A potential difference of 10.0 volts exists between two points, A and B, within an electric field. What is the

Two 5.0 mm × 5.0 mm electrodes are held 0.10 mm apart and are attached to 7.5 V battery. Without disconnecting the battery, a 0.

Musya8 [376]

Answer:

A) V = 7.5 V

B) E = 75,000 V/m

C) Q = 16.6 pC

D) V = 7.5 V

E) E = 24,000 V/m

F) Q = 52 pC

Explanation:

Given:

- The Area of plate A = ( 5 x 5 ) mm^2

- The distance between plates d = 0.10 mm

- The thickness of Mylar added t = 0.10 mm

- Voltage supplied by battery V = 7.5 V

Solution:

A) What is the capacitor's potential difference before the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

B) What is the capacitor's electric field before the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 1 (air) E = 7.5 / 0.10*10^-3

E = 75,000 V/m

C) What is the capacitor's charge Q before the Mylar is inserted?

C = k*A*ε / d

k = 1 (air) C = ( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 2.213 pF

Q = C*V

Q = 7.5*(2.213)

Q = 16.6 pC

D) What is the capacitor's potential difference after the Mylar is inserted?

- The potential difference across the two plates is equal to the voltage provided by the battery V = 7.5 V which remains constant throughout.

E) What is the capacitor's electric field after the Mylar is inserted?

- The Electric Field E between the capacitor plates is given by:

E = V / k*d

k = 3.13 E = 7.5 / (3.13)0.10*10^-3

E = 24,000 V/m

F) What is the capacitor's charge after the Mylar is inserted?

C = k*A*ε / d

k = 3.13 C = 3.13*( 0.005^2 * 8.85*10^-12 ) / 0.0001

C = 6.927 pF

Q = C*V

Q = 7.5*(6.927)

Q = 52 pC

6 0

2 years ago

At 1 atm pressure, the heat of sublimation of gallium is 277 kj/mol and the heat of vaporization is 271 kj/mol. to the correct n

Andreyy89

Below are the choices that can be found elsewhere:

a. 268 kJ
b. 271 kJ 
c. 9 kJ 
d. 6 kJ

So the key thing to realize here is what the information given to you actually means. Sublimation is going from a sold to a gas. Vaporization is going from a liquid to a gas. Hence you can create two equations from the information that you have:

Ga (s) --> Ga (g) delta H = 277 kJ/mol 

Ga (l) --> Ga (g) delta H = 271 kJ/mol 

From these two equations, you can then infer how to get the melting equation be simply finding the difference between the sublimation (two steps) and vaporization (one step). 

Ga (s) --> Ga (l) delta H = 6 kJ/mol 

At this point, all you need to do is a bit of stoichiometry. You start with 1.50 mol and multiply by the amount of energy per mole (6 kJ/mol). 

*ANSWER* 
9 kJ/mol (C)

7 0

2 years ago

You are given 6 identical resistors, and allowed to combine them in any possible configuration to produce different equivalent r

tatyana61 [14]

Answer:your gay

Explanation:

gay

7 0

2 years ago

Springfield's "classic rock" radio station broadcasts at a frequency of 102.1 mhz. what is the length of the radio wave in meter

Mila [183]

The frequency of the radio wave is:
$f=102.1 MHz = 102.1 \cdot 10^6 Hz$

The wavelength of an electromagnetic wave is related to its frequency by the relationship
$\lambda= \frac{c}{f}$
where c is the speed of light and f the frequency. Plugging numbers into the equation, we find
$\lambda= \frac{3 \cdot 10^8 m/s}{102.1 \cdot 10^6 Hz}= 2.94 m$
and this is the wavelength of the radio waves in the problem.

7 0

2 years ago

: Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at

ser-zykov [4K]

Complete Question

Two containers have a substantial amount of the air evacuated out of them so that the pressure inside is half the pressure at sea level. One container is in Denver at an altitude of about 6,000 ft and the other is in New Orleans (at sea level). The surface area of the container lid is A=0.0155 m. The air pressure in Denver is PD = 79000 Pa. and in New Orleans is PNo = 100250 Pa. Assume the lid is weightless.

Part (a) Write an expression for the force FNo required to remove the container lid in New Orleans.

Part (b) Calculate the force FNo required to lift off the container lid in New Orleans, in newtons.

Part (c) Calculate the force Fp required to lift off the container lid in Denver, in newtons.

Part (d) is more force required to lift the lid in Denver (higher altitude, lower pressure) or New Orleans (lower altitude, higher pressure)?

Answer:

a

The expression is $F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

b

$F_{No}= 7771.125 \ N$

c

$F_p = 2.2*10^{6} N$

d

From the value obtained we can say the that the force required to open the lid is higher at Denver

Explanation:

The altitude of container in Denver is $d_D = 6000 \ ft = 6000 * 0.3048 = 1828.8m$

The surface area of the container lid is $A = 0.0155m^2$

The altitude of container in New Orleans is sea-level

The air pressure in Denver is $P_D = 79000 \ Pa$

The air pressure in new Orleans is $P_{ro} = 100250 \ Pa$

Generally force is mathematically represented as

$F_{No} = \Delta P A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

So the $\Delta P$ which is the difference in pressure within and outside the container is

$\Delta P = P_{No} - \frac{P_{sea}}{2}$

Therefore

$F_{No} = A [P_{No} - \frac{P_{sea}}{2}]$

Now substituting values

$F_{No} = 0.0155 [100250 - \frac{101000}{2}]$

$F_{No}= 7771.125 \ N$

The force to remove the lid in Denver is

$F_p = \Delta P_d A$

So we are told the pressure inside is is half the pressure the at sea level so the the pressure acting on the container would

The pressure at sea level is a constant with a value of

$P_{sea} = 101000 Pa$

At sea level the air pressure in Denver is mathematically represented as

$P_D = \rho g h$

=> $g = \frac{P_D}{\rho h}$

Let height at sea level is h = 1

The air pressure at height $d_D$

$P_d__{D}} = \rho gd_D$

=> $g = \frac{P_d_D}{\rho d_D}$

Equating the both

$\frac{P_D}{\rho h} = \frac{P_d_D}{\rho d_D}$

$P_d_D = P_D * d_D$

Substituting value

$P_d__{D}} = 1828.2 * 79000$

$P_d__{D}} = 1.445*10^{8} Pa$

So

$\Delta P_d = P_{d} _D - \frac{P_{sea}}{2}$

=> $\Delta P_d = 1.445 *10^{8} - \frac{101000}{2}$

$\Delta P_d = 1.44*10^{8}Pa$

So

$F_p = \Delta P_d A$

$= 1.44*10^8 * 0.0155$

$F_p = 2.2*10^{6} N$

6 0

3 years ago