Answer:
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Explanation:
Answer:
a) Time = 2.67 s
b) Height = 35.0 m
Explanation:
a) The time of flight can be found using the following equation:
(1)
Where:
: is the final position in the horizontal direction = 80 m
: is the initial position in the horizontal direction = 0
: is the initial velocity in the horizontal direction = 30 m/s
a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)
t: is the time =?
By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:

b) The height of the hill is given by:
Where:
: is the final position in the vertical direction = 0
: is the initial position in the vertical direction =?
: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)
g: is the acceleration due to gravity = 9.81 m/s²
Hence, the height of the hill is:
I hope it helps you!
The block moves with constant velocity: for Newton's second law, this means that the resultant of the forces acting on the block is zero, because the acceleration is zero.
We are only concerned about the horizontal direction, and there are only two forces acting along this direction: the force F pushing the block and the frictional force

acting against the motion. Since their resultant must be zero, we have:

The frictional force is

where

is the coefficient of kinetic friction

is the weight of the block.
Substituting these values, we find the magnitude of the force F:
Answer:
Use the method on the image and solve it.
Answer:
The tangential speed of the tack is 8.19 m/s.
Explanation:
The wheel rotates 3.37 times a second that means wheel complete 3.37 revolutions in a second. Therefore, the angular speed ω of the wheel is given as follows:

Use the relation of angular speed with tangential speed to find the tangential speed of the tack.
The tangential speed v of the tack is given by following expression
v = ω r
Here, r is the distance to the tack from axis of rotation.
Substitute 21.174 rad/s for ω, and 0.387 m for r in the above equation to solve for v.
v = 21.174 × 0.387
v = 8.19m/s
Thus, The tangential speed of the tack is 8.19 m/s.