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Ratling [72]
3 years ago
15

What is the most commonly used method of food preservation today? refrigeration drying foods smoking raw meats salt curing

Physics
2 answers:
dmitriy555 [2]3 years ago
6 0
Refrigerator was what is commonly used today. We do dry foods and salt cure but that is not done on a daily basis
klasskru [66]3 years ago
3 0

Answer: Refrigeration

Explanation: Food preservation is the technique of preserving food items from spoiling. It is done by preventing the growth of microorganisms and preventing the rancidity of fats.

In ancient times, the methods used for preservation of foods were drying foods which remove water and thus reduces the bacterial growth, smoking raw meats also hampers the bacterial activity and salt curing leads to osmosis and thus the bacteria dies.

The latest method is refrigeration which makes the food colder and the rate of deterioration gets slow. Bacterial action reduces with refrigeration, as bacteria cannot survive at low temperatures.

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Life cycle of a medium mass star
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Which factors affect heat transfer between a warm and a cool substance?
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8 0
3 years ago
Magnet A has twice the magnetic field strength of magnet B and pulls on magnet B with a force of 100 N. The amount of force that
son4ous [18]

The force exerted by the magnetic in terms of the magnetic field is,

F\propto B

Where B is the magnetic fied strength and F is the force.

Thus, if the magnetic A has twice magnetic field strength than the magnet B,

Then,

B_A=2B_B

Thus, the force exerted by the magnet B is,

\begin{gathered} F_B\propto B_B \\ F_B\propto\frac{B_A}{2} \\ F_B=\frac{F_A}{2} \\ F_B=\frac{100}{2} \\ F_B=50\text{ N} \end{gathered}

Thus, the force exerted by the magnet B on magnet A is 50 N.

The force exerted by the magnet A exerts on the magnet B is exactly 100 N as given.

Hence, the option B is the correct answer.

3 0
1 year ago
A man 2 m tall walks horizontally at a constant rate of 1 m/s toward the base of a tower 23 m tall. When the man is 10 m from th
Evgen [1.6K]

Answer:

\dfrac{d\theta}{dt}=0.038\ rad/s

Explanation:

Given that

\dfrac{dx}{dt}= -1\ m/s

From the diagram

tan\theta=\dfrac{21}{x}

By differentiating with time t

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

When x= 10 m

tan\theta=\dfrac{21}{10}

θ = 64.53°

Now by putting the value in equation

sec^2\theta \dfrac{d\theta}{dt}=-\dfrac{21}{x^2}\dfrac{dx}{dt}

sec^264.53^{\circ} \dfrac{d\theta}{dt}=-\dfrac{21}{10^2}\times (-1)

\dfrac{d\theta}{dt}=0.038\ rad/s

Therefore rate of change in the angle is 0.038\ rad/s

8 0
3 years ago
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