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3 years ago

Visual perception can be improved through perceptual skills development. False True

Engineering

1 answer:

bogdanovich [222]3 years ago

4 0

Answer:

It's true

Explanation:

I took the quiz a few days ago and got it right!

Hope this helps:)

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A household refrigerator that has a power input of 450 W and a COP of 1.5 is to cool five large watermelons, 10 kg each, to 8°C.

Andreyy89

Answer:

6222.22 sec

Explanation:

Given data the power input to the refrigerator is 450 W

The COP of refrigerator is 1.5

Temperature $T_1=8^{\circ}C$

$T_2=28^{\circ}C$

mass of watermelon =10 kg

specific heat =4.2 KJ/kg°C

The amount of heat removed from 5 watermelon

$Q=mc_pdt=5\times 10\times 4.2\times (28-8)=4200 KJ$

We know that $COP=\frac{Q_1}{W}$

$1.5=\frac{Q_1}{450}$

$Q_1=675$ W=0.675 KW

so time required to cool the watermelon is

$t=\frac{Q_1}{Q_2}=\frac{4200}{0.675}=6222.22 sec$

4 0

3 years ago

7. An energy auditor is part of what career field?

Rashid [163]

An energy auditor is part of power operations

6 0

2 years ago

Help please all of the numbers b4 the equal sign are wrong

Nostrana [21]

Answer:

3/5" = 12'
1 3/4" = 35'
1 1/4" = 25'
9/10" = 18'
2 13/20" = 53'

Explanation:

One number is wrong; they all lack units.

The basic ratio is 1" = 20', so you can divide feet by 20 to find inches.

3/5" = 12'
1 3/4" = 35'
1 1/4" = 25'
9/10" = 18'
2 13/20" = 53'

Perhaps you want decimal inches:

0.60" = 12'
1.75" = 35'
1.25" = 25'
0.90" = 18'
2.65" = 53'

7 0

3 years ago

If p = .8 and n = 50, then we can conclude that the sampling distribution of pˆ p ^ is approximately a normal distribution.

defon

Answer:

It is true that the sampling distribution of the proportions is approximately a normal distribution.

Explanation:

Solution

Given that:

N = 50

P= 0.8

Thus

p ^ =√p(1-p)/n

p ^=√0.8 (1-0.8)/50

p ^= 0.0566

Therefore,we conclude that the sampling distribution of the proportions is approximately a normal distribution.

4 0

3 years ago

2.5 kg of air at 150 kPa and 12°C is contained in a gas-tight, frictionless piston-cylinder device. The air is now compressed to

Korolek [52]

Answer:

Work input =283.47 KJ

Explanation:

Given that

$P_1=150\ KPa$

$P_2=600\ KPa$

T=12°C=285 K

m= 2.5 kg

Given that this is the constant temperature process.

e know that work for isothermal process

$W=P_1V_1\ln \dfrac{P_1}{P_2}$

$W=mRT\ln \dfrac{P_1}{P_2}$

So now putting the values

$W=mRT\ln \dfrac{P_1}{P_2}$

$W=2.5\times 0.287\times 285\ln \dfrac{150}{600}$

W=-283.47 KJ

Negative sign indicates that work is done on the system.

So work input =283.47 KJ

8 0

3 years ago