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raketka [301]
2 years ago
11

The best way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is?

Physics
2 answers:
tigry1 [53]2 years ago
8 0
Answer choices are:

1. To use the big cooking pot from the stove and place directly into the refrigerator 
2. Placing the food into a shallow pan and stirring every few minutes 
3. To use a 5-gallon bucket and stirring every few minutes 
<span>4. By stacking several pans on top of the other cooled pans
</span>
Best answer choice:

2. Placing the food into a shallow pan and stirring every few minutes.

There is another way to cool soft and thick foods (such as beans, sauce or chili) when using the refrigerator is by having them to be placed in one container in which they are in a water bath, to be heated off. Hot food can be placed directly in the refrigerator or it can be rapidly chilled in an ice or cold water bath before refrigerating but putting hot food in the fridge right away is the safest. <span>Large amounts of </span>food<span> should be divided </span>into<span> small portions and </span>put in<span> shallow containers for quicker cooling </span>in the refrigerator because b<span>acteria </span>can<span> grow rapidly on </span>food<span> if it is left out at room temperature for more than 2 hours.</span>
AnnZ [28]2 years ago
3 0
<span>The best way to cool soft and thick foods when using the refrigerator is by having them to be placed and poured on a pan or another way is by having them to be placed in one container in which they are in a water bath, to be heated of.</span>
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during a cold winter day, wind at 42 km/h is blowing parallel to a 6-m-high and 10-m-high wall of a house. If the air outside is
Arturiano [62]

Answer: 9.08KW and 16.21KW

Explanation:

The convection over a flat surface with a length of 10 m and a width of 6m.

The mean temperature is (5oC + 12oC)/2 = 8.5oC.

Then find the following properties of air at this temperature from Table A-15:

k = 0.02428 W/m(oC, v= 1.413x10-5 m2/s, and Pr = 0.7340.

find the Reynolds number. Re= VL/v

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This means that the flow becomes turbulent over the plate and we can use the Nusselt number equation for combined laminar and turbulent flow.

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We then use this Nusselt number to find the heat transfer coefficient and the heat transfer.

Check the screen shot for the calculation

Ans

9.08 kW

Then if the wind velocity were doubled, the Re number would be doubled and we would repeat the calculations above, starting with this revised Reynolds number..

Ans

16.21 kW

4 0
2 years ago
Consider a vacuum-filled parallel plate capacitor (no dielectric material between the plates). What is the ratio of conduction c
Lorico [155]

Answer: 1798

Explanation:

Given that there is no dielectric material between the plates,

the permittivity of free space = 9 × 10^-12 f/m

The frequency F = 10 MHz

The ratio of conduction current JC to the displacement current Jd is also known as loss tangent.

Please find the attached file for the solution

8 0
2 years ago
A 0.3 mm long invertebrate larva moves through 20oC water at 1.0 mm/s. You are creating an enlarged physical model of this larva
AleksandrR [38]

Answer:

Explanation:

For the problem, we should have same reynolds number

ρvd/mu = constant

1000×1×10⁻³×0.3×10⁻³/1.002×10⁻³ = 1400×0.5×d/600

d = 25.66 cm

5 0
2 years ago
Calculate the number of moles in each of the following masses: 0.039 g of palladium 0.0073 kg of tantalum
marysya [2.9K]

Answer:

<em>The number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

Explanation:

Number of mole = reacting mass/molar mass

n = R.m/m.m......................... Equation 1

Where n = number of moles, R.m = reacting mass, m.m = molar mass.

For palladium,

R.m = 0.039 g and m.m = 106.42 g/mol

Substituting theses values into equation 1

n = 0.039/106.42

n = 0.00037 mole

For tantalum,

R.m = 0.0073 and m.m = 180.9 g/mol

Substituting these values into equation 1

n = 0.0073/180.9

n = 0.0000404 mole

<em>Therefore the number of moles of palladium and tantalum are 0.00037 mole and 0.0000404 mole respectively</em>

3 0
2 years ago
Two students walk in the same direction along a straight path, at a constant speed one at 0.90 m/s and the other at 1.90 m/s. a.
creativ13 [48]

Answer: a) 456.66 s ; b) 564.3 m

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v= distance/time so t=distance/v

The slower student time is: t=780m/0.9 m/s= 866.66 s

For the faster students t=780 m/1,9 m/s= 410.52 s

Therefore the time difference is 866.66-410.52= 456.14 s

In order to calculate the distance that faster student should  walk

to arrive 5,5 m before that slower student, we consider the follow expressions:

distance =vslower*time1

distance= vfaster*time 2

The time difference is 5.5 m that is equal to 330 s

replacing in the above expression we have

time 1= 627 s

time2 = 297 s

The distance traveled is 564,3 m

8 0
2 years ago
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