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3 years ago

5

Marc attaches a falling 500-kg object with a rope through a pulley to a paddle wheel shaft. He places the system in a well-insul

ated tank holding 25 kg of water. When the object falls, it causes the paddle wheel to rotate and churn the water. If the object falls a vertical distance of 100 m at constant speed, what is the temperature change of the water? (1 kcal = 4 186 J, the specific heat of water is 4 186 J/kgC, and g = 9.8 m/s2)

1 answer:

Klio2033 [76]3 years ago

5 0

Answer:

4.68227 °C

Explanation:

$m_o$ = Mass of object = 500 kg

$m_w$ = Mass of water = 25 kg

c = Specific heat of water at 20°C = 4186 J/kg°C

h = Height from which the object falls = 100 m

g = Acceleration due to gravity = 9.8 m/s²

The potential energy and heat will balance each other

$PE=Q\\\Rightarrowmc m_ogh=m_oc\Delta T\\\Rightarrow \Delta T=\frac{m_ogh}{m_oc}\\\Rightarrow \Delta T=\frac{500\times 9.8\times 100}{25\times 4186}\\\Rightarrow \Delta=4.68227\ ^{\circ}C$

The temperature change in the water is 4.68227 °C

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Answer:

B meet A 0.01 km east of flagpole

Explanation:

given data

distance A = 5.7 km west

velocity V1 = 8.9 km/h

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solution

we know A and B are 5.7 + 4.5 = 10.2 km apart

and we consider here B will run distance x km for meet

so time will be for B is

time B = distance / velocity

time B = x / 7 ...................1

and

for A distance for meet = ( 10.2 - x ) km

so time A = distance / velocity

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now equating equation 1 and 2

time A = time B

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x = 4.490

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3 years ago

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Answer:

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$\begin{aligned} q &= 7.35\; {\mu C} \times \frac{10^{-6}\; {\rm C}}{1\; {\mu C}} = 7.35\times 10^{-6}\; {\rm C}\end{aligned}$ .

An electric field of magnitude $E = 5.25\times 10^{5}\; {\rm N \cdot C^{-1}}$ would exert on this charge a force with a magnitude of:

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