Help me with # 3 & #4 please

20. A car battery with a 12-V emf and an internal resistance of 0.050 Ω is being charged with a current of 60 A. Note that in th

topjm [15]

Answer:

Part a)

V = 15 Volts

Part b)

P = 180 Watt

Part c)

Rate = 720 Watt

Explanation:

Part a)

When battery is in charging then the potential difference of the terminals of cell is given by

$\Delta V = EMF + iR$

here we know that

EMF = 12 volts

i = 60 A

r = 0.050 ohm

now we have

$\Delta V = 12 + (0.050)(60) = 15 Volts$

Part b)

Rate of thermal energy dissipated is the energy which is dissipating across the resistor

so here we have

$P = i^2 R$

$P = (60^2)(0.05)$

P = 180 Watt

Part c)

Rate at which Energy stored inside the cell is the rate of electrical energy that is converted into the chemical energy

$Rate = EMF \times i$

$Rate = (12)(60)$

$Rate = 720 Watt$

6 0

3 years ago

While refrigerant 410a is a near azeotropic refrigerant, it is still best when charging to remove the r-410a as a _______ from t

andreev551 [17]

While refrigerant 410a is a near azeotropic refrigerant, it is still best when charging to remove the r-410a as a liquid from the storage cylinder.
Azeotrope means a constant boiling mixture. it is a mixture of two or more liquids,by simple distillation whose proportions cannot be changed. A mixture behaving purely is azeotropic and the mixture which behave differently is called non-azeotropic.

7 0

4 years ago

How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Lapatulllka [165]

This question is incomplete, the complete question is;

A monatomic gas fills the left end of the cylinder in the following figure. At 300 K , the gas cylinder length is 14.0 cm and the spring is compressed by65.0 cm . How much heat energy must be added to the gas to expand the cylinder length to 16.0 cm ?

Answer:

the required heat energy is 16 J

Explanation:

Given the data in the question;

Lets consider the ideal gas equation;

PV = nRT

from the image, we calculate initial pressure;

Pi = ( 2000N/M × 0.06m) / 0.0008 m²

Pi = 15 × 10⁴ Pa

next we find Initial velocity

Vi = (0.0008 m²)(0.14) = 1.1 × 10⁻⁴ m²

now we find the number of moles

n = [(15 × 10⁴ Pa)(1.1 × 10⁻⁴ m²)] / 8.31 J/molK × 300K

N = 6.6 × 10⁻³ mol

next we calculate the final temperature;

Pf = ( 2000N/m × 0.08) / 0.0008 m²

Pf = 2 × 10⁵ Pa

Calculate the final Volume

Vf = (0.0008 m² × 0.16 m = 1.28 × 10⁻⁴ m³

we also determine the final temperature

$T_{f}$ = (2 × 10⁵ Pa × 1.28 × 10⁻⁴ m³) / 6.6 × 10⁻³ × 8.31 J/molK

$T_{f}$ = 466.8 K

so change in temperature ΔT

ΔT = 466.8 K - 300K = 166.8 K

we then calculate the change in thermal energy

ΔU = nCΔT

ΔU = ( 6.6 × 10⁻³ mol ) × 12.5 × 166.8K

ΔU = 13.761 J

C is the isochoric molar specific heat which is equal to 3R/2 for monoatomic

now we calculate the work done;

W = 1/2 × K( x $_{i\\}$ ² - x $_{f\\}$ ² )

W = 1/2 × ( 2000 N/m) ( 0.06² - 0.08² )

= - 2.8 J

and we then calculate the heat energy using the following expression;

Q = ΔU - W

we substitute

Q = 13.761 - (- 2.8 J)

Q = 13.761 + 2.8 J)

Q = 16 J

Therefore, the required heat energy is 16 J

5 0

3 years ago

A magnetic dipole <img src="https://tex.z-dn.net/?f=%5Cvec%7Bm%7D" id="TexFormula1" title="\vec{m}" alt="\vec{m}" align="absmidd

ser-zykov [4K]

We are asked to know the reason why magnetic dipole can be written as U = -m *B. The reason for this was that the negative sign is just showing or indicating the direction of the force. If it if negative, it means that the work done is opposite to the force field.

7 0

3 years ago

A skier is pulled by a towrope up a frictionless ski slope that makes an angle of 12 degrees with the horizontal. The rope moves

MArishka [77]

Answer:

Explanation:

Given,

Work done by the rope 900 m/s.
Angle of inclination of the slope = $\theta\ =\ 12^o$
Initial speed of the skier = v = 1.0 m/s
Length of the inclined surface = d = 8.0 m

part (a)

The rope is doing the work against the gravity on the skier to uplift up to the inclined surface. Therefore the work done by the rope is equal to the work done on the skier due to the gravity

$\therefore W_r\ =\ W_g\ =\ 900\ J$

In both cases the height attained by the skier is equal. and the work done by gravity does not depend upon the speed of the skier.

part (b)

Initial speed of the skier = v = 1.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 1.0}{8.0}\\\Rightarrow P\ =\ 112.5\ Watt$

Part (c)

Initial speed of the skier = v = 2.0 m/s.

Rate of the work done by the rope is power of the rope.

$Power\ =\ \dfrac{\Delta W}{\Delta t}\\\Rightarrow P\ =\ \dfrac{\Delta W}{\dfrac{d}{v}}\\\Rightarrow P\ =\ \dfrac{\Delta W\times v}{d}\\\Rightarrow P\ =\ \dfrac{900\times 2.0}{8.0}\\\Rightarrow P\ =\ 225\ Watt$

4 0

3 years ago