A 12.5-µF capacitor is connected to a power supply that keeps a constant potential difference of 24.0 V across the plates. A piece of material having a dielectric constant of 3.75 is placed between the plates, completely filling the space between them.
(a) How much energy is stored in the capacitor before and after the dielectric is inserted?
(b) By how much did the energy change during the insertion? Did it increase or decrease?

Answer:

5

Explanation:

Answer:

The atomic weight of the metal is 58.7 g/mol

Explanation:

Given;

density of the metal sample, ρ = 8.91 g/cm³

edge length of the face centered cubic cell, α = 352.4 pm = 352.4 x 10⁻¹⁰ cm

Volume of the unit cell of the metal;

V = α³

V = (352.4 x 10⁻¹⁰  cm)³

V = 4.376 x 10⁻²³ cm³

Mass of the metal in unit cell

mass = density x volume

mass = 8.91 g/cm³ x 4.376 x 10⁻²³ cm³

mass = 3.899 x 10⁻²² g

Atomic weight, based on 4 atoms per unit cell;

4 atoms = 3.899 x 10⁻²² g

6.022 x 10²³ atoms = ?

= (6.022 x 10²³atoms x 3.899 x 10⁻²² g) / (4 atoms)

= 58.699 g/mol

= 58.7 g/mol (this metal is Nickel)

Theerefore, the atomic weight of the metal is 58.7 g/mol

Answer:

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Explanation:

Given that

s = 9 cos(t) + 9 sin(t), t ≥ 0

Then acceleration and velocity is

v(t) = s′(t) = −9sin(t)+9cos(t)

a(t) = v′(t) = −9cos(t) −9sin(t)

Given that,

Distance = r

Electric field :

Electric field is equal to the multiplication of electric constant and charge divided by square of distance.

In mathematically form,

[tex]E=\dfrac{kq}{r^2}[/tex]

[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{q}{r^2}[/tex]

Electric force :

Electric force is equal to the multiplication of electric constant and both charges divided by square of distance.

In mathematically form,

[tex]E=\dfrac{kqQ}{r^2}[/tex]

[tex]E=\dfrac{1}{4\pi\epsilon_{0}}\dfrac{qQ}{r^2}[/tex]

We know that,

The relation between electric field and electric force is

[tex]E=\dfrac{F}{q}[/tex]

According to question,

If electron is released from rest then electron move towards the source point due to attractive force.

Hence, The electron move towards the source point due to attractive force.

Answer:

The potential at the center of the sphere is -924 V

Explanation:

Given;

radius of the sphere, R = 0.22 m

electric field at the surface of the sphere, E = 4200 N/C

Since the electric field is directed towards the center of the sphere, the charge is negative.

The Potential is the same at every point in the sphere, and it is given as;

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{q}{R}[/tex] -------equation (1)

The electric field on the sphere is also given as;

[tex]E = \frac{1}{4 \pi \epsilon _o} \frac{|q|}{R^2}[/tex]

[tex]|q |= 4 \pi \epsilon _o} R^2E[/tex]

Substitute in the value of q in equation (1)

[tex]V = \frac{1}{4 \pi \epsilon_o} \frac{-(4 \pi \epsilon _o R^2E)}{R} \ \ \ \ q \ is \ negative\ because \ E \ is\ directed \ toward \ the \ center\\\\V = -RE\\\\V = -(0.22* 4200)\\\\V = -924 \ V[/tex]

Therefore, the potential at the center of the sphere is -924 V

Answer:

because gravitational potential enegry is directly proportional to the height so more the height more the gravitational potential enegry. therefore gravitational potential enegry is greatest at high point than lower points.

Answer:

Explanation:

1 )

angular velocity of merry go round = 2π n

= 2π  x 3 / 60

ω = .1 x π radian / s

This will be the rotational speed of the whole system including that of Cora and Cameron . It will not depend upon their relative position with respect to

the centre of the merry go round .

So rotational speed of Cora = Rotational speed of Cameron

option ( d ) is correct .

2 )

Tangential speed  v = ω R where R is the distance from the centre of merry go round .

