Acceleration = (final velocity - initial velocity) / time
= (35-65)/10
= -3 m/s2
The shades are very different
Answer:
Explanation:
Let the velocity of projectile be v and angle of throw be θ.
The projectile takes 5 s to touch the ground during which period it falls vertically by 100 m
considering its vertical displacement
h = - ut +1/2 g t²
100 = - vsinθ x 5 + .5 x 9.8 x 5²
5vsinθ = 222.5
vsinθ = 44.5
It covers 160 horizontally in 5 s
vcosθ x 5 = 160
v cosθ = 32
squaring and adding
v²sin²θ +v² cos²θ = 44.4² + 32²
v² = 1971.36 + 1024
v = 54.73 m /s
Given:
F_gravity = 10 N
F_tension = 25 N
Let's find the net centripetal force exterted on the ball.
Apply the formula:

From the given figure, the force acting towards the circular path will be positive, while the force which points directly away from the center is negative.
Hence, the tensional force is positive while the gravitational force is negative.
Thus, we have:

Therefore, the net centripetal force exterted on the ball is 15 N.
ANSWER:
15 N