Ball is projected upwards with speed 20 m/s
now using kinematics
here when ball returns back to same position.
total displacement of ball will be zero.
So ball will return to initial position with same speed in opposite direction.
Answer:
volume of substance of weight of mercury is 13593 kilograms
Answer:
a) v = 13.8 m / s
, b) a = 95.49 m / s²
, c) a force that goes to the center of the carnival ride and d) μ = 0.10
Explanation:
For this exercise we will use the angular kinematics relationships and the equation that relate this to the linear kinematics
a) reduce the magnitudes to the SI system
w = 1.1 rev / s (2pi rad / 1rev) = 6.91 rad / s
The equation that relates linear and angular velocity is
v = w r
v = 6.91 2
v = 13.8 m / s
b) centripetal acceleration is given by
a = v² / r = w² r
a = 6.91² 2
a = 95.49 m / s²
c) this acceleration is produced by a force that goes to the center of the carnival ride
d) Here we use Newton's second law
fr -W = 0
fr = W
μ N = mg
Radial shaft
N = m a
N = m w² r
μ m w² r = m g
μ = g / w² r
μ = 9.8 / 6.91² 2
μ = 0.10
Answer: 39.5 N
Explanation:
2 x force x static friction coefficient = weight (weight=2μsF)
force = ?
static friction coefficient = 0.416
weight = 32.9N
substituting the values above into the equation we get
2 x F x 0.416 = 32.9
F = 32.9 / (2 x 0.416)
F = 39.5N
