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navik [9.2K]
3 years ago
15

In a device like the one shown below, the cylinder is allowed to fall a distance of 300 m. As a result, the temperature of the w

ater increases by 2.7°C. What would have been the increase in temperature if the cylinder were only allowed to fall 100 m?
Physics
1 answer:
Delvig [45]3 years ago
8 0

Answer:

Increase in the temperature of water would be 0.9 degree C

Explanation:

As we know by energy conservation

Change in the gravitational potential energy of the cylinder = increase in the thermal energy of the water

Here we know that the gravitational potential energy of the cylinder is given as

U = mgh

here we have

h = 300 m

now we can say

Mc\Delta T = (m \times 9.8 \times 300)

now if the cylinder falls from height h = 100 m

then we have

Mc\Delta T' = (m \times 9.8 \times 100)

now from above two equations

\frac{\Delta T'}{\Delta T} = \frac{100}{300}

\Delta T' = 2.7 \times \frac{1}{3} = 0.9 Degree C

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object made of rubber, such as the ballons are elecrical insulators, meaning that they resist electric charges flowing through them..

when the ballon has been rubbed enough times to gain a sufficient negative charge, it will be attracted to the wall

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Which graph accurately shows the relationship between kinetic energy and mass as mass increases
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Answer:

c

Explanation:

energy doesnt affect to mass of a object

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A computer with a resistance of 52.7 Ω has a power input of 210.0 W. Calculate the current in the computer. Round your answer to
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A cue ball of mass m1 = 0.325 kg is shot at another billiard ball, with mass m2 = 0.59 kg, which is at rest. The cue ball has an
Roman55 [17]

Answer:

v_{2f} = \frac{2vm_1}{m_2 + m_1}

Explanation:

If the collision is elastic and exactly head-on, then we can use the law of momentum conservation for the motion of the 2 balls

Before the collision

P_i = m_1v

After the collision

P_f = m_1v_{1f} + m_2v_{2f}

So using the law of momentum conservation

P_i = P_f

m_1v = m_1v_{1f} + m_2v_{2f}

We can solve for the speed of ball 1 post collision in terms of others:

v_{1f} = v - v_{2f}\frac{m_2}{m_1}

Their kinetic energy is also conserved before and after collision

m_1v^2/2 = m_1v_{1f}^2/2 + m_2v_{2f}^2/2

m_1v^2 = m_1v_{1f}^2 + m_2v_{2f}^2

From here we can plug in v_{1f} = v - v_{2f}\frac{m_2}{m_1}

m_1v^2 = m_1\left(v - v_{2f}\frac{m_2}{m_1}\right)^2 + m_2v_{2f}^2

m_1v^2 = m_1\left(v^2 - 2vv_{2f}\frac{m_2}{m_1} + v_{2f}^2\frac{m_2^2}{m_1^2}\right) + m_2v_{2f}^2

m_1v^2 = m_1v^2 - 2vv_{2f}m_2 + v_{2f}^2\frac{m_2^2}{m_1} + m_2v_{2f}^2

v_{2f}^2(m_2 + \frac{m_2^2}{m_1}) - 2vm_2v_{2f} = 0

v_{2f}(1 + \frac{m_2}{m_1}) = 2v

v_{2f} = \frac{2v}{1 + \frac{m_2}{m_1}} = \frac{2v}{\frac{m_1 + m_2}{m_1}} = \frac{2vm_1}{m_2 + m_1}

8 0
3 years ago
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A straight fin is made from copper (k = 388 W/m-K) and is 0.5 cm in diameter and 30 cm long. The temperature at the base of the
erastovalidia [21]

Answer:

The rate of transfer of heat is 0.119 W

Solution:

As per the question:

Diameter of the fin, D = 0.5 cm = 0.005 m

Length of the fin, l =30 cm = 0.3 m

Base temperature, T_{b} = 75^{\circ}C

Air temperature, T_{infty} = 20^{\circ}

k = 388 W/mK

h = 20\ W/m^{2}K

Now,

Perimeter of the fin, p = \pi D = 0.005\pi \ m

Cross-sectional area of the fin, A = \frac{\pi}{4}D^{2}

A = \frac{\pi}{4}(0.5\times 10^{-2})^{2} = 6.25\times 10^{- 6}\pi \ m^{2}

To calculate the heat transfer rate:

Q_{f} = \sqrt{hkpA}tanh(ml)(T_{b} - T_{infty})

where

m = \sqrt{\frac{hp}{kA}} = \sqrt{\frac{20\times 0.005\pi}{388\times 6.25\times 10^{- 6}\pi}} = 41.237

Now,

Q_{f} = \sqrt{20\times 388\times 0.005\pi\times 6.25\times 10^{- 6}\pi}tanh(41.237\times 0.3)(75 - 20) = 0.119\ W

5 0
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