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Agata [3.3K]
3 years ago
15

What is the amount of heat, in calories, given off from a 5 g piece of aluminum when it cools from 80°C to 20°C? The specific he

at capacity of aluminum is 0.215 cal/g∙°C. Show your work.
Physics
1 answer:
vagabundo [1.1K]3 years ago
8 0

Answer:

Q (heat) = S * m * (T2 - T1)  where Q is heat gained (loss), S the specific heat capacity of the substance and T2, T1 are the final and initial temps

Q = .215 cal  / (g deg C) * 5 g * (20 - 80) deg C = -64.5 cal

Since the question specifies the heat emitted, then  64.5 cal is the

heat emitted (loss).

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zhenek [66]

just analyze it in this way:

20cos30*=10( radical 3 )

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7 0
3 years ago
F an object has a mass of 200 kg and a weight of 1000 N, what is g?
jeka94

Answer:

g = 5 m/s square

Explanation:

Weight(W), Mass(m), Gravity(g)

W = mg

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7 0
2 years ago
find the time taken, if the speed of a train increased from 72 km/hr to 90 km/hr for 234 km. leave your answer in seconds
Airida [17]

Answer:

Time taken = 10400 s

Explanation:

Given:

Initial speed of the train, u=72\textrm{ km/h}=72\times \frac{5}{18}=20\textrm{ m/s}

Final speed of the train, v=90\textrm{ km/h}=90\times \frac{5}{18}=25\textrm{ m/s}

Displacement of the train, S=234\textrm{ km}=234\times 1000=234000\textrm{ m}

Using Newton's equation of motion,

v - u = at\\a=\frac{v-u}{t}

Now, using Newton's equation of motion for displacement,

v^{2}-u^{2}=2aS

Now, plug in the value of a=\frac{v-u}{t} in the above equation. This gives,

v^{2}-u^{2}=2\times \frac{v-u}{t}\times S\\(v+u)(v-u)=\frac{2(v-u)S}{t}\\t=\frac{2(v-u)S}{(v+u)(v-u)}\\t=\frac{2S}{v+u}

Now, plug in 234000 m for S, 25 m/s for v and 20 m/s for u. Solve for t.

t=\frac{2S}{v+u}\\t=\frac{2\times 234000}{25+20}\\t=\frac{468000}{45}=10400\textrm{ s}

Therefore, the time taken by the train is 10400 s.

3 0
3 years ago
What is the distance of earths surface from the surface of sun
storchak [24]
Max: 152 million km
min 146 million km
4 0
3 years ago
Read 2 more answers
A geologist sees a fault along which blocks of rock in the footwall have moved higher relative to blocks of rock in the hanging
Luda [366]

Answer: Normal fault

Explanation:

The type of fault that is explained above is a normal fault. We should note that normal faults typically takes place in a divergent boundary in a scenario where the crusts may have been pulled apart.

Since the crust is pulled apart in this case, it leads to the downward movement of the hanging wall which leads to the football being above the hanging wall.

6 0
3 years ago
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