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Pepsi [2]
3 years ago
4

A battery has emf E and internal resistance r = 1.50 Ω. A 14.0 Ω resistor is connected to the battery, and the resistor consumes

electrical power at a rate of 92.0 J/s. What is the emf of the battery?
Physics
1 answer:
irinina [24]3 years ago
3 0

Answer:

E = 39.68 V

Explanation:

EMF ( Electromotive force) : This is defined as the magnitude of the potential difference of both the external circuit and the inside of the cell. The S.I unit is Volt.

The Expression for E.M.F is given as,

E = I(R+r) ................... Equation 1

Where E = EMF, I = current, R = External Resistance, r = internal resistance.

Also,

P = I²R

I = √(P/R) ..................... Equation 2

Where P = power, R = External resistance.

Given: P = 92 J/s, R = 14 Ω.

Substitute into equation 2

I = √(92/14)

I = √(6.57)

I = 2.56 A.

Also Given: r = 1.5 Ω.

Substitute into equation 1

E = 2.56(1.5+14)

E = 2.56(15.5)

E = 39.68 V.

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Two facing surfaces of two large parallel conducting plates separated by 8.5 cm have uniform surface charge densities such that
elena-s [515]

Answer:

positive plate

E = 5.764 KV / m

W = 490eV or 7.85 * 10^-17 J

E_p = 4.74 *10^(-12) eV

E_k = 490 eV

Explanation:

part a

The potential difference between two plates = 490 V

Distance between two plates = 8.5 cm

Answer: The positive plate is at higher potential because of convention.

part b

Electric Field between the plates

E = V / d

E = 490 / 0.085 = 5.764 KV / m

Answer: Electric Field between the plates E = 5.764 KV / m

part c

Work done by electric field

W = V*q

W = 490 * 1.602*10^-19

W = 7.85 * 10^-17 J

or W = 490 eV

Answer: Work done by electric field W = 490eV or 7.85 * 10^-17 J

part d

Potential Energy of an electron gained:

E_p = m_e * g * d / (1.602*10^-19)

E_p =  9.109*10^-31* 9.81 * 0.085 / (1.602*10^-19)

E_p = 4.74 *10^(-12) eV

Very very small E_p approximately 0

Answer: Potential Energy of an electron gained E_p = 4.74 *10^(-12) eV or 0.

part e

Kinetic Energy of an electron gained:

W - E_p = E_k

E_k = 490eV - 4.74*10^(-12)eV

E_k = 490 eV

Answer: Kinetic Energy of an electron gained E_k = 490 eV

7 0
4 years ago
Mr. White claims that he invented a heat engine with a maximum efficiency of 90%. He measured the temperature of the hot reservo
Studentka2010 [4]

Answer:

The error he made was that he didn't convert the unit of temperature to Kelvin.

The correct efficiency is 24%

Explanation:

Parameters given:

Temperature of hot reservoir = 100°C = 373 K

Temperature of cold reservoir = 10°C = 273 K

The efficiency of a heat engine is given as:

E = 1 - (Qc/Qh) = 1 - (Tc/Th)

Where

Qc = Output heat;

Qh = Input heat;

Tc = Temperature of the cold reservoir;

Th = Temperature of the hot reservoir.

=> E = 1 - (283/373)

E = 1 - 0.76

E = 0.24

In percentage,

E = 0.24 * 100 = 24%

Hence, the efficiency of the engine is actually 24%.

The error he made was that he didn't convert the temperature to Kelvin. If we leave the temperatures in °C, we have that:

E = 1 - (10/100)

E = 1 - 0.1 = 0.9

In percentage,

E = 0.9 * 100 = 90%

7 0
3 years ago
According to newtons first law,if no net force acts on a object,the object continues in motion with?
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The object keps going cause of wind friction
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You serve a volleyball with a mass of 2.1 kg. The ball leaves your hand with a speed of 30 m/s. The ball has what energy, kineti
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Both kinetic and potential. Kinetic as it is moving and Potential due to its relative position to the ground, in this case it is in the air, elevated from the ground.
4 0
3 years ago
A vertical wall (8.7 m x 3.2 m) in a house faces due east. A uniform electric field has a magnitude of 210 N/C. This field is pa
Dominik [7]

Answer:

\phi=4344.72Nm^2/c

Explanation:

From the question we are told that:

Dimension of Wall:

 (L*B)=(8.7 m * 3.2 m)

Electric field B=210 N/C

Angle \theta =42 \textdegree North

Generally the equation for electric Flux is mathematically given by

 \phi=EAcos\theta

 \phi=210*(8.7*3.2)*cos 42

 \phi=4344.72Nm^2/c

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