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Ket [755]
2 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]2 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]2 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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Linear momentum (mass x speed) has to be conserved.

-- Momentum before the jump:

(boy's mass) x (boy's speed) = (25 kg) x (4.0 m/s) = 100 kg-m/s

(cart's mass) x (cart's speed) = (15 kg) x (zero) = zero

Total momentum before the jump:  (100 kg-m/s) + (zero) = (100 kg-m/s)

-- Momentum after the jump:

(mass of boy+cart) x (speed of boy+cart) = (40 kg) x (speed)

-- Momentum after the jump = momentum before the jump

(40 kg) x (speed) = 100 kg-m/s

Divide each side by  40 kg:

Speed = (100 kg-m/s) / (40 kg)

<em>Speed = 2.5 m/s</em>  (d)

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3 years ago
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Answer:

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Explanation:

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3 years ago
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3 years ago
Tính hiệu suất nhiệt của động cơ nhiệt biết nhiệt lượng ở nguồn nóng 420,4kJ/kg và nhiệt lượng ở nguồn lạnh 218kJ/kg.
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2 years ago
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MA_775_DIABLO [31]

Answer:

<u>Resolving</u><u> </u><u>horizontally</u><u>.</u> :

\sum d_{x} =  - (17.0 \cos 20.0) - (11.0 \cos 35.0) + (30.0 \cos 50.0) + 0 \\  { \underline{d _{x} =  -  5.702 \: m}} \\  \\  \sum d _{y} = (17.0 \sin 20.0) + 12.0 - (11.0  \sin 35.0) - (30.0 \sin 50.0) \\ { \underline{d _{y}  =  - 11.476 \: m}}

therefore, for resultant:

d =  \sqrt{ {d _{y} }^{2} + d _{x}  {}^{2}  }

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d =  \sqrt{ {( - 5.702)}^{2} +  {( - 11.476)}^{2}  }  \\  \\ d =  \sqrt{164.211}  \\  \\ { \boxed{ \boxed{ \bf{d = 12.8 \: m}}}}

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