Answer:
a) a = 0.477 m/s^2
b) u = 0.04862
Explanation:
Given:-
- The rotational speed of the turntable N = 33 rev/min
- The watermelon seed is r = 4.0 cm away from axis of rotation.
Find:-
(a) Calculate the acceleration of the seed, assuming that it does not slip. (b) What is the minimum value of the coefficient of static friction between the seed and the turntable if the seed is not to slip
Solution:-
- First determine the angular speed (w) of the turntable.
w = 2π*N / 60
w = 2π*33 / 60
w = 3.456 rad/s
- The watermelon seed undergoes a centripetal acceleration ( α ) defined by:
α = w^2 * r
α = 3.456^2 * 0.04
α = 0.477 m / s^2
- The minimum friction force (Ff) is proportional to the contact force of the seed.
- The weight (W) of the seed with mass m acts downwards. The contact force (N) can be determined from static condition of seed in vertical direction.
N - W = 0
N = W = m*g
- The friction force of the (Ff) is directed towards the center of axis of rotation, while the centripetal force acts in opposite direction. The frictional force Ff = u*N = u*m*g must be enough to match the centripetal force exerted by the turntable on the seed.
Ff = m*a
u*m*g = m*a
u = a / g
u = 0.477 / 9.81
u = 0.04862
Answer:
C
Explanation:
this is because i need more space
It occurs as detrital grains in sedimentary rocks. It forms under extreme pressure.
Speed = Distance/ Time
Speed = 400 / 4
Speed = 100 km/hr.
100 km per hour.
Answer:
It is showing the wavelength.
Explanation: Hope it helps you:)))
have a good day
