A person drops a brick from the top of a building. The height of the building is 400 m and the mass of the brick is 2.00 kg. Wha
1 answer:
This question involves the conservation of energy. There are two energy in this case, potential energy and kinetic energy. Let's divid the energy into three status.
1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
1. Before dropping, all potential energy
2.dropping, potential energy transformed to kinetic energy
3. before hitting the ground, all Kinetic energy.
Recall the formula for both energy, which are U=mgh, and K=1/2mv^2
Since the energy is conserved in this case ( b/c otherwise it will say in the problem), the amount of energy at the beginning should equal to the energy at the end. Therefore we have, mgh=1/2mv^2
plug the number in and solve for velocity.
2x400x10=1/2 x 2 x v^2
v^2=8000
v=
v=40
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Answer:
the high temperature needed to operate this refrigerator is C) 137.4° C
Explanation:
Hello!
The carnot refrigeration cycle is one in which a machine absorbs heat from an enclosure and expels it to the surroundings, the equation that defines the COP performance coefficient for this cycle is:
COP=performance coefficient =2.1
T1= Low temperature
T2=high temperature
Now use algebra to find the high temperature
If we replace the values:
note = remember that the temperature must be in absolute units, for which we must add 273.15 to the low temperature to find the temperature in Kelvin
T1 = 5 + 273.15 = 278.15K
In celsius
T2=410.60-273.15=137.4° C
the high temperature needed to operate this refrigerator is C) 137.4° C