Mass have no effect for the projectile motion and u want to know the height "h"
first,
find the vertical and horizontal components of velocity
vertical component of velocity = 12 sin 61
horizontal component of velocity = 12 cos 61
now for the vertical motion ;
S = ut + (1/2) at^2
where
s = h
u = initial vertical component of velocity
t = 0.473 s
a = gravitational deceleration (-g) = -9.8 m/s^2
h=[12×sin 610×0.473]+[−9.8×(0.473)2]
u can simplify this and u will get the answer
h=.5Gt2
H=1.09m
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Answer:
520000 or 520000 pa
Force = 520N
Area of contact = 0.001
Pressure: 520000 or 520000
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
