Explanation:
The given data is as follows.
height (h) = 98.0 m, speed (v) = 73.0 m/s,
Formula of height in vertical direction is as follows.
h = ,
or, t =
Now, formula for the required distance (d) is as follows.
d = vt
=
=
= 326.5 m
Thus, we can conclude that 326.5 m is the horizontal distance from the target from where should the pilot release the canister.
Answer:
Accelerations of both the sides is 0.6125 , A moves downwards whereas B moves upwards.
=6.125 rad/
Tension on side A = 4.5 × g= 44.1 m/
Tension on side B= 2.0 × g= 19.6 m/
Explanation:
As both, the blocks A and B are attached due to the constraint they can only possess a single acceleration a.
Observe the figure attached, let the tension with Block A be and the tension attached with Block B be
.
Tensions will be only be due to the weight of the blocks as no other force is present.
= 4.5 × g= 44.1 m/
= 2.0 × g= 19.6 m/
Now, lets make a torque equation about the center of the wheel and find the alpha
×R-
×R= MI( Moment of Inertia of Wheel)× Alpha
where, R= Radius of the wheel=0.100m and
Alpha()= Angular acceleration of the wheel
MI of the wheel= 0.400 kg/
=6.125 rad/
Acceleration = R ×
= 0.1 * 6.125
=0.6125
Accelerations of both the sides is 0.6125 , A moves downwards whereas B moves upwards.
