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andreyandreev [35.5K]
3 years ago
10

Identify which type of source is being described.

Physics
2 answers:
zavuch27 [327]3 years ago
7 0

Answer:

Identify which type of source is being described.

✔ Primary

sources present original research, whereas

✔ secondary

sources summarize scientific information.

Second Part----

Indicate whether each of the following is a primary source or a secondary source.

An article in Journal of the American Medical Association:  

✔ primary source

An article in Time magazine:  

✔ secondary source

An article in The New York Times newspaper:  

✔ secondary source

Explanation:

Just did the problem.

frez [133]3 years ago
3 0

Answer:

Primary, secondary

Explanation:

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friction

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Which diagram best represents the field around a positively charged particle?
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How do we know that an object has accelerated?
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Could someone help me with this<br> you will get 20pts
jarptica [38.1K]

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left column Right column

Option 1 - Option 4

Option 2 - Option 1

Option 3 - Option 3

Option 4 - Option 2

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Read 2 more answers
A cube of ice at an initial temperature of -15.00°C weighing 12.5 g total is placed in 85.0 g of water at an initial temperature
Leona [35]

<u>Answer:</u> The specific heat of ice is 2.11 J/g°C

<u>Explanation:</u>

When ice is mixed with water, the amount of heat released by water will be equal to the amount of heat absorbed by ice.

Heat_{\text{absorbed}}=Heat_{\text{released}}

The equation used to calculate heat released or absorbed follows:

Q=m\times c\times \Delta T=m\times c\times (T_{final}-T_{initial})

m_1\times c_1\times (T_{final}-T_1)=-[m_2\times c_2\times (T_{final}-T_2)]       ......(1)

where,

q = heat absorbed or released

m_1 = mass of ice = 12.5 g

m_2 = mass of water = 85.0 g

T_{final} = final temperature = 22.24°C

T_1 = initial temperature of ice = -15.00°C

T_2 = initial temperature of water = 25.00°C

c_1 = specific heat of ice = ?

c_2 = specific heat of water = 4.186 J/g°C

Putting values in equation 1, we get:

12.5\times c_1\times (22.24-(-15))=-[85.0\times 4.186\times (22.24-25)]

c_1=2.11J/g^oC

Hence, the specific heat of ice is 2.11 J/g°C

3 0
4 years ago
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