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Vlad [161]
3 years ago
5

The electric field everywhere on the surface of a thin, spherical shell of radius 0.770 m is of magnitude 860 N/C and points rad

ially toward the center of the sphere.
(a) What is the net charge within the sphere's surface?
(b) What can you conclude about the nature and distribution of the charge inside the spherical shell?
Physics
1 answer:
11111nata11111 [884]3 years ago
5 0

Answer:

(a) Q = 7.28\times 10^{14}

(b) The charge inside the shell is placed at the center of the sphere and negatively charged.

Explanation:

Gauss’ Law can be used to determine the system.

\int{\vec{E}} \, d\vec{a} = \frac{Q_{enc}}{\epsilon_0}\\E4\pi r^2 = \frac{Q_{enc}}{\epsilon_0}\\(860)4\pi(0.77)^2 = \frac{Q_{enc}}{8.8\times 10^{-12}}\\Q_enc = 7.28\times 10^{14}

This is the net charge inside the sphere which causes the Electric field at the surface of the shell. Since the E-field is constant over the shell, then this charge is at the center and negatively charged because the E-field is radially inward.

The negative charge at the center attracts the same amount of positive charge at the surface of the shell.

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Airida [17]

Answer:

  • The work made by the gas is 7475.69 joules
  • The heat absorbed is 7475.69 joules

Explanation:

<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

dW =  P \ dv

We can solve this by integration:

\Delta W = \int\limits_{s_1}^{s_2}\,dW = \int\limits_{v_1}^{v_2} P \ dv

but, first, we need to find the dependence of Pressure with Volume. For this, we can use the ideal gas law

P \ V = \ n \ R \ T

P = \frac{\ n \ R \ T}{V}

This give us

\int\limits_{v_1}^{v_2} P \ dv = \int\limits_{v_1}^{v_2} \frac{\ n \ R \ T}{V} \ dv

As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

\Delta W= \ n \ R \ T  \left [ ln (V) \right ]^{v_2}_{v_1}

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

Now, lets use the value from the problem.

The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

So:

\Delta W = \ 1 mol \ 8.314 \frac{m^3 \ Pa}{K \ mol} \ 300.15 \ K  ln (\frac{20 atm}{1 atm})

\Delta W = 7475.69 joules

<h3>Heat</h3>

We know that, for an ideal gas, the energy is:

E= c_v n R T

where c_v its the internal energy of the gas. As the temperature its constant, we know that the gas must have the energy is constant.

By the first law of thermodynamics, we know

\Delta E = \Delta Q - \Delta W

where \Delta W is the Work made by the gas (please, be careful with this sign convention, its not always the same.)

So:

\Delta E = 0

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7 0
3 years ago
A sealed test tube traps 25.0 cm3 of air at a pressure of 1.00 atm and temperature of 18°C. The test tube’s stopper has a diamet
puteri [66]

Answer:

180° C

Explanation:

First we start by finding the area of the stopper.

A = πd²/4, where d = 1.5 cm = 0.015 m

A = 3.142 * 0.015² * ¼

A = 1.767*10^-4 m²

Next we find the force on the stopper

F = (P - P•)A, where

F = 10 N

P = pressure inside the tube,

P• = 1 atm

10 = (P - 101325) * 1.767*10^-4

P - 101325 = 10/1.767*10^-4

P - 101325 = 56593

P = 56593 + 101325

P = 157918 Pascal

Now, remember, in an ideal gas,

P1V1/T1 = P2V2/T2, where V is constant, then we have

P1/T1 = P2/T2, and when we substitute the values, we have

101325/(273 + 18) = 157918/ T2

101325/291 = 157918/ T2

T2 = (157918 * 291)/101325

T2 = 453 K

T2 = 453 - 273 = 180° C

3 0
3 years ago
Which jovian planet should have the most extreme seasonal changes? a. Saturn b. Neptune c. Jupiter d. Uranus
Maksim231197 [3]

Answer:

D). Uranus.

Explanation:

Jovian planets are described as the planets which are giant balls of gases and located farthest from the sun which primarily include Jupiter, Saturn, Uranus, and Neptune.

As per the question, 'Uranus' is the jovian planet that would have the most extreme seasonal changes as its tilted axis leads each season to last for about 1/4 part of its 84 years orbit. The strong tilted axis encourages extreme changes in the season on Uranus. Thus, <u>option D</u> is the correct answer.

3 0
3 years ago
If a 3.5 gram ping pong ball were traveling to the right horizontally at 12 m/s, and a larger 12 g super ball were thrown direct
algol [13]

Answer:

v = 14.32 m/s

Explanation:

According to the principle of conservation of linear momentum, both the momentum and kinetic energy of the system are conserved. Since the two balls are in the same direction of motion before collision, then;

m_{1} u_{1} + m_{2} u_{2} = (m_{1} + m_{2}) v

0.035 × 12 + 0.120 × 15 = (0.035 + 0.120) v

0.420 + 1.800 = (0.155) v

2.22 = 0.155 v

⇒ v = \frac{2.22}{0.155}

      = 14.323

The velocity of the balls after collision is 14.32 m/s.

3 0
3 years ago
A mass of 100 kg is pulled by a 392 N force in the +X direction along a rough surface (uk=0.4) with uniform velocity v=20 m/s. W
alexdok [17]

Answer:

The total work done will be zero.

Explanation:

Given that,

Mass = 100 kg

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Velocity = 20 m/s

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F'=F-\mu mg

F'=392-0.4\times100\times9.8

F'=0

The net force is zero.

Hence, The total work done will be zero by all forces on the object.

5 0
3 years ago
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