the density of mercury is 13.6 gram per centimetre cubic find the volume of 2720 grams of mercury in metres cubic
1 answer:
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Answer:
Approximately 0.0898 W/m².
Explanation:
The intensity of light measures the power that the light delivers per unit area.
The source in this question delivers a constant power of . If the source here is a point source, that
of power will be spread out evenly over a spherical surface that is centered at the point source. In this case, the radius of the surface will be 9.6 meters.
The surface area of a sphere of radius is equal to
. For the imaginary 9.6-meter sphere here, the surface area will be:
.
That power is spread out evenly over this 9.6-meter sphere. The power delivered per unit area will be:
.
Answer:
Second Trial satisfy principle of conservation of momentum
Explanation:
Given mass of ball A and ball B
Let mass of ball and
Final velocity of ball
Final velocity of ball
initial velocity of ball
Initial velocity of ball
Momentum after collision
Momentum before collision
Conservation of momentum in a closed system states that, moment before collision should be equal to moment after collision.
Now,
Plugging each trial in this equation we get,
First Trial
momentum before collision moment after collision
Second Trial
moment before collision moment after collision
Third Trial
momentum before collision moment after collision
Fourth Trial
momentum before collision moment after collision
We can see only Trial- 2 shows the conservation of momentum in a closed system.
Explanation:
The given value of P is as follows.
P = 1.06F + 22.18,
or, P' = 1.06
As p' is defined and non-zero. Hence, only critical points are boundary points.
For F = 10, the value of P will be calculated as follows.
P =
= 32.78
For F = 70, the value of P will be calculated as follows.
P =
= 96.38
Therefore, the minimum value of P is 32.78 and maximum value of P is 96.38.
Answer:
The temperature of the steam during the heat rejection process is 42.5°C
Explanation:
Given the data in the question;
the maximum temperature T in the cycle is twice the minimum absolute temperature T
in the cycle
T = 0.5T
now, we find the efficiency of the Carnot cycle engine
η = 1 - T
/T
η = 1 - T
/0.5T
η = 0.5
the efficiency of the Carnot heat engine can be expressed as;
η = 1 - W
/Q
where W is net work done, Q
is is the heat supplied
we substitute
0.5 = 60 / Q
Q = 60 / 0.5
Q = 120 kJ
Now, we apply the first law of thermodynamics to the system
W = Q
- Q
60 = 120 - Q
Q = 60 kJ
now, the amount of heat rejection per kg of steam is;
q = Q
/m
we substitute
q = 60/0.025
q = 2400 kJ/kg
which means for 1 kilogram of conversion of saturated vapor to saturated liquid , it takes 2400 kJ/kg of heat ( enthalpy of vaporization)
q = h
= 2400 kJ/kg
now, at h = 2400 kJ/kg from saturated water tables;
T = 40 + ( 45 - 40 ) (
)
T = 40 + (5) × (0.5)
T = 40 + 2.5
T = 42.5°C
Therefore, The temperature of the steam during the heat rejection process is 42.5°C