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g100num [7]
3 years ago
15

A boat is travelling down stream in a river at 20 m/s south. A passenger on the boat walks toward the back of the boat at 10 m/s

north. Relative to the ground, at what velocity is the passenger moving?
A) 10 m/s north
B) 10 m/s south
C) 30 m/s north
D) 30 m/s south
Physics
1 answer:
artcher [175]3 years ago
8 0

The question is unclear regarding the boat's velocity. Is it 20 m/s south relative to the water, or relative to the earth? (It is a river, after all...)

There's also the possibility that the boat's velocity relative to the river is 0. Take the south direction to be negative and north to be positive, and denote by v_{A/B} the velocity of a body A relative to a body B. Under these conditions,

v_{B/E}=v_{B/W}+v_{W/E}\iff-20\,\dfrac{\mathrm m}{\mathrm s}=0+-20\,\dfrac{\mathrm m}{\mathrm s}

(B for boat, E for earth, W for water) so that the passenger's velocity relative to the earth is

v_{P/E}=v_{P/B}+v_{B/W}+v_{W/E}

(P for passenger)

v_{P/E}=10\,\dfrac{\mathrm m}{\mathrm s}+0-20\,\dfrac{\mathrm m}{\mathrm s}=-10\,\dfrac{\mathrm m}{\mathrm s}

or 10 m/s south.

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3 0
3 years ago
You wish to cool a 1.83 kg block of tin initially at 88.0°C to a temperature of 57.0°C by placing it in a container of kerosene
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Answer:

0.273 liters are needed to accomplish this task without boiling.

Explanation:

The minimum boiling point of kerosene is 150\,^{\circ}C. According to this question, we need to determine the minimum volume of liquid such that heat received is entirely sensible, that is, with no phase change.

If we consider a steady state process and that energy interactions with surrounding are negligible, then we get the following formula by the Principle of Energy Conservation:

\rho_{k}\cdot V_{k}\cdot c_{k}\cdot (T-T_{k,o}) = m_{t}\cdot c_{t}\cdot (T_{t,o}-T) (1)

Where:

\rho_{k} - Density of kerosene, measured in kilograms per cubic meter.

V_{k} - Volume of kerosene, measured in cubic meters.

c_{k}, c_{t} - Specific heats of the kerosene and tin, measured in joule per kilogram-Celsius.

T_{k,o}, T_{t,o} - Initial temperatures of kerosene and tin, measured in degrees Celsius.

T - Final temperatures of the kerosene-tin system, measured in degrees Celsius.

Please notice that the block of tin is cooled at the expense of the temperature of the kerosene until thermal equilibrium is reached.

From (1), we clear the volume of kerosene:

V_{k} = \frac{m_{t}\cdot c_{t}\cdot (T_{t,o}-T)}{\rho_{k}\cdot c_{k}\cdot (T-T_{k,o})}

If we know that m_{t} = 1.83\,kg, c_{t} = 218\,\frac{J}{kg\cdot ^{\circ}C}, T_{t,o} = 88\,^{\circ}C, T_{k,o} = 24.0\,^{\circ}C, T = 57\,^{\circ}C, c_{k} = 2010\,\frac{J}{kg\cdot ^{\circ}C} and \rho_{k} = 820\,\frac{kg}{m^{3}}, then the volume of the liquid needed to accomplish this task without boiling is:

V_{k} = \frac{(1.83\,kg)\cdot \left(218\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (88\,^{\circ}C-57\,^{\circ}C)}{\left(820\,\frac{kg}{m^{3}} \right)\cdot \left(2010\,\frac{J}{kg\cdot ^{\circ}C} \right)\cdot (57\,^{\circ}C-24\,^{\circ}C)}

V_{k} = 2.273\times 10^{-4}\,m^{3}

V_{k} = 0.273\,L

0.273 liters are needed to accomplish this task without boiling.

3 0
3 years ago
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