d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
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Answer:
a.
b.1058 N
Explanation:
We are given that
Mass of each dog,M=18.5 kg
Mass of sled with rider,m=250 kg
a.Average force,F=185 N


By Newton's second law



b.By Newton's second law

Substitute the values

Hence, the force in the coupling between the dogs and the sled=1058 N
Answer:
(A) 60 J
Explanation:
At state 1
KE₁=100 J
At state 2
KE₂ = 0
U₂=80 J
Given that surface is rough so friction force will act in opposite to the direction of motion
Lets take work done by friction = Wfr
From work power energy
Work done by all forces = Change in kinetic energy
Wfr + U₂=ΔKE
Wfr+80 = 100
Wfr= 20 J
Now when book slides from top position then
Wfr+ U = KEf - KEi
-20 + 80 = KEf-0
KEf= 60 J
(A) 60 J