Tangential speed of Cora = .1 x π x 1

= .314 m /s

Tangential speed of Cameron = .1 x π x 2

= .628 m /s .

So tangential speed of Cameron is twice that of Cora .

Option ( e ) is correct .

Answer:

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Explanation:

Answer:

Q = 23.49 C

Explanation:

We have,

Electric current drawn by the air conditioner is 16.2 A

Time, t = 1.45 s

It is required to find the electric charge passes through the circuit during this period. We know that electric current is defined as the electric charge flowing per unit time. So,

[tex]I=\dfrac{q}{t}\\\\q=It\\\\q=16.2\times 1.45\\\\q=23.49\ C[/tex]

So, the charge of 23.49 C is passing through the circuit during this period.

Answer:

208 V

Explanation:

resistor voltage = Vr = 124 V

capacitor voltage Vc = 167 V

the total voltage in the RC circuit is the resultant voltage of the resistor and the capacitor

total voltage i= [tex]\sqrt{Vr^{2} + Vc^{2} }[/tex]

==> [tex]\sqrt{124^{2} + 167^{2} } =[/tex] 208 V

Answer:

[tex]I = 0.083 kg m^2[/tex]

Explanation:

Mass of the bucket, m = 23 kg

Radius of the pulley, r = 0.050 m

The bucket is released from rest, u = 0 m/s

The time taken to fall, t = 2 s

Speed, v = 8.0 m/s

Moment of Inertia of the pulley, I = ?

Using the equation of motion:

v = u + at

8 = 0 + 2a

a = 8/2

a = 4 m/s²

The relationship between the linear and angular accelerations is given by the equation:

[tex]a = \alpha r[/tex]

Angular acceleration, [tex]\alpha = a/r[/tex]

[tex]\alpha = 4/0.050\\\alpha = 80 rad/s^2[/tex]

Since the bucket is falling, it can be modeled by the equation:

mg - T = ma

T = mg - ma = m(g-a)

T = 23(9.8 - 4)

The tension, T = 133.4 N

The equation for the pulley can be modeled by:

[tex]T* r = I * \alpha\\133.4 * 0.050 = I * 80\\6.67 = 80 I\\I = 6.67/80\\I = 0.083 kg m^2[/tex]

Answer:

(a) max. height = 3.641 m

(b) flight time = 1.723 s

(c) horizontal range = 31.235 m

(d) impact velocity = 20 m/s

Above values have been given to third decimal.  Adjust significant figures to suit accuracy required.

Explanation:

This problem requires the use of kinematics equations

v1^2-v0^2=2aS .............(1)

v1.t + at^2/2 = S ............(2)

where

v0=initial velocity

v1=final velocity

a=acceleration

S=distance travelled

SI units and degrees will be used throughout

Let

theta = angle of elevation = 25 degrees above horizontal

v=initial velocity at 25 degrees elevation in m/s

a = g = -9.81 = acceleration due to gravity (downwards)

(a) Maximum height

Consider vertical direction,

v0 = v sin(theta) = 8.452 m/s

To find maximum height, we find the distance travelled when vertical velocity = 0, i.e. v1=0,

solve for S in equation (1)

v1^2 - v0^2 = 2aS

S = (v1^2-v0^2)/2g = (0-8.452^2)/(2*(-9.81)) = 3.641 m/s

(b) total flight time

We solve for the time t when the vertical height of the object is AGAIN = 0.

Using equation (2) for vertical direction,

v0*t + at^2/2 = S    substitute values

8.452*t + (-9.81)t^2 = 3.641

Solve for t in the above quadratic equation to get t=0, or t=1.723 s.

So time for the flight = 1.723 s

(c) Horiontal range

We know the horizontal velocity is constant (neglect air resistance) at

vh = v*cos(theta) = 25*cos(25) = 18.126 m/s

Time of flight = 1.723 s

Horizontal range = 18.126 m/s * 1.723 s = 31.235 m

(d) Magnitude of object on hitting ground, Vfinal

By symmetry of the trajectory, Vfinal = v = 20, or

Vfinal = sqrt(v0^2+vh^2) = sqrt(8.452^2+18.126^2) = 20 m/s

Answer:

number of electrons = 2.18*10^18 e

Explanation:

In order to calculate the number of electrons that move trough the second wire, you take into account one of the Kirchoff's laws. All the current that goes inside the junction, has to go out the junction.

