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Ket [755]
3 years ago
5

What was one of the main benefits to the use of the icebox for cooling food? Select all that apply. provided a source for cold w

ater made it easier to prepare food slowed microbial growth in food allowed people to store a few days of food in their homes
Physics
2 answers:
cupoosta [38]3 years ago
4 0

The correct answers are:

  • It slowed microbial growth in food.
  • It allowed people to store a few days of food in their homes.

Hope I helped! If so, please feel free to rate my answer and consider giving it the Brainliest.


Alika [10]3 years ago
3 0

Explanation:

When we store food in a cooler device like refrigerator then it helps in decreasing the growth of microbes in the food. As a result, food readily does not get affected by bacteria and thus it does not get spoiled soon.

An ice box also helps in the storage of food for a few days in homes as it also slow down the growth of bacteria.

Thus, we can conclude that the main benefits to the use of the icebox for cooling food are as follows.

  • slowed microbial growth in food.
  • allowed people to store a few days of food in their homes.
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Ff=μN=μmg=0.255*20=5.1 N

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Your chances of getting into a collision when talking on a cell phone _________: A. Double B. Triple C. Quadruple D. Remain the
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C. Quadruple

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A __ is used to measure air pressure.
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Each plate of a parallel‑plate capacitor is a square of side 4.19 cm, 4.19 cm, and the plates are separated by 0.407 mm. 0.407 m
alexandr1967 [171]

Answer:

The electric field strength inside the capacitor is 49880.77 N/C.

Explanation:

Given:

Side length of the capacitor plate (a) = 4.19 cm = 0.0419 m

Separation between the plates (d) = 0.407 mm = 0.407\times 10^{-3}\ m

Energy stored in the capacitor (U) = 7.87\ nJ=7.87\times 10^{-9}\ J

Assuming the medium to be air.

So, permittivity of space (ε) = 8.854\times 10^{-12}\ F/m

Area of the square plates is given as:

A=a^2=(0.0419\ m)^2=1.75561\times 10^{-3}\ m^2

Capacitance of the capacitor is given as:

C=\dfrac{\epsilon A}{d}\\\\C=\frac{8.854\times 10^{-12}\ F/m\times 1.75561\times 10^{-3}\ m^2 }{0.407\times 10^{-3}\ m}\\\\C=3.819\times 10^{-11}\ F

Now, we know that, the energy stored in a parallel plate capacitor is given as:

U=\frac{CE^2d^2}{2}

Rewriting in terms of 'E', we get:

E=\sqrt{\frac{2U}{Cd^2}}

Now, plug in the given values and solve for 'E'. This gives,

E=\sqrt{\frac{2\times 7.87\times 10^{-9}\ J}{3.819\times 10^{-11}\ F\times (0.407\times 10^{-3})^2\ m^2}}\\\\E=49880.77\ N/C

Therefore, the electric field strength inside the capacitor is 49880.77 N/C

8 0
3 years ago
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