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2 years 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]2 years 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 close coiled helical spring of round steel wire 10 mm diameter having 10 complete turns with a mean radius of 60 mm is subject

kow [346]

Answer:

The deflection of the spring is 34.56 mm.

Explanation:

Given that,

Diameter = 10 mm

Number of turns = 10

$Radius_{mean} = 60\ mm$

$Diameter_{mean} = 120\ mm$

Load = 200 N

We need to calculate the deflection

Using formula of deflection

$\delta=\dfrac{8pD^3n}{Cd^4}$

Put the value into the formula

$\delta=\dfrac{8\times200\times(120)^3\times10}{80\times10^{3}\times10^4}$

$\delta =34.56\ mm$

Hence, The deflection of the spring is 34.56 mm.

4 0

3 years ago

An AC power source has an rms voltage of 120 V and operates at a frequency of 60.0 Hz. If a purely inductive circuit is made fro

Dmitrij [34]

Answer:

(a) 17634.24 Ω

(b) 0.0068 A

Explanation:

(a)

The formula for inductive inductance is given as

X' = 2πFL................... Equation 1

Where X' = inductive reactance, F = frequency, L = inductance

Given: F = 60 Hz, L = 46.8 H, π = 3.14

Substitute into equation 1

X' = 2(3.14)(60)(46.8)

X' = 17634.24 Ω

(b)

From Ohm's law,

Vrms = X'Irms

Where Vrms = Rms Voltage, Irms = rms Current.

make Irms the subject of the equation

Irms = Vrms/X'...................... Equation 2

Given: Vrms = 120 V, X' = 17634.24 Ω

Substitute into equation 2

Irms = 120/17634.24

Irms = 0.0068 A

5 0

2 years ago

If the sun had twice the mace how would that affect the gravitational force of the sun

daser333 [38]

Answer: Gravity is the force that keeps planets in orbit around the Sun. Gravity alone holds us to Earth's surface.

Planets have measurable properties, such as size, mass, density, and composition. A planet's size and mass determines its gravitational pull.

A planet's mass and size determines how strong its gravitational pull is.

Models can help us experiment with the motions of objects in space, which are determined by the gravitational pull between them.

Explanation:

5 0

2 years ago

Write down Newton's second law in terms of momentum and acceleration. Write down this law in the form of differential equation (

svetoff [14.1K]

Explanation:

According to Newton's second law of motion, the rate of change of momentum is directly proportional to the applied unbalanced force. The mathematical expression is given by:

$F=\dfrac{d(mv)}{dt}$

Where

F is the applied force

m is the mass of the object

v is the velocity with which it is moving

$F=m\dfrac{dv}{dt}$

Momentum of a particle is given by the product of mass and velocity as :

$p=mv$

Hence, this is the required solution.

3 0

2 years ago

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a un

Kamila [148]

Answer:

a) 6738.27 J

b) 61.908 J

c) $\frac{4492.18}{v_{car} ^{2} }$

Explanation:

The complete question is

A flywheel is a mechanical device used to store rotational kinetic energy for later use. Consider a flywheel in the form of a uniform solid cylinder rotating around its axis, with moment of inertia I = 1/2 mr2.

Part (a) If such a flywheel of radius r1 = 1.1 m and mass m1 = 11 kg can spin at a maximum speed of v = 35 m/s at its rim, calculate the maximum amount of energy, in joules, that this flywheel can store?

Part (b) Consider a scenario in which the flywheel described in part (a) (r1 = 1.1 m, mass m1 = 11 kg, v = 35 m/s at the rim) is spinning freely at its maximum speed, when a second flywheel of radius r2 = 2.8 m and mass m2 = 16 kg is coaxially dropped from rest onto it and sticks to it, so that they then rotate together as a single body. Calculate the energy, in joules, that is now stored in the wheel?

Part (c) Return now to the flywheel of part (a), with mass m1, radius r1, and speed v at its rim. Imagine the flywheel delivers one third of its stored kinetic energy to car, initially at rest, leaving it with a speed vcar. Enter an expression for the mass of the car, in terms of the quantities defined here.

moment of inertia is given as

$I$ = $\frac{1}{2}$ $mr^{2}$

where m is the mass of the flywheel,

and r is the radius of the flywheel

for the flywheel with radius 1.1 m

and mass 11 kg

moment of inertia will be

$I$ = $\frac{1}{2}$ $*11*1.1^{2}$ = 6.655 kg-m^2

The maximum speed of the flywheel = 35 m/s

we know that v = ωr

where v is the linear speed = 35 m/s

ω = angular speed

r = radius

therefore,

ω = v/r = 35/1.1 = 31.82 rad/s

maximum rotational energy of the flywheel will be

E = $Iw^{2}$ = 6.655 x $31.82^{2}$ = 6738.27 J

b) second flywheel has

radius = 2.8 m

mass = 16 kg

moment of inertia is

$I$ = $\frac{1}{2}$ $mr^{2}$ = $\frac{1}{2}$ $*16*2.8^{2}$ = 62.72 kg-m^2

According to conservation of angular momentum, the total initial angular momentum of the first flywheel, must be equal to the total final angular momentum of the combination two flywheels

for the first flywheel, rotational momentum = $Iw$ = 6.655 x 31.82 = 211.76 kg-m^2-rad/s