance = 3.25 m

We need to calculate the distance

Using formula of electric field

$E_{1}=E_{2}$

$\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(d-x)^2}$

$\dfrac{q_{1}}{q_{2}}=\dfrac{(x)^2}{(d-x)^2}$

$\sqrt{\dfrac{q_{1}}{q_{2}}}=\dfrac{x}{d-x}$

$x=(d-x)\times\sqrt{\dfrac{q_{1}}{q_{2}}}$

Put the value into the formula

$x=(3.25-x)\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x+x\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})=3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}$

$x=\dfrac{3.25\times\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}}}{(1+\sqrt{\dfrac{3.8\times10^{-6}}{3.2\times10^{-6}}})}$

$x=1.69\ m$

Hence, The distance is 1.69 m.

5 0

3 years ago

03: A mass with a 60 g vibrate at the end of a spring. The amplitude of the motion is 0.394 ft

Flauer [41]

Answer:

a) T = 1.69 s, b) k = 0.825 N / m, c) v = 1.46 feet/s, d) a = 5.41 ft / s²,

e) v = - 1,319 ft / s, a = - 2.70 ft / s², f) K = 4.8 10⁻³ J, U = 1.49 10⁻³ J

Explanation:

In a mass-spring system with simple harmonic motion, the angular velocity is

w = $\sqrt{\frac{k}{m} }$

a) find the period

angular velocity, frequency, and period are related

w = 2π f = 2π / T

f = 1 / T

T = 1 / f

T = 1 / 0.59

T = 1.69 s

b) the spring constant

w = 2π f

w = 2π 0.59

w = 3.70 rad / s

w² = k / m

k = w² m

k = 3.70² 0.060

k = 0.825 N / m

c) the maximum speed

simple harmonic movement is described by the expression

x = A cos (wt + Ф)

speed is defined by

v = $\frac{dx}{dt}$

v = -A w sin (wt + fi)

the speed is maximum when the cosine is ± 1

v = A w

v = 0.394 3.70

v = 1.46 feet/s

d) maximum acceleration

a = $\frac{dv}{dt}$

a = - A w² cos wt + fi

the acceleration is maximum when the cosine is ±1

a = A w²

a = 0.394 3.70²

a = 5.41 ft / s²

e) velocity and acceleration for x = 6 cm

let's reduce the cm to feet

x = 6 cm (1 foot / 30.48 cm) = 0.1969 foot

Before doing this part we must find the phase angle (Ф), the most common way to start the movement is to move the spring a small distance and release it, so its initial speed is zero for t = 0 s

let's use the expression for the velocity

v = -A w sin (0 + Фi)

0 = - A w sin Ф

so sin Ф = 0 which implies that Фi = 0

the equation of motion is

x = A cos wt

x = 0.394 cos 3.70t

we substitute

0.1969 = 0.394 cos 370t

3.70 t = cos⁻¹ (0.1969 / 0.394)

let's not forget that the angle is in radians

3.70, t = 1.047

t = 1.047 / 3.70

t = 0.2826 s

we substitute this time in the equation for velocity and acceleration

v = - Aw sin wt

v = - 0.394 3.70 sin 3.70 0.2826

v = - 1,319 ft / s

a = - A w² cos wt

a = - 0.394 3.70² cos 3.70 0.2826

a = - 2.70 ft / s²

f) the kinetic and potential energy at this point

K = ½ m v²

let's slow down to the SI system

v = 1.319 ft / s (1 m / 3.28 ft) = 0.402 m / s

K = ½ 0.060 0.402²

K = 4.8 10⁻³ J

U = ½ k x²

U = ½ 0.825 0.06²

U = 1.49 10⁻³ J

5 0

2 years ago

Two pulses are moving along a string. One pulse is

raketka [301]

Answer:

The answer is the 3rd option!

6 0

2 years ago

A lab cart with a mass of 15 kg is moving with constant velocity, v, along a straight horizontal track. A student drops a 2 kg m

lbvjy [14]

The equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ (option 3) represents the horizontal momentum of a 15 kg lab cart moving with a constant velocity, v, and that continues moving after a 2 kg object is dropped into it.

The horizontal momentum is given by:

$p_{i} = p_{f}$

$m_{1}v_{1}_{i} + m_{2}v_{2}_{i} = m_{1}v_{1}_{f} + m_{2}v_{2}_{f}$

Where:

m₁: is the mass of the lab cart = 15 kg

m₂: is the <em>mass </em>of the object dropped = 2 kg

$v_{1}_{i}$ : is the initial velocity of the<em> lab cart </em>

$v_{2}_{i}$ : is the <em>initial velocit</em>y of the <em>object </em>= 0 (it is dropped)

$v_{1}_{f}$ : is the final velocity of the<em> lab cart </em>

$v_{2}_{f}$ : is the <em>final velocity</em> of the <em>object </em>

Then, the horizontal momentum is:

$15v_{1}_{i} + 2*0 = 15v_{1}_{f} + 2v_{2}_{f}$

When the object is dropped into the lab cart, the final velocity of the lab cart and the object <u>will be the same</u>, so:

$15v_{1}_{i} + 2*0 = v_{f}(15 + 2)$

Therefore, the equation $15v_{i} + 2*0 = (15 + 2)v_{f}$ represents the horizontal momentum (option 3).

Learn more about linear momentum here:

brainly.com/question/2141713?referrer=searchResults

brainly.com/question/2400186?referrer=searchResults

I hope it helps you!

4 0

2 years ago