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1 year ago

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

2 answers:

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
4. By stacking several pans on top of the other cooled pans

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. Large amounts of food should be divided into small portions and put in shallow containers for quicker cooling in the refrigerator because bacteria can grow rapidly on food if it is left out at room temperature for more than 2 hours.

AnnZ [28]1 year ago

3 0

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.

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A certain rigid aluminum container contains a liquid at a gauge pressure of P0 = 2.02 × 105 Pa at sea level where the atmospheri

MaRussiya [10]

Answer:

$dz=19217687.07\ m$

Explanation:

Given:

initial gauge pressure in the container, $P_0=2.02\times 10^{5}\ Pa$

atmospheric pressure at sea level, $P_a=1.01\times 10^5\ Pa$

initial volume, $V_0=4.4\times 10^{-4}\ m^3$

maximum pressure difference bearable by the container, $dP_{max}=2.26\times 10^{5}\ Pa$

density of the air, $\rho_a=1.2\ kg.m^{-3}$

density of sea water, $\rho_s=1.2\ kg.m^{-3}$

The relation between the change in pressure with height is given as:

$\frac{dP_{max}}{dz} =\rho_a.g_n$

where:

dz = height in the atmosphere

$g_n$ = standard value of gravity

Now putting the respective values:

$\frac{2.26\times 10^{5}}{dz} =1.2\times 9.8$

$dz=19217.687\ km$

$dz=19217687.07\ m$

Is the maximum height above the ground that the container can be lifted before bursting. (Since the density of air and the density of sea water are assumed to be constant.)

7 0

1 year ago

What visible signs indicate a precipitation reaction when two solutions are mixed?

Illusion [34]

Formation of an insoluble solid

Explanation:

One of the remarkable visible signs that indicates a precipitation reaction when two solutions are mixed is the formation of an insoluble solid. The insoluble solid formed is the precipitate.

Precipitates usually forms in single replacement reactions and double replacement or double decomposition reactions.
They form when two soluble compounds react. One of the product is an insoluble solid in the solution called the precipitate.
The solubility table helps to predict whether precipitates forms in a reaction.

Learn more:

precipitate brainly.com/question/8896163

#learnwithBrainly

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1 year ago

Consider a spring that does not obey Hooke’s law very faithfully. One end of the spring is fixed. To keep the spring stretched o

IRINA_888 [86]

Answer:

a) $W=-0.0103125\ J$

b) $W=0.0059375\ J$

c) Compressing is easier

Explanation:

Given:

Expression of force:

$F=kx-bx^2+cx^3$

where:

$k=100\ N.m^{-1}$

$b=700\ N.m^{-2}$

$c=12000\ N.m^{-3}$

$x$ when the spring is stretched

$x$ when the spring is compressed

hence,

$F=100x-700x^2+12000x^3$

From the work energy equivalence the work done is equal to the spring potential energy:

here the spring is stretched so, $x=-0.05\ m$

Now,

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (-0.05)-700\times (-0.05)^2+12000\times (-0.05)^3}{-0.05}$

$j=-8.25\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times -8.25\times (-0.05)^2$

$W=-0.0103125\ J$

When compressing the spring by 0.05 m

we have, $x=0.05\ m$

The spring constant at this instant:

$j=\frac{F}{x}$

$j=\frac{100\times (0.05)-700\times (0.05)^2+12000\times (0.05)^3}{0.05}$

$j=4.75\ N.m^{-1}$

Now work done:

$W=\frac{1}{2} j.x^2$

$W=0.5\times 4.75\times (0.05)^2$

$W=0.0059375\ J$

Since the work done in case of stretching the spring is greater in magnitude than the work done in compressing the spring through the same deflection. So, the compression of the spring is easier than its stretching.

8 0

1 year ago

A ball is falling at terminal velocity. Terminal velocity occurs when the ball is in equilibrium and the forces are balanced. Wh

Greeley [361]

Answer:

A free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N.

Explanation:

This is because at terminal velocity, the ball stops accelerating and the net force on the ball is zero. For the net force to be zero, equal and opposite forces must act on the ball, so that their resultant force is zero. That is F₁ + F₂ = 0 ⇒ F₁ = -F₂

Since F₁ = 20 N, then F₂ = -F₁ = -20 N

So, if F₁ points upwards since it is positive, then F₂ points downwards since it is negative.

So, a free body diagram with 2 forces: the first pointing downward labeled F Subscript g Baseline 20 N and the second pointing upward labeled F Subscript air Baseline 20 N best describes the ball falling at terminal velocity.

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2 years ago

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A split highway has a number of lanes for traffic. For traffic going in one direction, the radius for the inside of the curve is

Arte-miy333 [17]

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1 year ago