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natita [175]
3 years ago
6

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 d

egrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.
Physics
1 answer:
melisa1 [442]3 years ago
3 0

This question is incomplete

Complete Question

The table below shows different items that are found sitting on the counter in a kitchen. The temperature of the kitchen is 72 degrees Fahrenheit. Classify the objects according to whether their molecules will speed up or slow down after being left in the kitchen for a period of time.

Drag the objects to the correct category.

Object Temperature (Fahrenheit)

ice cubes 26°F

glass of tea 60°F

cooked piece of meat 160°F

butter 55°F

pot of water 75°F

bowl of soup 140°F

Answer:

Molecules that would speed up

ice cubes 26°F

glass of tea 60°F

butter 55°F

Molecules that would slow down

cooked piece of meat 160°F

pot of water 75°F

bowl of soup 140°F

Explanation:

Molecules that are found or contained in a substances have the tendency to react in such as what that they are are sped up or slowed down. This is due to their exposure to various kinds of changes in temperature, which could be a hot temperature , cold temperature e.t.c

When a substance is exposed to a hot temperature, the molecules of that substance speed up depending on how hot the temperature is while when a substance is cooled down or exposed to cold temperature, the molecules tend to slow down.

In the question above, the temperature of the kitchen is 72 degrees Fahrenheit.

Molecules that would speed up after been left in the Kitchen for a while:

a) Ice cubes 26°F: This is because 72°F is a warm took temperature, so the ice cubes would melt causing the molecules to speed up.

b)Glass of tea 60°F: This glass of tea is at a cool temperature of 60°F, when it is kept in the Kitchen which has a temperature of 72°F for while, the temperature of the glass of tea would increase due to an increase in its rate of reaction causing the molecules of the glass of water to speed up.

c) Butter 55°F: This is because 72°F is a warm temperature, so the butter would melt, increasing its rate of reaction and causing the molecules to speed up

Molecules that would slow down

a) Cooked piece of meat 160°F : 160°F is a very hot temperature and when it is left in a the Kitchen(72°F) , the cooked meat would begin to cool down and drop in temperature from 160°F, causing the molecules to slow down.

b) Pot of water 75°F: The temperature of the kitchen and the temperature of the pot of water is the same, hence the molecules of the water would slow down.

c) Bowl of soup 140°F : This bowl of soup is very hot and when it is left in the Kitchen(72°F) , the bowl of soup would begin to cool down and drop in temperature from 140°F, slowing down the rate of reaction and causing the molecules to slow down.

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a) m=20000Kg

b) v=0.214m/s

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We will separate the problem in 3 parts, part A when there were no coals on the car, part B when there is 1 coal on the car and part C when there are 2 coals on the car. Inertia is the mass in this case.

For each part, and since the coals are thrown vertically, the horizontal linear momentum p=mv must be conserved, that is, p=m_Av_A=m_Bv_B=m_Cv_C, were each velocity refers to the one of the car (with the eventual coals on it) for each part, and each mass the mass of the car (with the eventual coals on it) also for each part. We will write the mass of the hopper car as m_h, and the mass of the first and second coals as m_1 and m_2 respectively

We start with the transition between parts A and B, so we have:

m_Av_A=m_Bv_B

Which means

m_hv_A=(m_h+m_1)v_B

And since we want the mass of the first coal thrown (m_1) we do:

m_hv_A=m_hv_B+m_1v_B

m_hv_A-m_hv_B=m_1v_B

m_1=\frac{m_hv_A-m_hv_B}{v_B}=\frac{m_h(v_A-v_B)}{v_B}

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m_Bv_B=m_Cv_C

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(m_h+m_1)v_B=(m_h+m_1+m_2)v_C

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v_C=\frac{(m_h+m_1)v_B}{(m_h+m_1+m_2)}

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