Answer:
D. 1.8 × 102 newtons radially inward
Explanation:
The magnitude of the centripetal force is given by:

where
m is the mass of the object
v is the tangential speed
r is the radius of the circular trajector
In this problem, we have m = 4.0 kg, v = 6.0 m/s and r = 0.80 m, therefore substituting into the equation we get

The centripetal force is the force that keeps the object in a circular trajectory, so it is a force that is always directed inward (towards the centre of the circular path) and radially. Therefore, the correct answer is
D. 1.8 × 102 newtons radially inward
B is your answer oscillating
Answer:
It is likely to be option B: 50 degrees and 50 degrees
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Answer:
Explanation:
Let 100 m/s be the velocity of projection.
So horizontal component
= 100 cos42
= 74.31 m /s
Vertical component = - 100 sin 42 . in upward direction
66.91 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = - 66.91 t + .5 x 9.8 x t²
4.9 t² - 66.91 t - 2.1 = 0
t = 13.685 s
Horizontal distance covered
= 13.685 x 74.31
= 1016.93 m
If angle of projction is 40°
So horizontal component
= 100 cos40
= 76.60 m /s
Vertical component = - 100 sin 42 . in upward direction
64.27 m/s
Net displacement = 2.1 downwards ( + ve )
Using s = ut + 1/2 gt²
2.1 = -76.60 t + .5 x 9.8 x t²
4.9 t² - 76.60 t - 2.1 = 0
t = 15.659 s
Horizontal distance covered
= 15.659 x 76.60
= 1199.49 m
So horizontal range is increased , if angle of projection is increased .