The answer is A. waterfall
To be considered as an engine , it should be a Man-made objects that could be used to produce power that creates motions.
From all the options above could be used to produce such power, but the waterfall is not made by mandkind
Answer:
112.58
Explanation:
The Coefficient of Performance of any system is denoted by COP=Q/W, where Q is the useful heat supplied or removed and W is the work required by the system. According to the first law of thermoddynamics Qh= Qc + W, where Qh is the heat transfered to the hot reservoir and Qc is the heat collected from the cold reservoir. Substituting the values for W and apllying the limitation for maximum theoretical efficiency we end up with the eqution shown below.
The Coefficient of Performance of air conditioner or COP is denoted by
COP(cool) = Tc/(Th- Tc)
where Tc: the lowest temperature
Th: the highest temperature
converting the values to Kelvin and adding them in the above equation
COP(cool) = (25+273)/((34+273)-(25+273))
= 298/(307-298)
= 298/9 = 33.11
From the question, it is stated that COP=SEER/3.4
hence, SEER= COP * 3.4
SEER= 33.11 * 3.4 = 112.58
Saturn, it is known as the planet with rings around it
Your position in meters will, measured relative to the starting point of the car behind you, be
x1(t) = 10 + 23.61 t - 1/2 4.2 t^2
his position will be
x2(t) = 16.67 t
Hence at any time the separation s(t) will be
s(t) = x1(t) - x2(t) = 10 + 6.94 t -2.1 t^2
Now I assume you mean that you will decelerate UNTIl you are driving at the legal speed limit (60 km/h). That will take you:
16.67 m/s = 23.61m/s - 4.2 m/s^2 * t
t = 1.65 seconds
What is the separation at that time? If it is still greater than zero, there will be no collision:
s(1.65) = 10 + 6.94 *1.65 -2.1 (1.65)^2 = 15.73 meters.
Hence you will NOT collide. The 1.65 s you calculated was the time needed to brake to the speed of 60 km/h.
Answer:(a) With our choice for the zero level for potential energy of the car-Earth system when the car is at point B ,
U
B
=0
When the car is at point A, the potential energy of the car-Earth system is given by
U
A
=mgy
where y is the vertical height above zero level. With 135ft=41.1m, this height is found as:
y=(41.1m)sin40.0
0
=26.4m
Thus,
U
A
=(1000kg)(9.80m/s
2
)(26.4m)=2.59∗10
5
J
The change in potential energy of the car-Earth system as the car moves from A to B is
U
B
−U
A
=0−2.59∗10
5
J=−2.59∗10
5
J
(b) With our choice of the zero configuration for the potential energy of the car-Earth system when the car is at point A, we have U
A
=0. The potential energy of the system when the car is at point B is given by U
B
=mgy, where y is the vertical distance of point B below point A. In part (a), we found the magnitude of this distance to be 26.5m. Because this distance is now below the zero reference level, it is a negative number.
Thus,
