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aksik [14]
3 years ago
15

Please post detailed answers to the following questions. Please use complete sentences. What are some of the challenges that you

might face if you wanted to observe and study stars from Earth? NEED TWO PARAGHRAPHS
Physics
1 answer:
Degger [83]3 years ago
8 0
As what Douglas Adams says, "<span>Space is big. Really big. You just won't believe how vastly, hugely, mind-bogglingly big it is. I mean, you may think it's a long way down the road to the chemist, but that's just peanuts to space."

The fact that the stars are very far from Earth, one can't fully understand the stars as it would require hundred of years to closely travel from earth to the stars because it is located million miles away from the Earth. Aside from its location, the stars are also hard to closely study because it emits lights that are harm for people when exposed longer. </span>
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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
uranmaximum [27]

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
Two auto technicians are discussing the stages of engine operation. Auto Technician A says during the intake stage of engine ope
pav-90 [236]

Neither technician is correct.

Please don't touch my car.

7 0
3 years ago
A spherical shell with a net charge of 3Q surrounds a point charge of -q at the center of the shell. The charges on the inner an
aleksley [76]

Answer:

1) The charge on the outer shell is +4·Q

2) The charge on the inner shell is +Q

Explanation:

1) The given parameters of the spherical shell are;

The net charge on the spherical shell = 3·Q

The point charge surrounded by the spherical shell = -Q

Let 'x' represent the charge on the outer shell, and let 'y', represent the charge on the inner shell, we have;

The net charge, 3·Q = -q + x

∴ x = 3·Q + Q = 4·Q

The charge on the outer shell, x = 4·Q

2) The net charge in the shell is zero, therefore, the charge on the inner shell, 'y', is given as follows;

-Q + y = 0

∴ y = +Q

The charge on the inner shell, y = +Q

5 0
3 years ago
A quantity of a gas has an absolute pressure of 400 kPa and an absolute temperature of 110 degrees kelvin. When the temperature
shepuryov [24]

You can make sure there's no change in volume by keeping
your gas in a sealed jar with no leaks.  Then you can play with
the temperature and the pressure all you want, and you'll know
that the volume is constant.

For 'ideal' gases,

       (pressure) times (volume) is proportional to (temperature).

And if volume is constant, then

                 (pressure) is proportional to (temperature) .

So if you increase the temperature from 110K to 235K,
the pressure increases to  (235/110)  of where it started.

     (400 kPa) x (235/110)  =  854.55 kPa. (rounded)

Obviously, choice-b is the right one, but
I don't know where the .46 came from.
 
4 0
3 years ago
ryan hypothesizes that darker colors heat up faster. he places a thermometer inside a red wool sock, a green cotton glove, and a
telo118 [61]
Light are transfer through waves in the atmosphere and yes it  true that the darker the color is the more heat it could absorb thus it is also explain that the lighter the color is the less heat or light its absorb its because the light is bounces back through other form of light and it lessens the amount of heat in a substance. In Ryan's procedure the possible wrong that he done is the present of a green cotton glove. Green color are one of the color that bounces light and could not support the hypothesis of Ryan and the possible temperature he could get is not the accurate one.
3 0
3 years ago
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