Then, if you assume that the current of the wire 1 and 3 go inside the junction, then, all this current have to go out trough the second junction:

[tex]i_1+i_3=i_2[/tex]                 (1)

i1 = 0.40 A

i2 = 0.75 A

you solve the equation i3 from the equation (1):

[tex]i_3=i_2-i_1=0.75A-0.40A=0.35A[/tex]

Next, you take into account that 1A = 1C/s = 6.24*10^18

Then, you have:

[tex]0.35A=0.35\frac{C}{s}=0.35*\frac{6.24*10^{18}e}{s}=2.18*10^{18}\frac{e}{s}[/tex]

The number of electrons that trough the wire 3 is 2.18*10^18 e/s

Answer:

E = - dV/dx

Explanation:

Las superficies equipòtenciales son superficie donde el potencial eléctrico es constante por lo cual nos podemos desplazaren ella sin realizar nigun trabajo.

El campo electrico es el campo que existen algún punto en el espacio creado por alguna ddistribucion de carga.

De los antes expuesto las dos magnitudes están relacionadas

         E = - dV/dx

por lo cual el potenical es el gradiente del potencial eléctrico.

Como el campo eléctrico sobre un superficie equipotenciales constante, podemos colocar una punta de prueba con un potencial dado y seguir la linea que de una diferencia de potencial constar, lo cual permite visualizar las forma de cada linea equipotencial

Answer:

Explanation:

In a conductor, electric current can flow freely, in an insulator it cannot.

Metals such as copper typify conductors, while most non-metallic solids are said to be good insulators, having extremely high resistance to the flow of charge through them.

Most atoms hold on to their electrons tightly and are insulators.

Answer:

[tex]\frac{50}{\pi }[/tex]Hz

Explanation:

In alternating current (AC) circuits, voltage (V) oscillates in a sine wave pattern and has a general equation as a function of time (t) as follows;

V(t) = V sin (ωt + Ф)            -----------------(i)

Where;

V = amplitude value of the voltage

ω = angular frequency = 2 π f        [f = cyclic frequency or simply, frequency]

Ф = phase difference between voltage and current.

Now,

From the question,

V(t) = 230 sin (100t)              ---------------(ii)

By comparing equations (i) and (ii) the following holds;

V = 230

ω = 100

Ф = 0

But;

ω = 2 π f = 100

2 π f = 100             [divide both sides by 2]

π f = 50

f = [tex]\frac{50}{\pi }[/tex]Hz

Therefore, the frequency of the voltage is [tex]\frac{50}{\pi }[/tex]Hz

Answer: Use this F=Ma.

Explanation: So your answer will be

F=1 Kg+9.8 ms-2

So the answer will be

F=9.8N

How'd I do this?

I just used Newton's second law of motion.

I'll also put the derivation just in case.

Applied force α (Not its alpha, proportionality symbol) change in momentum

Δp α p final- p initial

Δp α mv-mu (v=final velocity, u=initial velocity and p=v*m)

or then

F α m(v-u)/t

So, as we know v=final velocity & u= initial velocity and v-u/t =a.

So F α ma, we now remove the proportionality symbol so we'll add a proportionality constant to make the RHS & LHS equal.

So, F=kma (where k is the proportionality constant)

k is 1 so you can ignore it.

So, our equation becomes F=ma

Answer:

0.25 J

Explanation:

Given that

Force on the spring, F = 30 N

Natural length of the spring, l1 = 20 cm = 0.2 m

Final length of the spring, l2 = 35 cm = 0.35 m

Extension of the spring, x = 0.35 - 0.2 = 0.15 m

The other extension is 40 - 35 cm = 5 cm = 0.05 m

Work done = ?

