Answer:
"0.758315 sec" is the appropriate choice.
Explanation:
The given values are:
Angular speed,
= 29 rad/s
or,
=
Tire rotates,
= 3.5 times
Now,
The time interval will be:
⇒
⇒
⇒
⇒
A tennis player tosses a tennis ball straight up and then catches it after 2.00 s at the same height as the point of release. (a
1 answer:
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Answer:
In first case speed of fish will be m/s.
In the second case the speed of fish m/s
Explanation:
Given data :-
Mass of bigger fish ( m₁ ) = 5 kg.
Mass of small fish ( m₂ ) = 1 kg.
Speed of large fish ( v₁ ) = 1 m/s
Mass of bigger fish after eating smaller one = 5 + 1 = 6 kg.
Case - 1
Momentum of bigger fish before eating the smaller fish = m₁* v₁ = 5 * 1 = 5 kg.m/s
Momentum of bigger fish after eating the larger fish = ( m₁ + m₂)*v
v = speed of bigger fish immediately after lunch.
Using the conservation of momentum.
m₁* v₁ = ( m₁ + m₂)*v
5 = 6 * v
v = m/s.
Case -2
Speed of small fish = 4 m/s
Momentum of bigger fish before lunch = 5 kg.m/s
Momentum of smaller fish before lunch = 4*1 = 4 kg.m/s
Net momentum before lunch = 5 - 4 = 1 kg.m/s
Momentum of bigger fish after eating the larger fish = 6 * V
Using the conservation of momentum.
1 = 6 * V
V = m/s.
Answer:
(A) 180 kg·m²
(B) 0.111 rad/s²
(C) The number of revolutions the merry-go-round will complete until it finally stops is 3.142 or π rev
Explanation:
The equation of the moment of inertia of a solid cylinder is presented as follows;
Where:
I = Moment of inertia of the merry-go-round
M = Mass of the merry-go-round
R = Radius of the merry-go-round
Therefore, I = 1/2×90×2² = 180 kg·m²
(B) For the angular acceleration we have;
Therefore, since the force × radius = The torque, we have, angular acceleration is found as follows
F × R = τ
10.0 × 2.0 = 20 = I×α = 180×α
α = 20/180 = 0.111 rad/s².
angular acceleration = 0.111 rad/s².
(C) Here we have ω₀ = 20 rev/ min = 20×2×π rad/min = π·40/60 rad/s
2/3·π rad/s
ω = ω₀ - α×t
∴ t = ω₀/α = (2/3·π rad/s)/(0.111 rad/s²) = 18.85 s
Hence we have
θ = ω₀·t + 1/2·α·t², plugging in the values, we have;
θ = 2/3·π×18.85 - 1/2·0.111·18.85²
θ = 19.74 rad
Therefore, since 2·π radian = 1 revolution
The number of revolutions the merry-go-round will complete until it stops is 19.74/(2·π) = 3.142 or π revolutions.