In your own words, describe two conditions that are necessary for an eclipse to occur.
1 answer:
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Answer:
Specific heat capacity, c = 468.75 J/Kg°C
Explanation:
Given the following data;
Power = 1.5 kW to Watts = 1.5 * 1000 = 1500 Watts
Time = 5 seconds
Mass = 0.2 kg
Initial temperature = 20°C
Final temperature = 100°C
To find specific heat capacity;
First of all, we would have to determine the energy consumption of the kettle;
Energy = power * time
Energy = 1500 * 5
Energy = 7500 Joules
Next, we would calculate the specific heat capacity of water.
Heat capacity is given by the formula;
Where;
- Q represents the heat capacity or quantity of heat.
- m represents the mass of an object.
- c represents the specific heat capacity of water.
- dt represents the change in temperature.
dt = T2 - T1
dt = 100 - 20
dt = 80°C
Making c the subject of formula, we have;
Substituting into the equation, we have;
<em>Specific heat capacity, c = 468.75 J/Kg°C</em>
Answer:
<em>Details in the explanation</em>
Explanation:
<u>Vertical Launch</u>
When an object is thrown vertically in free air (no friction), it moves upwards at its maximum speed while the acceleration of gravity starts to brake it. At a given time and height, the object stops in mid-air and starts to fall back to the launching point until reaching it with the same speed it was launched.
We are given an expression for the height of an object in function of time t
<em>Please note we have deleted the second 'squared' from the formula since it's incorrect and won't describe the motion of vertical launch.</em>
We now have to evaluate h for the following times, assuming h comes in feet
At t=1 sec
The object is at a height of 48 feet
At t=2 sec
The object is at a height of 64 feet. This is the maximum height the object will reach, as we'll see below
At t=3 sec
The object is at a height of 48 feet. We can clearly see it's returning from the maximum height and is going down
At t=4 sec
The object is at ground level and has returned to the launch point.