Packages having a mass of 6 kgkg slide down a smooth chute and land horizontally with a speed of 3 m/sm/s on the surface of a co
1 answer:
Answer:
t = 1.02 s
Explanation:
The computation of the time required is shown below:
The package speed for belt is
= 3 - 1
= 2 m/s
Moreover, the decelerative force would be acted on the block i.e u.m.g
So, the decelerative produced
= 0.2 × 9.81
= 1.962 m/s^2
And, final velocity = 0
v = u - at
here
V = 0 = final velocity
u = 2 m/s
so,
0 = 2 - 1.962 × t
t = 1.02 s
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Answer:
So No of slices to be consumed by each person = n = 65
Explanation:
Energy released by one slice = E1
h = 8850 m ; m = 79 kg ,η= 10.5%
We know that potential energy given as
u = m g h
u = 79 x 9.81 x 8850
we know from the defination of efficiency that, η= E(out)/E(in)
Now amount of PE has to be compensated, In our case, E(out) =u
Let n be the number of bread slices to be consumed.
n = E(in)/E1
n=64.76
So No of slices to be consumed by each person = n = 65