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.
The particular temperature at which vaporisation occurs is known as the boiling point of liquid. Volume of water increases when it boils at 100° C. 1 cm3 of water at 100 ° C becomes 1760 cm3 of steam at 100 ° C.
Hope it helps!!!!!!!!!!!!!!!!!!!!! ~~~~~~~~~~~~~~~~~~~
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C because the stack of paper was divided into 4
or one hailstone we have;
Force = Mass X acceleration = 0.005kg x 9.8.} This is when the hailstone is not inclined at an angle.
When the hailstone is inclined at an angle of 45, then the component of force along the glass window will be F =0.005kg x 9.8 x sin45= 0.005kg x 9.8 x 0.707= 0.0346N.
Therefore, total force for the 500 hailstones would be 500x0.0346N=17.32N
This force is acting on an area equal to 0.600m2
Pressure = Force per unit area = 17.32N/0.600m2 = 28.9Pa
