Answer:
Yes, it is possible.
It is called humidity. The quantity is reduced as the temperature go down. That’s why it condenses in a fridge or in a air conditioner unit. To create vapor there is various ways such as heating (evaporation) or fogging of liquid water. It will stop when the vapor pressure is reached and it depends of the temperature. The vapor pressure of water is 101.32 kPa at 100°C, which is the atmospheric pressure at sea level. It means that at that pressure and that temperature and above the water is totally converted in gas in the boiling process.
We start to see more off the dark side (waning) or more of the light side (waxing) the more of the dark side we see, the less moon there seems to be. The more light side, the more moon.
Answer: B.Heat is never converted completely into mechanical energy.
Explanation:
According to the second principle of thermodynamics:
<em>"The amount of entropy in the universe tends to increase over time"
</em>
However the first formulation of this law (by Sadi Carnot) states:
<em>"There is an upper limit to the efficiency of conversion of heat to work, in a heat engine
"</em>
This means the heat cannot be completely transformed into mechanical energy in a machine. That is why a machine's effieciency is always less than 100%
Complete question is;
a. Two equal sized and shaped spheres are dropped from a tall building. Sphere 1 is hollow and has a mass of 1.0 kg. Sphere 2 is filled with lead and has a mass of 9.0 kg. If the terminal speed of Sphere 1 is 6.0 m/s, the terminal speed of Sphere 2 will be?
b. The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1. The masses remain 1.0 kg and 9.0 kg, The terminal speed (in m/s) of Sphere 2 will now be
Answer:
A) V_t = 18 m/s
B) V_t = 10.39 m/s
Explanation:
Formula for terminal speed is given by;
V_t = √(2mg/(DρA))
Where;
m is mass
g is acceleration due to gravity
D is drag coefficient
ρ is density
A is Area of object
A) Now, for sphere 1,we have;
m = 1 kg
V_t = 6 m/s
g = 9.81 m/s²
Now, making D the subject, we have;
D = 2mg/((V_t)²ρA))
D = (2 × 1 × 9.81)/(6² × ρA)
D = 0.545/(ρA)
For sphere 2, we have mass = 9 kg
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρA))]
V_t = 18 m/s
B) We are told that The cross sectional area of Sphere 2 is increased to 3 times the cross sectional area of Sphere 1.
Thus;
Area of sphere 2 = 3A
Thus;
V_t = √[2 × 9 × 9.81/(0.545/(ρA) × ρ × 3A))]
V_t = 10.39 m/s
Answer:
a) final angular speed w2 = 414rpm
b) time to stop t = 312.5s
c) angular distance from start to stop s = s = 1171.88 rev
Explanation:
Average angular speed wa = (w1+w2)/2 = Angular distance/time = s/t
Given;
w1 = 450 rpm = 450/60 rev/s = 7.5 rev/s
s = 180 rev
t = 25s
a)
s/t = (w1+w2)/2
w2 = 2s/t - w1
Substituting the values;
w2 = 2×180/25 - 7.5 = 6.9 rev/s
w2 = 6.9×60rpm = 414rpm
b)
Angular deceleration a = (w1-w2)/t = (7.5-6.9)/25
a = 0.024 rev/s^2
time t for w2 = 0
t = (w1-w2)/a = (7.5-0)/0.024
t = 312.5 s
c)
Angular distance from start to stop s;
s = w1t - 0.5at^2
s = 7.5×312.5s - 0.5×0.024×312.5^2
s = 1171.88 rev