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.
Answer:
1) Mass that needs to be converted at 100% efficiency is 0.3504 kg
2) Mass that needs to be converted at 30% efficiency is 1.168 kg
Explanation:
By the principle of mass energy equivalence we have
where,
'E' is the energy produced
'm' is the mass consumed
'c' is the velocity of light in free space
Now the energy produced by the reactor in 1 year equals
Thus the mass that is covertred at 100% efficiency is
Part 2)
At 30% efficiency the mass converted equals
Answer:
W = 1562.5 J
Explanation:
Path 1:
W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J
Path 2:
W₂ = F₂*d₂ = 130 N * 10 m = 1300 J
Path 3:
W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J (opposite to the motion)
We get
W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J
That's actually a lot easier than finding total <u><em>distance.</em></u>
To find displacement, you only have to know where the trip started from, and where it ended. It doesn't matter what route was followed to actually travel from the start to the finish. The displacement is the straight-line distance and direction between those two points.
