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Fofino [41]
3 years ago
13

A refrigerator operates on average for 10.0 hours an day. If the power rating is the refrigerator is 709 w how much electrical e

nergy does the fridge use in one day
Engineering
1 answer:
Varvara68 [4.7K]3 years ago
6 0

Answer:

61257.6 kW per day

Explanation:

The rating of the refrigerator is given as : 709 w

This means the power consumption is 709 joules per second

The operating time is given as 10 hours.

Change hours to second as

1 hour = 3600 seconds

10 hours = 36000 seconds

Apply the rating as;

709 w = 1 s

?         = 36000 seconds

perform cross product to get;

709 * 36000= 25524000 w

25524000 = 25524kW

For a day, 24 hours will be;

25524 kW = 10 hours

?                 =24 hours

={24*25524}/10 = 61257.6 kW per day

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What is the difference between a Datum and a Datum Feature? a) A Datum and Datum Feature are synonymous. b) A Datum is theoretic
aliina [53]

Difference between Datum and Datum feature is<em> 'Datum is theoretical and Datum feature is real'. </em>

Option: (b)

<u>Explanation:</u>

A Datum is a perfect plane, line, point or surface but only occurs theoretically.

However a Datum Feature is fully based on a tangible surface, axis or point on a part where that theoretical datum is located.

The reason behind in this is they are not equal to each other because the 'part surface' is never 100% perfect.

The important functional features of the Datum is controlled during measurements.

7 0
3 years ago
Acoke can with inner diameter(di) of 75 mm, and wall thickness (t) of 0.1 mm, has internal pressure (pi) of 150 KPa and is suffe
NemiM [27]

Answer:

All 3 principal stress

1. 56.301mpa

2. 28.07mpa

3. 0mpa

Maximum shear stress = 14.116mpa

Explanation:

di = 75 = 0.075

wall thickness = 0.1 = 0.0001

internal pressure pi = 150 kpa = 150 x 10³

torque t = 100 Nm

finding all values

∂1 = 150x10³x0.075/2x0,0001

= 0.5625 = 56.25mpa

∂2 = 150x10³x75/4x0.1

= 28.12mpa

T = 16x100/(πx75x10³)²

∂1,2 = 1/2[(56.25+28.12) ± √(56.25-28.12)² + 4(1.207)²]

= 1/2[84.37±√791.2969+5.827396]

= 1/2[84.37±28.33]

∂1 = 1/2[84.37+28.33]

= 56.301mpa

∂2 = 1/2[84.37-28.33]

= 28.07mpa

This is a 2 d diagram donut is analyzed in 2 direction.

So ∂3 = 0mpa

∂max = 56.301-28.07/2

= 14.116mpa

6 0
3 years ago
A fluid of specific gravity 0.96 flows steadily in a long, vertical 0.71-in.-diameter pipe with an average velocity of 0.90 ft/s
KengaRu [80]

Answer:

0.00650 Ib s /ft^2

Explanation:

diameter ( D ) = 0.71 inches = 0.0591 ft

velocity = 0.90 ft/s ( V )

fluid specific gravity = 0.96 (62.4 )  ( x )

change in pressure ( P ) = 0 because pressure was constant

viscosity =  (change in p - X sin∅ ) D^{2} / 32 V

              = ( 0 - 0.96( 62.4) sin -90 ) * 0.0591 ^2  / 32 * 0.90

              = - 59.904 sin (-90) * 0.0035 / 28.8

              = 0.1874 / 28.8

  viscosity = 0.00650 Ib s /ft^2

8 0
3 years ago
Read 2 more answers
When the Moon is in the position shown, how would the Moon look to an observer on the North Pole?
kirill115 [55]

Answer:

cant see the moon sorry dude

5 0
3 years ago
A large tank is filled to capacity with 500 gallons of pure water. Brine containing 2 pounds of salt per gallon is pumped into t
Nataly [62]

Answer:

A) A(t) = 10(100 - t) + c(100 - t)²

B) Tank will be empty after 100 minutes.

Explanation:

A) The differential equation of this problem is;

dA/dt = R_in - R_out

Where;

R_in is the rate at which salt enters

R_out is the rate at which salt exits

R_in = (concentration of salt in inflow) × (input rate of brine)

We are given;

Concentration of salt in inflow = 2 lb/gal

Input rate of brine = 5 gal/min

Thus;

R_in = 2 × 5 = 10 lb/min

Due to the fact that the solution is pumped out at a faster rate, thus it is reducing at the rate of (5 - 10)gal/min = -5 gal/min

So, after t minutes, there will be (500 - 5t) gallons in the tank

Therefore;

R_out = (concentration of salt in outflow) × (output rate of brine)

R_out = [A(t)/(500 - 5t)]lb/gal × 10 gal/min

R_out = 10A(t)/(500 - 5t) lb/min

So, we substitute the values of R_in and R_out into the Differential equation to get;

dA/dt = 10 - 10A(t)/(500 - 5t)

This simplifies to;

dA/dt = 10 - 2A(t)/(100 - t)

Rearranging, we have;

dA/dt + 2A(t)/(100 - t) = 10

This is a linear differential equation in standard form.

Thus, the integrating factor is;

e^(∫2/(100 - t)) = e^(In(100 - t)^(-2)) = 1/(100 - t)²

Now, let's multiply the differential equation by the integrating factor 1/(100 - t)².

We have;

So, we ;

(1/(100 - t)²)(dA/dt) + 2A(t)/(100 - t)³ = 10/(100 - t)²

Integrating this, we now have;

A(t)/(100 - t)² = ∫10/(100 - t)²

This gives;

A(t)/(100 - t)² = (10/(100 - t)) + c

Multiplying through by (100 - t)²,we have;

A(t) = 10(100 - t) + c(100 - t)²

B) At initial condition, A(0) = 0.

So,0 = 10(100 - 0) + c(100 - 0)²

1000 + 10000c = 0

10000c = -1000

c = -1000/10000

c = -0.1

Thus;

A(t) = 10(100 - t) + -0.1(100 - t)²

A(t) = 1000 - 10t - 0.1(10000 - 200t + t²)

A(t) = 1000 - 10t - 1000 + 20t - 0.1t²

A(t) = 10t - 0.1t²

Tank will be empty when A(t) = 0

So, 0 = 10t - 0.1t²

0.1t² = 10t

Divide both sides by 0.1t to give;

t = 10/0.1

t = 100 minutes

6 0
3 years ago
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