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
Places with continental climates typically have hot summers and cold winters. As compared to places with mild climates, places with continental climates tend to experience the two extremes when it comes to the seasons. Summers can get very hot, and winters can get very cold.
This is what I got:
Net force in the Y direction:
ΣFy = T1 - T2
F = ma
ma = T1 - T2
Isolate for T2
ma - T1 = -T2
Multiply by -1
T1 - ma = T2
100 - (3)(2) = T2
100 - 6 = T2
T2 = 94 N
<span>10 hertz
Hertz is the frequency of oscillation which is the number of oscillations per second. So if something takes 0.10 s per oscillation, divide 1 second by the period to get the frequency. So
1 / 0.10s = 10 1/s = 10 Hertz
Therefore the object is vibrating at 10 hertz.</span>