a ball is thrown straight up with an initial speed 40m/s.a) how high does it go? b) what will be its velocity be in 7 secs?c) ho
1 answer:
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Answer:
1.195 m
2.8375 s
2.21433 rad/s
Explanation:
d = Distance = 2.39 m
N = Number of cycles = 8
t = Time to complete 8 cycles = 22.7 s
Radius would be equal to the distance divided by 2
The radius is 1.195 m
Time period would be given by
Time period of the motion is 2.8375 s
Angular speed is given by
The angular speed of the motion is 2.21433 rad/s
Answer:
Each thruster has to applied a force of 294.5N in tangential direction
Explanation:
Mass of the ring ,M =2.1×105kg
Mass of the ship ,m = 3.5×104kg
Radius of the ring R = 86.0 m
distance of ship from center of the ring, r =31.0 m
Let force applied by each thruster be F
Time taken to reach gravity ,t = 3hrs = 3600× 3 =10800sec
The movement of ring make the object kept at the edge feel a force of centrifuge in outward direction.
Centrifugal force = weight of the object on earth
Assume the ring is moving with angular speed ω
Centripetalforce of the object kept at ring
m₁R ω²=m₁g (m₁=mas of object)
⇒Rω² = g
⇒ω = √g/R
The ring start from 0 angular speed with constant angular acceleration
Let the constant angular acceleration be ∝
∝ = ω / t
(ω = √g/R)
∝ =
Consider Torque on the ring and ship system
T = FR + FR = 2FR
Moment of inertia of ring ship system
I = I(ring)+I(ship)+I(ship)
= MR² + mr² + mr² = MR² + 2mr²
angular acceleration of the ring ship system
∝ =
Now we have ,
∝ = , ∝ =
equating both values
We have,
where,
m = 2.1 × 10⁵kg, R = 86m, m = 3.5 × 10₄kg,
r = 31m, g = 9.8m/s² , t = 10800sec
Each thruster has to applied a force of 294.5N in tangential direction