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Monica [59]
3 years ago
8

Think of the last five things you read. Now think of a type of reading you havent done recenty

Physics
1 answer:
GarryVolchara [31]3 years ago
6 0

Answer:

1) Half Girlfriend

2) I am Malala

3) Diamond fire

4)I too had a love story

5)Your Dream are Now Mine

Explanation:

I don't have a same pattern to read but mostly of romantic and fiction. I read less autobiographies but when I read I am Malala , it was an inspiring one.My favorite type of reading is romantic types and least favorite is non fiction especially History

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39 g aluminum spoon (specific heat 0.904 J/g·°C) at 24°C is placed in 166 mL (166 g) of coffee at 83°C and the temperature of th
tatuchka [14]

<u>Answer:</u> The final temperature of the solution is 80.14^oC

<u>Explanation:</u>

The amount of heat released by coffee will be absorbed by aluminium spoon.

Thus, \text{heat}_{absorbed}=\text{heat}_{released}

To calculate the amount of heat released or absorbed, we use the equation:  

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

Also,

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 aluminium = 39 g

m_2 = mass of coffee = 166 g

T_{final} = final temperature = ?

T_1 = temperature of aluminium = 24^oC

T_2 = temperature of coffee = 83^oC

c_1 = specific heat of aluminium = 0.904J/g^oC

c_2 = specific heat of coffee= 4.1801J/g^oC

Putting all the values in equation 1, we get:

39\times 0.904\times (T_{final}-24)=-[166\times 4.1801\times (T_{final}-83)]

T_{final}=80.14^oC

Hence, the final temperature of the solution is 80.14^oC

4 0
4 years ago
If viewed at the correct time and in the right conditions, asteroids, comets, and meteors are all visible to the nked eye.
UkoKoshka [18]

Answer:

True

Explanation:

If it is at right conditions and correct time (night time most likely) they will all be visible

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Explanation:

hope this helps ( not sure tho )

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Which imaging technique uses radiation that is emitted from the material being imaged?
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MRI

Explanation:

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