<span>It's another energy balance equation, though: energy to start with is the same as energy that you end with. Suppose that we start a distance r0 from the Earth and end a distance r1 from the Moon, then the energy balance gives:
1 v02 - G M / r0 - G m / (D - r0) = 1 v12 - G M / (D - r1) - G m / r1
...where m is the moon's mass.
One simple limit takes D ? ? and 1 v02 ? G M / r0 (the escape velocity equation), to yield:
1 v12 ? G M / r1
v1 ? ?( 2 G M / r1 ) = 2377 m/s.</span>
Answer:
Geothermal energy can heat, cool, and generate electricity: Geothermal energy can be used in different ways depending on the resource and technology chosen—heating and cooling buildings through geothermal heat pumps, generating electricity through geothermal power plants, and heating structures through direct-use
Explanation:
Answer:
The final equilibrium T_{f} = 25.7[°C]
Explanation:
In order to solve this problem we must have a clear concept of heat transfer. Heat transfer is defined as the transmission of heat from one body that is at a higher temperature to another at a lower temperature.
That is to say for this case the heat is transferred from the iron to the water, the temperature of the water will increase, while the temperature of the iron will decrease. At the end of the process a thermal balance is found, i.e. the temperature of iron and water will be equal.
The temperature of thermal equilibrium will be T_f.
The heat absorbed by water will be equal to the heat rejected by Iron.
Heat transfer can be found by means of the following equation.
where:
Qiron = Iron heat transfer [kJ]
m = iron mass = 200 [g] = 0.2 [kg]
T_i = Initial temperature of the iron = 300 [°C]
T_f = final temperature [°C]
Cp_iron = 437 [J/kg*°C]
Cp_water = 4200 [J/kg*°C]
MULTIMETER combines the functions of ammeter, voltmeter and ohmmeter as a minimum.
Answer:
<em>The terminal velocity at sea level is 7.99 m/s</em>
<em>The terminal velocity at an altitude of 5000 m is 10.298 m/s </em>
<em></em>
Explanation:
mass of sphere m = 10 kg
radius of sphere r = 0.5 m
air density at sea level p = 1.22 kg/m^3
drag coefficient Cd = 0.8
terminal velocity = ?
Area of the sphere A = = 4 x 3.142 x = 3.142 m^2
terminal velocity is gotten from the relationship
where g = acceleration due to gravity = 9.81 m/s^2
imputing values into the equation
= <em>7.99 m/s</em>
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If at an altitude of 5000 m where air density = 0.736 kg/m^3, then we replace value of air density in the relationship as 0.736 kg/m^3
= <em>10.298 m/s </em>