Given Information:
Angular displacement = θ = 51 cm = 0.51 m
Radius = 1.8 cm = 0.018 m
Initial angular velocity = ω₁ = 0 m/s
Angular acceleration = α = 10 rad/s
²
Required Information:
Final angular velocity = ω₂ = ?
Answer:
Final angular velocity = ω₂ = 21.6 rad/s
Explanation:
We know from the equations of kinematics,
ω₂² = ω₁² + 2αθ
Where ω₁ is the initial angular velocity that is zero since the toy was initially at rest, α is angular acceleration and θ is angular displacement.
ω₂² = (0)² + 2αθ
ω₂² = 2αθ
ω₂ = √(2αθ)
We know that the relation between angular displacement and arc length is given by
s = rθ
θ = s/r
θ = 0.51/0.018
θ = 23.33 radians
finally, final angular velocity is
ω₂ = √(2αθ)
ω₂ = √(2*10*23.33)
ω₂ = 21.6 rad/s
Therefore, the top will be rotating at 21.6 rad/s when the string is completely unwound.
Answer: assuming that the billiard balls are of identical weight the impacted billiard ball will move forward at around 0.5m/s (not considering energy conservation). The ball impacting the 2nd one would stop because most of its Kinetic energy would have been transferred into the not moving ball.
Explanation: hope this helps!
Answer:
The workdone is 
Explanation:
From the question we are told that
The height of the cylinder is 
The face Area is 
The density of the cylinder is 
Where 
is the density of freshwater which has a constant value
Now
Let the final height of the device under the water be 
Let the initial volume underwater be 
Let the initial height under water be 
Let the final volume under water be 
According to the rule of floatation
The weight of the cylinder = Upward thrust
This is mathematically represented as
So 
=> 
Now the work done is mathematically represented as
![= \rho_w g A [\frac{h^2}{2} ] \left | h_f} \atop {h}} \right.](https://tex.z-dn.net/?f=%3D%20%20%20%5Crho_w%20g%20A%20%5B%5Cfrac%7Bh%5E2%7D%7B2%7D%20%5D%20%5Cleft%20%7C%20h_f%7D%20%5Catop%20%7Bh%7D%7D%20%5Cright.)
![= \frac{g A \rho}{2} [h^2 - h_f^2]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%20%5Bh%5E2%20-%20h_f%5E2%5D)
![= \frac{g A \rho}{2} (h^2) [1 - \frac{h_f^2}{h^2} ]](https://tex.z-dn.net/?f=%3D%20%5Cfrac%7Bg%20A%20%5Crho%7D%7B2%7D%20%28h%5E2%29%20%20%5B1%20%20-%20%5Cfrac%7Bh_f%5E2%7D%7Bh%5E2%7D%20%5D)
Substituting values
Answer:
Work done by a tug boat, W = 1.735 x 10⁸ J
Explanation:
Given,
The of each tugboat, F = 1.5 x 10⁶ N
The angle of each tugboat forms with the resultant force, θ = 19°
The displacement of the supertanker, s = 710 m
The individual tugboat will be responsible for the displacement, d = 710/2
= 355 m
The displacement component in each tugboat direction = 355 · sin θ meter
Therefore, the work done by each tugboat is
W = F x S joules
Substituting the values in the above equation
W = 1.5 x 10⁶ x 355 · sin θ
= 1.735 x 10⁸ J
Hence, the work done by each tugboat is, W = 1.735 x 10⁸ J
