Acceleration is given by:
where
v is the final velocity
u is the initial velocity
t is the time interval
Let's apply the formula to the different parts of the problem:
A)
Let's convert the quantities into SI units first:
t = 4.0 min = 240 s
So the acceleration is
B)
As before, let's convert the quantities into SI units first:
t = 94 s
So the acceleration is
C)
For this part we have to use a different formula:
where we have
v = 0 is the final velocity
u = 89.2 m/s is the initial velocity
a is the acceleration
d = 75 m is the distance covered
Solving for a, we find
Answer:
Part A: D
Part B: W = Qh - Qc
Part C: e = 1 - Qc/Qh
Explanation:
The heat engine is the engine that transforms heat (Q) in work (W), and by the second law of the thermodynamics, its efficiency can not be 100%, it means that some heat must be dissipated.
Part A:
The engine works with two sources of heat, one hot (Qh) at a hot temperature (Th) and another cold (Qc) at a cold temperature (Tc). It is necessary so, the hot source will give energy to the fluid of the engine, and the cold source will be the source where these heat will dissipate and the fluid will return to its original temperature. So,
Qh > Qc, and Th > Tc
Part B:
The ideal heat engine is the one that can use the most amount of heat to transform it at work. It is characterized by Qh/Qc = Th/Tc.
The work is the useful energy, so it is the total heat (Qh) less the heat dissipated (Qc):
W = Qh - Qc
Part C:
The effiency is the useful energy divided by the total energy. Because W = Qh - Qc:
e = W/Qh
e = (Qh - Qc)/Qh
e = Qh/Qh - Qc/Qh
e = 1 - Qc/Qh
Answer:
Part a)
the efficiency will be 100%
Part b)
for above condition internal resistance of the cell must be ZERO
Explanation:
Electric current supplied by the battery is given as
now the total power supplied by the cell is given as
power supplied to external load is given as
Part a)
If output power is adjusted to have maximum value then value of internal resistance must be ZERO
for same condition power given by the cell
so the efficiency will be 100%
Part b)
for above condition internal resistance of the cell must be ZERO
Yes this is not gonna work
Answer:
The second system must be set in motion seconds later
Explanation:
The oscillation time, T, for a mass, m, attached to spring with Hooke's constant, k, is:
One oscillation takes T secondes, and that is equivalent to a 2π phase. Then, a difference phase of π/2=2π/4, is equivalent to a time t=T/4.
If the phase difference π/2 of the second system relative to the first oscillator. The second system must be set in motion seconds later