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:
The correct answer is c. More total rainfall from a slower moving storm.
When the the forward speed of the hurricanes and tropical storms slows down they tend to increase the rainfall. Because of the slow movement the storm can be for few days over a given region and produce rainfall without stopping, thus create major flooding, pilling up of the coastal water, and produce persistent strong winds even though they have decreased in their forward speed.
Explanation:
Answer:
Explanation:
The football players collide in a completely inelastic collision, in other words they have the same velocity after the collision, this velocity has a magnitude V=1.6m/s.
We need to use the conservation of momentum Law, the total momentum is the same before and after the collision, at the initial point the receiver does not have any speed
(1)
We solve in order to find the receiver mass:
We assign the variables: T as tension and x the angle of the string
The <span>centripetal acceleration is expressed as v²/r=4.87²/0.9 and (0.163x4.87²)/0.9 = </span><span>T+0.163gcosx, giving T=(0.163x4.87²)/0.9 – 0.163x9.8cosx.
</span>
<span>(1)At the bottom of the circle x=π and T=(0.163x4.87²)/0.9 – .163*9.8cosπ=5.893N. </span>
<span>(2)Here x=π/2 and T=(0.163x4.87²)/0.9 – 0.163x9.8cosπ/2=4.295N. </span>
<span>(3)Here x=0 and T=(0.163x4.87²)/0.9 – 0.163x9.8cos0=2.698N. </span>
<span>(4)We have T=(0.163v²)/0.9 – 0.163x9.8cosx.
</span><span>This minimum v is obtained when T=0 </span><span>and v verifies (0.163xv²)/0.9 – 0.163x9.8=0, resulting to v=2.970 m/s.</span>
