Answer:
v = 12.12 m/s
Explanation:
Given that,
Radius of the curvature, r = 30 m
To find,
The car's speed at the bottom of the dip.
Solution,
Let mg is the true weight of the passenger. When it is moving in the circular path, the centripetal force act on it. It is given by :
The normal reaction of the passenger is given by :
N = 1.5 mg
Let v is the car's speed at the bottom of the dip. It can be calculated as:
v = 12.12 m/s
So, the speed of the car at the bottom of the dip is 12.12 m/s. Hence, this is the required solution.
All of them at the same time if they start off at the same temperature and same volume
Answer:
it comes from your knowledge and the information you have to get the reason why that is the answer so you are putting together things that you already know what the new information you have
Answer:
Explanation:
Given that,
Hot reservoir temperature is
TH = 58°C
TH = 58 + 273 = 331 K
Cold reservoir temperature
TC = -17°C
TC = -17°C + 273 = 256K
In 24minutes 590 J of heat was removed from hot reservoir
t = 24mins
t = 24 × 60 = 1440 seconds
Then,
Power is given as
P = E / t
P = 590 / 1440
P = 0.41 W
Since this removed from the hot reservoir, then, QH = 0.41W
We want to find heat expelled to the cold reservoir QX
Efficiency is given as
η = 1 - TH/TC = 1 - QH/QC
1 - TH/TC = 1 - QH/QC
TH/TC = QH/QC
Make QC subject of formula
QC = QH × TC / TH
QC = 0.41 × 256 / 331
QC = 0.317 W
