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3 years ago

A certain lightning bolt moves 50.0 C of charge. How many units of fundamental charge is this?

1 answer:

7 0

Answer: There are number of electrons.
Explanation:
We are given 50 Coulombs of charge and we need to find the number of electrons that can hold this much amount of charge. So, to calculate that we will use the equation:

where,
n = number of electrons
Charge of one electron =
Q = Total charge = 50 C.
Putting values in above equation, we get:

Hence, there are number of electrons.

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"a horn emits a frequency of 1000 HZ. A 14 m/s wind is blowing toward a listener. What is the frequency of the sound heard by th

Semenov [28]

Answer:

$f_L = 1000 Hz$

Explanation:

given,

speed of wind, = 14 m/s

frequency of horn,f_s = 1000 Hz

speed of sound,V = 344 m/s

frequency heard by the listener

using Doppler effect

$f_L = \dfrac{v+v_L}{v+v_s}f_s$

f_L is the frequency of the sound heard by the listener

f_s is the frequency of sound emitted by the listener

V is the speed of sound

v_L is the speed of listener

v_s is the speed of source

now,

considering the frame of reference in which wind is at rest now, both listener and the source will be moving at 14 m/s

$f_L = \dfrac{v+14}{v+14}\times 1000$

now on solving we will get

$f_L = 1000 Hz$

hence, the frequency heard by the listener is equal to 1000 Hz

7 0

3 years ago

3. A -4.00-uC charge lies 20.0cm to the right of a 2.00-uC charge on the x axis. What is the force on the 2.00-uC charge?

Sladkaya [172]

<u>Answer:</u>

The force on the 2.00-uC charge is $36 \times 10^{10} \mathrm{N}$

<u>Explanation:</u>

We know that force between two charges is given by Coulomb’s law,

$\mathrm{F}=\mathrm{k} \frac{q 1 q 2}{r * r}$

Where k = Coulomb’s constant =

$9.0 \times 10^{9} \mathrm{Nm}^{2} \mathrm{C}^{-2}$

And q1 and q2 are the charges given to be = -4.00-uC and 2.00-uC charges

And r = distance between the charges = 20 cm = 0.2 m

Substituting the given values in the formula we get force applied on $2.00\ \mu C$ charge,

F = $36 \times 10^{10} \mathrm{N}$ attractive force which is the required answer.

3 0

3 years ago

A series circuit has a voltage supply of 12 V and a resistor of 2.4 kΩ. How much power is dissipated by the resistor?

adell [148]

Answer:

(B) 0.06W

Explanation:

power absorbed by the resistor is given by $\frac{V^2}{R}=I^2R$

Where V = voltage

R= resistance

I = current through the circuit

we have given V =12 Volt and resistance =2.4 K $\Omega$

current $I=\frac{V}{R}=\frac{12}{2.4\times 1000}=5\times 10^{-3}A=5mA$

power using voltage and resistance equation

$p=\frac{12^2}{2.4\times 1000}=60mW$ =0.06W

using current equation $P=I^2R=5^2\times 12=60mW$ = 0.06W

4 0

3 years ago

Two horizontal forces act on a 1.4 kg chopping block that can slide over a friction-less kitchen counter, which lies in an xy pl

kogti [31]

Answer:

Part a)

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

Explanation:

As per Newton's II law we know that

F = ma

so we will have

$a = \frac{F}{m}$

so we will have

$a = \frac{F_1 + F_2}{m}$

Part a)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i - 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i - 0.7 \hat j}{1.4}$

$a = (0.64\hat i - 0.5 \hat j)m/s^2$

Part b)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (-3\hat i + 4\hat j)}{1.4}$

$a = \frac{0.9 \hat i + 7.3 \hat j}{1.4}$

$a = (0.64\hat i + 5.21 \hat j)m/s^2$

Part c)

$a = \frac{(3.9 \hat i + 3.3 \hat j) + (3\hat i - 4\hat j)}{1.4}$

$a = \frac{6.9 \hat i - 0.7 \hat j}{1.4}$

$a = (4.92\hat i - 0.5 \hat j)m/s^2$

3 0

3 years ago

While riding in a moving car, you toss an egg directly upward. Does the egg tend to land behind you, in front of you, or back in

bazaltina [42]

Answer:

Explanation:

The movement of a body can be analyzed using New's first law. In an inertial frame (without acceleration) every body is kept at rest or moving at constant speed until there is an external force that changes this state

Let's analyze these cases in the framework of this first law

a) If the vehicle is going at constant speed the two bodies (the egg and the hands) do not change movement so he had returned to the hands

b) If the vehicle accelerates the passenger goes faster, but the egg that is not subject to anything does not change the movement, so it falls behind the passenger

c) If the vehicle slows down, the passenger reduces its speed and the distance traveled in time, but the egg that is not attached follows its movement and falls in front of the passenger.

4 0

3 years ago