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
Answer:
Option D, The equator gets more direct sunlight throughout the yea
Explanation:
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Answer:
Work done = 422.45 kJ
Explanation:
given,
weight of equipment = 6 kN
coefficient of kinetic friction = 0.05
distance up to which it is pulled = 1000 m
constant acceleration = 0.2 m/s²
Work done by the camper = ?
actual acceleration acting a'
m a = m a' - μ mg
a' = a + μ g
a' = 0.2 + 0.05 x 9.8
a' = 0.69 m/s²
Work done = Force x distance
F = m a'

F = 422.44897 N
Work done = F x d
Work done = 422.44897 x 1000
Work done = 422449 J
Work done = 422.45 kJ