Considering Hooke's Law of Elasticity.

We're told that the spring stretches through 15 cm, and thereafter asked to find the work done in stretching it through 5 cm

The question is solved in the attachment below

Answer:

141178.8

Explanation:

use : density x velocity²

1.2 x 343² = 141178.8 pa

Answer:

Explanation:

given:

radius (r) = 1.88 km

velocity (v) = 136.3 km/hr

required:

banking angle ∡ ?

first:

convert 1.88 km to m = 1.88km * 1000m / 1km

r = 1880 m

convert velocity v = 136.3 km/hr to m/s = 136.3 km/hr * (1000 m/ 3600s)

v = 37.86 m/s

now.. calculate the angle

Ф = inv tan (v² / r * g)            we know that gravity = 9.8 m/s²

Ф = inv tan (37.86² / (1880 * 9.8))

Ф = 4.4°

Answer:

-1748*10^N/C

Explanation:

See attached file

Answer:

same direction = 80 (n)

opposite direction = 20 (n) going one direction

Explanation:

same direction means they are added to each other

and opposite means acting on eachother

Answer:

The speed of the rod is 1.383 m/s

Explanation:

Given;

length of the conducting rod, L = 31 cm = 0.31 m

induced emf on the rod, emf = 0.75V

magnetic field around the rod, B = 1.75 T

Apply the following Faraday's equation for electromagnetic induction in a moving rod to determine the speed of the rod.

emef = BLv

where;

B is the magnetic field

L is length of the rod

v is the speed of the rod

v = emf / BL

v = (0.75) / (1.75 x 0.31)

v = 1.383 m/s

Therefore, the speed of the rod is 1.383 m/s

Answer:

12.5 rad/s²

Explanation:

Angular Acceleration: This can be defined as the ratio of linear acceleration and radius. The S.I unit is rad/s²

From the question,

a = αr................... Equation 1

Where a = linear acceleration, α = angular acceleration, r = radius.

But,

a = (v-u)/t.............. Equation 2

Where v = final velocity, u = initial velocity, t = time.

Substitute equation 2 into equation 1

(v-u)/t = αr

make α the subject of the equation

α = (v-u)/tr................. Equation 3

Given: v = 30 m/s, u = 0 m/s, t = 6 s, r = 0.4 m

Substitute into equation 3

α = (30-0)/(0.4×6)

α = 30/2.4

α = 12.5 rad/s²

Answer:

[tex]r_f=\frac{1}{6}r_i[/tex]

Explanation:

To find the new separation of the charges, you first take into account the formula for the electric force, when the force are separated a distance of ri.

You use the following expression:

[tex]F_i=k\frac{q_Aq_B}{r_i^2}[/tex]          (1)

k: Coulomb's constant

qA: charge of A particle

qB: charge of B particle

When the charges are separated to a new distance rf, the new force is 36Fi, if the charges have not changed, you have:

[tex]F_f=36F_i=k\frac{q_Aq_B}{r_f^2}[/tex]         (2)

To find the new separation you replace the expression for Fi of the equation (1) into the equation (2) and solve for rf in terms of ri:

[tex]36F_i=36k\frac{q_Aq_B}{r_i^2}=k\frac{q_Aq_B}{r_f^2}\\\\\frac{36}{r_i^2}=\frac{1}{r_f^2}\\\\r_f=\frac{1}{6}r_i[/tex]

The new separation of the charges is 1/6 times of the initial separation

Answer:

Radius r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Explanation:

Area of a circle;

A = πr^2

A = area

r = radius

Making r the subject of formula;

r = √(A/π) ........1

Given;

A = 1300 cm^2

Substituting into the equation 1;

r = √(1300/π)

r = 20.34214472564 cm

r = 20.34 cm

The radius that can produces such a disk is 20.34 cm

Answer:

a. 70 rad/s²

b. 5000 rev

Explanation:

As we know,

[tex]\omega = 20000\frac{rev}{min}\frac{2 \pi rad}{1 \ rev}\frac{1 \ min}{60 \ sec}[/tex]

then,

[tex]\omega=2100 \ rad/s[/tex]

a...

