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larisa86 [58]
1 year ago
10

How often must you check the temperature of food that is being held with temperature control?.

Physics
1 answer:
loris [4]1 year ago
6 0

You need to check the temperature of food being stored in a temperature-controlled environment every four hours. The process of changing a space's temperature is called temperature control.

Cooking food alone may not be enough to avoid food poisoning, though, if the bacteria in food are allowed to grow to large numbers. When the temperature is between 5°C and 63°C, bacteria can grow. The risk zone is the range between 5°C and 63°C.

Temperature control is a process where the passage of heat energy into or out of a space or substance is adjusted to achieve the desired temperature. This process involves measuring or otherwise detecting changes in the temperature of the space (and all of the objects contained therein) or of the substance.

Learn more about temperature here

brainly.com/question/11244611

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A muon has a rest mass energy of 105.7 MeV, and it decays into an electron and a massless particle. If all the lost mass is conv
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Answer:

The electron’s velocity is 0.9999 c m/s.

Explanation:

Given that,

Rest mass energy of muon = 105.7 MeV

We know the rest mass of electron = 0.511 Mev

We need to calculate the value of γ

Using formula of energy

K_{rel}=(\gamma-1)mc^2

\dfrac{K_{rel}}{mc^2}=\gamma-1

Put the value into the formula

\gamma=\dfrac{105.7}{0.511}+1

\gamma=208

We need to calculate the electron’s velocity

Using formula of velocity

\gamma=\dfrac{1}{\sqrt{1-(\dfrac{v}{c})^2}}

\gamma^2=\dfrac{1}{1-\dfrac{v^2}{c^2}}

\gamma^2-\gamma^2\times\dfrac{v^2}{c^2}=1

v^2=\dfrac{1-\gamma^2}{-\gamma^2}\times c^2

Put the value into the formula

v^2=\dfrac{1-(208)^2}{-208^2}\times c^2

v=c\sqrt{\dfrac{1-(208)^2}{-208^2}}

v=0.9999 c\ m/s

Hence, The electron’s velocity is 0.9999 c m/s.

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3 years ago
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Answer:

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Explanation:

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Answer:

-1m/s

Explanation:

We can calculate the speed of block A after collision

According to collision theory:

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Substitute the given values

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5Va = -5

Va = -5/5

Va = -1m/s

Hence the velocity of ball A after collision is -1m/s

Note that the velocity of block B is zero before collision since it is stationary

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