A certain lightning bolt moves 50.0 c of charge. how many fundamental units of charge (qe) is this?

Answers

Answer 1

Answer: The lightning bolt that moves 50.0 C of charge is converted to qe units. The conversion factor is 1 coulumb is equivalent to 6.24150975·10^18 e. Thus, converting 50 C to e is:

50.0 C * (6.24150975x10^18 e)/1C
equals 3.120754875x10^20 e

Answer 2

Answer:

Answer: There are $3.125\tims 10^(20)$ 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:

$nq_e=Q$

where,

n = number of electrons

$q_e$ Charge of one electron = $1.6* 10^(-19)C$

Q = Total charge = 50 C.

Putting values in above equation, we get:

$n* 1.6* 10^(-19)C=50C\n\nn=(50C)/(1.6* 10^(-19)C)\n\nn=3.125* 10^(20)C$

Hence, there are $3.125\tims 10^(20)$ number of electrons.

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A large rectangular tub is filled to a depth of 2.60 m with olive oil, which has density 915 kg/m3 . If the tub has length 5.00 m and width 3.00 m, calculate (a) the weight of the olive oil, (b) the force of air pressure on the sur- face of the oil, and (c) the pressure exerted upward by the bottom of the tub.

Answers

Answer:

weight is 3.50 x 10^5 N

force is 1.52 * 10^6 N

pressure is 1.25 * 10^5 Pa

Explanation:

given data

Given data

depth = 2.60 m

density = 915 kg/m3

length = 5.00 m

width = 3.00 m

to find out

weight of the olive oil, force of air pressure and the pressure exerted upward

solution

we know density = mass / volume

mass = density* width *length *depth

mass = (915)(3)(5)(2.60)

mass = 3.57 x 10^4 Kg

so weight = mg = 3.57 x 10^4 (9.81) = 3.50 x 10^5 N

weight is 3.50 x 10^5 N

and

we know force = pressure * area

area = 3 *5 = 15 m²

and we know atmospheric Pressure is about 1.01 * 10^5 Pa

so force = 1.01 * 10^5 (15) = 1.52 x 10^6 N

force is 1.52 * 10^6 N

and

we know Fup - Fdown = Weight

Fup = 1.52 * 10^6 + 3.50 * 10^5

Fup = 1.87 * 10^6 N

so pressure = Fup / area

pressure = 1.87 * 10^6 / 15

pressure is 1.25 * 10^5 Pa

The position of a particular particle as a function of is given by r=(8.5t i+5.6j-2tk )m (a) Determine the particle’s velocity and acceleration as a function of time.(b) Find the instantaneous velocity and acceleration of the particle at t=3.0 s.

Answers

Answer:

use google to find answer

Explanation:

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The radius of Earth is 6370 km in the Earth reference frame. The cosmic ray is moving at 0.880Co relative to Earth.a. In the reference frame of a cosmic ray how wide does Earth seem along the flight direction?
b. In the reference frame of a cosmic ray how wide does Earth seem perpendicular to the flight direction?
Express your answer with the appropriate units.

Answers

Answer:

6052114.67492 m

$12.742* 10^(6)\ m$

Explanation:

v = Velocity of cosmic ray = 0.88c

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

d = Width of Earth = Diameter of Earth = $12.742* 10^(6)\ m$

When the cosmic ray is moving towards Earth then in the frame of the cosmic ray the width of the Earth appears smaller than the original

This happens due to length contraction

Length contraction is given by

$d_e=d\sqrt{1-(v^2)/(c^2)}\n\Rightarrow d_e=12.742* 10^(6)\sqrt{1-(0.88^2c^2)/(c^2)}\n\Rightarrow d_e=6052114.67492\ m$

The Earth's width is 6052114.67492 m

Contraction only occurs in the cosmic ray's frame of reference in the direction of the ray. But in perpendicular direction the width remains unchanged.

Hence, the width is $12.742* 10^(6)\ m$

Two point charges are placed on the x axis.The firstcharge, q1= 8.00 nC, is placed a distance 16.0 mfromthe origin along the positive x axis; the second charge,q2= 6.00 nC, is placed a distance 9.00 mfrom the originalong the negative x axis.[Give the x and y components of the electric fieldas an ordered pair. Express your answer innewtons per coulomb to three significant figures.Keep in mind that an x component that points tothe right is positive and a y component thatpoints upward is positive.]

Answers

Answer:

E = (0, 0.299) N

Explanation:

Given,

Charge $q_1\ =\ 8.0\ nC$
Charge $q_2\ =\ 6.0\ nC$
Distance of the first charge from the origin = (16m, 0)
Distance of the second charge from the origin = (-9, 0)
Point where the electric field required = (0, 12m)

Let $\theta_1\ and\ theta_2$ be the angle of the electric fields by first and second charge at the point A.

$\therefore sin\theta_1\ =\ (12)/(20)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(20)\ \right )\n\Rightarrow \theta_1\ =\ 36.87^o\n\n\therefore sin\theta_1\ =\ (12)/(9)\n\Rightarrow \theta_1\ =\ sin^(-1)\left ((12)/(9)\ \right )\n\Rightarrow \theta_1\ =\ 53.13^o\n$

Electric field by charge $q_1$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 8* 10^(-9))/(20^2)\n\Rightarrow F_1\ =\ 0.18\ N/C$

Electric field by the charge $q_2$ at point A,

$F_1\ =\ (kq_1)/(r_1^2)\n\Rightarrow F_1\ =\ (9* 10^9* 6.0* 10^(-9))/(16^2)\n\Rightarrow F_1\ =\ 0.24\ N/C$

Now,

Net electric field in horizontal direction at point A $F_x\ =\ F_(1x)\ +\ F_(2x)\n\Rightarrow F_x\ =\ F_1cos\theta_1\ +\ F_2cos\theta_2\n\Rightarrow F_x\ =\ 0.18*( -cos36.87^o)\ +\ 0.24* cos53.13^o\n\Rightarrow F_x\ =\ -0.144\ +\ 0.144\ N/C\n\Rightarrow F_x\ =\ 0\ N/C$

Net electric field in vertical direction at point A.

$F_y\ =\ F_(1y)\ +\ F_(2y)\n\Rightarrow F_y\ =\ F_1sin\theta_1\ +\ F_2sin\theta_2\n\Rightarrow F_y\ =\ 0.18* sin36.87^o\ +\ 0.24* sin53.13^o\n\Rightarrow F_y\ =\ 0.180\ +\ 0.192\n\Rightarrow F_y\ =\ 0.299\ N/C$

Hence, the net electric field at point A,

$F\ =\ ( 0, 0.299 )\ N/C.$

A lead ball is dropped into a lake from a diving board 6.10 mm above the water. After entering the water, it sinks to the bottom with a constant velocity equal to the velocity with which it hit the water. The ball reaches the bottom 4.50 ss after it is released. How deep is the lake?

Answers

Answer:

D=1.54489 m

Explanation:

Given data

S=6.10 mm= 0.0061 m

To find

Depth of lake

Solution

To find the depth of lake first we need to find the initial time ball takes to hit the water.To get the value of time use below equation

$S=v_(1)t+(1/2)gt^(2) \n 0.0061m=(0m/s)t+(1/2)(9.8m/s^(2) )t^(2)\n t^(2)=(0.0061m)/(4.9m/s^(2) )\n t=\sqrt{1.245*10^(-3) }\n t=0.035s$

So ball takes 0.035sec to hit the water

As we have found time Now we need to find the final velocity of ball when it enters the lake.So final velocity is given as

$v_(f)=v_(i)+gt\nv_(f)=0+(9.8m/s^(2) )(0.035s)\n v_(f)=0.346m/s$

Since there are (4.50-0.035) seconds left for (ball) it to reach the bottom of the lake

So the depth of lake given as:

$D=|vt|\nD=|0.346m/s*4.465s|\nD=1.54489m$

Answer: d = 1.54m

The depth of the lake is 1.54m

Explanation:

The final velocity of the ball just before it hit the water can be derived using the equation below;

v^2 = u^2 + 2as ......1

Where ;

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance travelled.

Since the initial velocity is zero, and the acceleration is due to gravity, the equation becomes:

v^2 = 2gs

v = √2gs ......2

g = 9.8m/s^2

s = 6.10mm = 0.0061m

substituting into equation 2

v = √(2 × 9.8× 0.0061)

v = 0.346m/s

The time taken for the ball to hit water from the time of release can be given as:

d = ut + 0.5gt^2

Since u = 0

d = 0.5gt^2

Making t the subject of formula.

t = √(2d/g)

t = √( 2×0.0061/9.8)

t = 0.035s

The time taken for the ball to reach the bottom of the lake from the when it hits water is:

t2 = 4.5s - 0.035s = 4.465s

And since the ball falls for 4.465s to the bottom of the lake at the same velocity as v = 0.346m/s. The depth of the lake can be calculated as;

depth d = velocity × time = 0.346m/s × 4.465s

d = 1.54m

The depth of the lake is 1.54m

The erg is a unit of work in units of centimeters (cm), grams (g), and seconds (s), and 1 erg=1 g⋅cm^2/s^2 . Recall that the SI unit of force is the newton (N) and is equal to kg⋅m/s^2 . You push an object 0.032 m by exerting 0.010 N of force. The work is the force times the distance.How much work have you done, expressed in ergs?

A: 3,200 ergs

B: 32 ergs

C: 0.32 ergs

D: 0.00032 ergs

Answers

For the given problem, the amount of work done expressed in ergs is 3200 ergs.

Answer: Option A

Explanation:

The work done on an objects are the force acting on it to move the object to a particular distance. So, work done on the object will be directly proportional to the force acting on it and the displacement.

Here, the force acting on the object is given as 0.010 N and the displacement of the object is 0.032 m. So, the work done on the object is

$\text { Work done }=\text { Force } * \text { displacement }$

$\text { Work done }=0.010 \mathrm{N} * 0.032 \mathrm{m}=0.00032 \mathrm{Nm}$

It is known that $1 N=1 \mathrm{kg} \mathrm{ms}^(-2)$

So, the work done can be expressed in $k g m s^(-2)$ as,

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2)$

It is known that $1 \mathrm{erg}=1 \mathrm{g} \mathrm{cm}^(2) / \mathrm{s}^(2)$ , so the conversion of units from Nm to erg will be done as follows:

$\text { Work done }=0.00032 \mathrm{kgm}^(2) \mathrm{s}^(-2) * \frac{1000 \mathrm{g}}{1 \mathrm{kg}} * \frac{100 * 100 \mathrm{cm}^(2)}{m^(2)}=3200 \mathrm{g} \mathrm{cm}^(2) \mathrm{s}^(-2)$

Thus, work done in ergs is 3200 ergs.

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