(a) 6.04 rev/s
The speed of the ball is given by:
where
is the angular speed
r is the distance of the ball from the centre of the circle
In situation 1), we have
r = 0.600 m
So the speed of the ball is
In situation 2), we have
r = 0.900 m
So the speed of the ball is
So, the ball has greater speed when rotating at 6.04 rev/s.
(b) 
The centripetal acceleration of the ball is given by
where
v is the speed
r is the distance of the ball from the centre of the trajectory
For situation 1),
v = 30.6 m/s
r = 0.600 m
So the centripetal acceleration is
(c) 
For situation 2 we have
v = 34.1 m/s
r = 0.900 m
So the centripetal acceleration is
Answer:
Not between significant digits.
Explanation:
A zero not significant when it's not between significant digits.
A)
It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:
Through Definition of Velocity, comes:
B)
Using the Velocity Hourly Equation in vertical direction, we have:
The angle of impact is given by:
If you notice any mistake in my english, please let me know, because i am not native.
Answer:
In combination, the equatorial bulge and the effects of the surface centrifugal force due to rotation mean that sea-level gravity increases from about 9.780 m/s2 at the Equator to about 9.832 m/s2 at the poles, so an object will weigh approximately 0.5% more at the poles than at the Equator.
Answer:
The gauge pressure in Pascals inside a honey droplet is 416 Pa
Explanation:
Given;
diameter of the honey droplet, D = 0.1 cm
radius of the honey droplet, R = 0.05 cm = 0.0005 m
surface tension of honey, γ = 0.052 N/m
Apply Laplace's law for a spherical membrane with two surfaces
Gauge pressure = P₁ - P₀ = 2 (2γ / r)
Where;
P₀ is the atmospheric pressure
Gauge pressure = 4γ / r
Gauge pressure = 4 (0.052) / (0.0005)
Gauge pressure = 416 Pa
Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa
