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sattari [20]
3 years ago
10

Mercury has one of the lowest specific heats. This fact added to its liquid state at most atmospheric temperatures make it effec

tive for use in thermometers. If 275 J of energy are added to 0.450 kg of mercury, the mercury's temperature will increase by 4.09 K. What is the specific heat of mercury.
Physics
1 answer:
UNO [17]3 years ago
5 0

The specific heat of mercury is 149.4 J/(kgK)

Explanation:

When a substance is supplied with an amount of energy Q, its temperature increases according to the equation:

\Delta T=\frac{Q}{mC_s}

where

\Delta T is the increase in temperature

m is the mass of the sample

C_s is its specific heat capacity

For the sample of mercury in this problem we have

Q = 275 J

m = 0.450 kg

\Delta T = 4.09 K

Therefore, by re-arranging the equation we find the mercury's specific heat:

C_s = \frac{Q}{m\Delta T}=\frac{275}{(0.450)(4.09)}=149.4 J/(kgK)

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Explain how neurons convey information using both electrical and chemical signals.
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A ball is hit with a paddle, causing it to travel straight upward. It takes 2.90 s for the ball to reach its maximum height afte
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Answer:

A. 28.42 m/s

B. 41.21 m

Explanation:

From the question given above, the following data were obtained:

Time (t) to reach the maximum height = 2.90 s

Initial velocity (u) =?

Maximum height (h) =?

A. Determination of the initial velocity of the ball.

Time (t) to reach the maximum height = 2.90 s

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) =?

v = u – gt (since the ball is going against gravity)

0 = u – (9.8 × 2.9)

0 = u – 28.42

Collect like terms

0 + 28.42 = u

u = 28.42 m/s

Thus, the initial velocity of the ball is 28.42 m/s

B. Determination of the maximum height reached by the ball.

Final velocity (v) = 0 m/s (at maximum height)

Acceleration due to gravity (g) = 9.8 m/s²

Initial velocity (u) = 28.42 m/s

Maximum height (h) =?

v² = u² – 2gh (since the ball is going against gravity)

0² = 28.42² – (2 × 9.8 × h)

0 = 807.6964 – 19.6h

Collect like terms

0 – 807.6964 = – 19.6h

– 807.6964 = – 19.6h

Divide both side by – 19.6

h = – 807.6964 / – 19.6

h = 41.21 m

Thus, the maximum height reached by the ball is 41.21 m

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3 years ago
Water with a volume flow rate of 0.001 m3/s, flows inside a horizontal pipe with diameter of 1.2 m. If the pipe length is 10m an
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Answer:

\triangle P=1.95*10^{-4}

Explanation:

Mass m=0.001

Diameter d=1.2m

Length l=10m

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Generally the equation for Friction factor is mathematically given by

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Where Re

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Therefore

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\triangle P=1.95*10^{-4}

 

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