⇒  [tex]\bar{\alpha}=\frac{\omega-\omega_{0}}{\Delta t}[/tex]

On substituting the values, we get

⇒      [tex]=\frac{2100}{30}[/tex]

⇒      [tex]=70 \ rad/s^2[/tex]

b...

⇒  [tex]\theta=\theta_{0}=\omega_{0}t+\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\alpha t^2[/tex]

       [tex]=\frac{1}{2}\times 70\times (30)^2[/tex]

       [tex]=31500 \ rad[/tex]

       [tex]=31500 \ rad\frac{1 \ rev}{2\pi rad}[/tex]

       [tex]=5000 \ rev[/tex]

(a) The average angular acceleration will be 70 rad/s².

(b) 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

Angular acceleration is defined as the pace of change of angular velocity with reference to time. It is denoted by α. Its unit is rad/s².

The given data in the problem is;

n is the revolution of centrifugal rotor =  20000 rpm

t is the time interval= the 30s

(α) is the Angular acceleration=?

n₁ is the revolution when the acceleration is constant =?

(a) The average angular acceleration will be  70 rad/s².

The value of the angular velocity is given by

[tex]\rm \omega_f = \frac{2\pi N}{60} } \\\\ \rm \omega_f = \frac{2 \times 3.14 \times 20000}{60} \\\\ \rm \omega_f= 2100\ rad/sec.[/tex]

The formula for angular acceleration is guven by;

[tex]\rm \alpha =\frac{ \omega_f-\omega_i}{dt} \\\\ \rm \alpha =\frac{ 2100-0}{3}\\\\ \rm \alpha =70\ rad/sec^2[/tex]

Hence the average angular acceleration will be 70 rad/s².

(b 5000 revolutions have the centrifuge rotor turned during its acceleration period.

[tex]\rm \theta= \theta_0+\frac{1}{2} \alpha t^2 \\\\ \rm \theta= \frac{1}{2} \times 70 \times (30)^2 \\\\ \rm \theta=31500\ rad[/tex]

As we know that the angular velocity is given by

[tex]\rm \omega = \frac{\theta}{t} \\\\ \rm \omega = \frac{31500}{30} \\\\ \rm \omega = 1050 \ rad/sec[/tex]

The relation of angular velocity and revolution will be

[tex]\rm n= \frac{ \omega \times 60}{2\pi} \\\\ \rm n= \frac{ 2100 \times 60}{2\times 3.14 } \\\\ \rm n = 20063.69 \ rev[/tex]

Hence 20063.69 revolutions have the centrifuge rotor turned during its acceleration period.

To learn more about angular acceleration refer to the link ;

https://brainly.com/question/408236

Answer:

the acceleration during the collision is: - 5  [tex]\frac{m}{s^2}[/tex]

Explanation:

Using the formula:

[tex]a=\frac{\Delta\,v}{\Delta\,t}[/tex]

we get:

[tex]a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}[/tex]

Answer:

It is calculated as Force × perpendicular distance.

Explanation:

Torque is a rotational force and twisting force that can cause an object to rotate in it's axis. This cause angular rotation.

The torque due to gravity on a body about its centre of mass is zero because the centre of mass is the that point of the body at which the force acts by the gravity that is mg.

But if the pivot point of a balance is not at the centre of mass of the balance, it will be FORCE × PERPENDICULAR DISTANCE because at that point, there is no centre in which the force act on a body by gravity. The distance and force will be use to calculate.

Answer:

Time taken to stop = 6 sec

Explanation:

Given:

Rotation of wheel = 1.10 × 10² rev/min

Final velocity (v) = 0

Angular acceleration (a) = - 1.92 rad/s²

Find:

Time taken to stop

Computation:

Initial velocity (u) = (1.10 × 10² × 2π rad) / 60 sec

Initial velocity (u) = 11.52 rad / sec

We know that,

V = U +at

t = (v-u)a

t = (0 - 11.52) / (-1.92)

Time taken to stop = 6 sec