Why freezer is made in the upper part of refrigrator
1 answer:
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Answer:
2500 J
Explanation:
We can solve the problem by using the first law of thermodynamics:
where
Uf is the final internal energy of the system
Ui is the initial internal energy
Q is the heat added to the system
W is the work done by the system
In this problem, we have:
Q = +1000 J (heat that enters the system)
W = +500 J (work done by the system)
Ui = 2000 J (initial internal energy)
Using these numbers, we can re-arrange the equation to calculate the final internal energy:
Answer:
A kilowatt (kW) is a unit of power.
Explanation:
The power of an object is given by :
Here,
E is the energy required
t is time
The SI unit of power is Watts and the SI unit of energy is Joule. the commercial unit of energy is kilowatt per hour.
Option (1) : A kilojoule (kJ) is a unit of power is incorrect.
Option (2) : A gigawatt (GW) is a unit of energy is incorrect.
Option (3) : A watt (W) is a unit of energy is incorrect.
Option (4) : A kilowatt x hour per year (kWh/yr) is a unit of energy is incorrect.
Option (4) : A kilowatt (kW) is a unit of power is correct.
Hence, the correct option is (d).
Refer to the diagram shown below.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.
In order for the balloon to strike the professor's head, th balloon should drop by 18 - 1.7 = 16.3 m in the time at the professor takes to walk 1 m.
The time for the professor to walk 1 m is
t = (1 m)/(0.45 m/s) = 2.2222 s
The initial vertical velocity of the balloon is zero.
The vertical drop of the balloon in 2.2222 s is
h = (1/2)*(9.8 m/s²)*(2.2222 s)² = 24.197 m
Because 24.97 > 16.3, the balloon lands in front of the professor, and does not hit the professor.
The time for the balloon to hit the ground is
(1/2)*(9.8)*t² = 18
t = 1.9166 s
The time difference is 2.2222 - 1.9166 = 0.3056 s
Within this time interval, the professor travels 0.45*0.3056 = 0.175 m
Therefore the balloon falls 0.175 m in front of the professor.
Answer:
The balloon misses the professor, and falls 0.175 m in front of the professor.