1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
lyudmila [28]
3 years ago
6

To make ice, a freezer that is a reverse Carnot engine extracts 37 kJ as heat at -17°C during each cycle, with coefficient of pe

rformance 5.8. The room temperature is 26.9°C. How much (a) energy per cycle is delivered as heat to the room and (b) work per cycle is required to run the freezer?
Physics
1 answer:
Dvinal [7]3 years ago
5 0

Answer:

a) 6.4 kJ

b) 43.4 kJ

Explanation:

a)

Q_{a} = Heat absorbed = 37 kJ

β  = Coefficient of performance = 5.8

W = Work done

Heat absorbed is given as

Q_{a} = β W

37 = (5.8) W

W = 6.4 kJ

b)

Q  = work per cycle required

Q  = Q_{a} + W

Q  = 37 + 6.4

Q  = 43.4 kJ

You might be interested in
PLEASE HELP
siniylev [52]
A person standing on the moon watching the earth rotate
3 0
3 years ago
Read 2 more answers
Which is not a family of the periodic table?
Vsevolod [243]

Answer:

anions

Explanation:

The other are families of the periodic table.

An anion is a negatively charged ion.

6 0
3 years ago
Read 2 more answers
4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o
goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a  capacitor chosen to provide a peak-to-peak ripple voltage of  (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through  a 10-to-1 step-down transformer. It uses a silicon diode  that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

 (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0
3 years ago
People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p
Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

g' = \omega^2 (\dfrac{r}{2})

g' =\dfrac{1}{2}(\omega^2 r)

g'=\dfrac{g}{2}

People at the halfway experience g/2

hence, the correct answer is option B

7 0
3 years ago
An electron passes location < 0.02, 0.04, -0.06 > m and 5 us later is detected at location < 0.02, 1.62,-0.79 > m (1
hram777 [196]

Answer:

(a)

Average velocity = < 0, 316000, 146000> m/s

(b) < 0, 2.844, 1.314 > m

Explanation:

r1 = < 0.02, 0.04, - 0.06 > m

r2 = < 0.02, 1.62, - 0.79 > m

time, t = 5 micro second = 5 x 10^-6 s

(a) Average velocity is defined as the ratio of total displacement to the total time taken.

Displacement = r = r2 - r1

r = < 0.02 - 0.02, 1.62 - 0.04, - 0.79 + 0.06 > m

r = < 0, 1.58, - 0.73 > m

So. Average velocity =  \frac{< 0, 1.58, - 0.73 >}{5 \times 10^{-6}}

Average velocity = < 0, 316000, 146000> m/s

Average velocity = < 0, 316000, 146000> m/s

(b)

Distance = velocity x time

Here time, t = 9 micro second

d =  < 0, 316000, 146000>  x 9 x 10^-6 m

d = < 0, 2.844, 1.314 > m

5 0
4 years ago
Other questions:
  • Which type of transformer is found outside a residential house
    5·1 answer
  • For each of the motions described below, determine the algebraic sign (+, -, or 0) of the velocity and acceleration of the objec
    14·1 answer
  • A boy flies a kite with the string at a 30 degree angle to the horizontal. The tension in the string is 4.5N .
    7·1 answer
  • When Isabella pushes a box of mass 5.25 kilograms with a force of 15.7 newtons, it accelerates by a rate of 2.50 meters/second2.
    12·2 answers
  • viewed from earth two stars form an angle of 76.04 degrees. StarA is 23.30 light years from earth star Bis 34.76 light years fro
    10·1 answer
  • Four charges are at the corners of a square centered at the origin as follows q at a a 2q at a a 3q at a a and 6q at a a A fifth
    13·1 answer
  • A circular loop (radius of 20 cm) is in a uniform magnetic field of 0.15 T. What angle(s) between the normal to the plane of the
    8·1 answer
  • Rahul runs a distance of 12
    14·1 answer
  • What distance is required for a train
    9·1 answer
  • A 70.0Kg cyclist Rides a 15.0kg
    9·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!