Answer:
Explanation:
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To objective is to find the:
(i) required heat exchanger area.
(ii) flow rate to be maintained in the evaporator.
Given that:
water temperature = 300 K
At a reasonable depth, the water is cold and its temperature = 280 K
The power output W = 2 MW
Efficiency 
= 3%
where;
However, from the evaporator, the heat transfer Q can be determined by using the formula:
Q = UA(L MTD)
where;
Also;
LMTD = 4.97
Thus, the required heat exchanger area A is calculated by using the formula:
where;
U = overall heat coefficient given as 1200 W/m².K
The mass flow rate:
Answer:
5) Displacement = +3.125 m
Displacement is in the same direction as the force vector.
6) Force = -53.89 N
Force is in an opposite direction relative to the displacement.
Explanation:
5) We are given;
Force; F = 160 N.
Workdone; W = +500 J
Now, formula for workdone is;
W = Force × displacement
Thus, displacement = Work/force
Displacement = 500/160
Displacement = +3.125 m
Thus, displacement is in the same direction as the force vector.
6) We are given;
Displacement; d = 18 m.
Workdone; W = -970 J
Like in the first answer above,
Workdone = Force × Displacement
Thus;
Force = Workdone/Displacement
Force = -970/18
Force = -53.89 N
Since force is negative and displacement is positive, it means force is in an opposite direction relative to the displacement.
Answer: 2.94×10^8 J
Explanation:
Using the relation
T^2 = (4π^2/GMe) r^3
Where v= velocity
r = radius
T = period
Me = mass of earth= 6×10^24
G = gravitational constant= 6.67×10^-11
4π^2/GMe = 4π^2 / [(6.67x10^-11 x6.0x10^24)]
= 0.9865 x 10^-13
Therefore,
T^2 = (0.9865 × 10^-13) × r^3
r^3 = 1/(0.9865 × 10^-13) ×T^2
r^3 = (1.014 x 10^13) × T^2
To find r1 and r2
T1 = 120min = 120*60 = 7200s
T2 = 180min = 180*60= 10800s
Therefore,
r1 = [(1.014 x 10^13)7200^2]^(1/3) = 8.07 x 10^6 m
r2 = [(1.014 x 10^13)10800^2]^(1/3) = 10.57 x 10^6 m
Required Mechanical energy
= - GMem/2 [1/r2 - 1/r1]
= (6.67 x 10^-11 x 6.0 x 10^24 * 50)/2 * [(1/8.07 × 10^-6 )- (1/10.57 × 10^-6)]
= (2001 x 10^7)/2 * (0.1239 - 0.0945)
= (1000.5 × 10^7) × 0.0294
= 29.4147 × 10^7 J
= 2.94 x 10^8 J.
Newton's three forces, normal, tension and friction, are present in a surprising number of physical situations
Newton's Laws, that describe the relationship between an obejct and the forces acting upon it, apply in almost every physical situation, from quantum mechanics to electricity.
The correct answer is:
Newton’s laws can explain the forces that occur between objects every day
