The angle turned through the flywheel of a generator during a time interval t is given by theta = at + bt^3 -ct^4, where a, b, a
1 answer:
Answer:
α =
Explanation:
The angle, θ, has been given as:
θ =
To obtain angular acceleration, α, we have to differentiate the angle twice, with respect to time, t.
Differentiating once will yield the angular velocity, ω:
ω = dθ/dt =
This can then be differentiated to obtain angular acceleration:
α = dω/dt =
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Answer:
The index of refraction of the liquid is 1.35.
Explanation:
It is given that,
Critical angle for a certain air-liquid surface,
Let n₁ is the refractive index of liquid and n₂ is the refractive index of air, n₂ =1
Using Snell's law for air liquid interface as :
So, the index of refraction of the liquid is 1.35. Hence, this is the required solution.
Answer:
Explanation:
Weight of the ladder is 355N
WL = 355N
The weight of the ladder acts at center of the ladder I.e at 4m from the bottom
Weight of firefighters is 870N
Wf = 870N
The fire fighter is at 6.3m from the bottom of the ladder.
N1 is the normal force exerted by the wall
N2 is the normal force exerted by the ground
Using Newton law
Check attachment
ΣFy = 0, since body is in equilibrium
N₂ - WL - Wf = 0
N₂ = WL + Wf
N₂ = 870 + 355
N₂ = 1225 N
This the normal force exerted by the ground on the wall.
Now to get N₁, let take moment about point A
So, before we take the moment we need to make sure that the forces are perpendicular to the plane(ladder), we need to resolve the weight of the ladder, firefighter and the normal of the wall to be perpendicular to the plane.
ΣMa = 0
Clockwise moment is equal to anti-clockwise moment
Moment Is the produce of force and perpendicular distance.
M = F×r
So,
WL•Cos50 × 4 + Wf•Cos50 × 6.3 —N₁•Sin50 × 8 = 0
355•Cos50 × 4 + 870•Cos50 × 6.3 =
N₁•Sin50 × 8
1825.52 + 3523.12 = 6.13N₁
6.13N₁ = 5348.64
N₁ = 5348.64/6.13
N₁ = 872.54 N.
The normal force exerted by the wall is 872.54N
