Let's cut through the weeds and the trash
and get down to the real situation:
A stone is tossed straight up at 5.89 m/s .
Ignore air resistance.
Gravity slows down the speed of any rising object by 9.8 m/s every second.
So the stone (aka Billy-Bob-Joe) continues to rise for
(5.89 m/s / 9.8 m/s²) = 0.6 seconds.
At that timer, he has run out of upward gas. He is at the top
of his rise, he stops rising, and begins to fall.
His average speed on the way up is (1/2) (5.89 + 0) = 2.945 m/s .
Moving for 0.6 seconds at an average speed of 2.945 m/s,
he topped out at
(2.945 m/s) (0.6 s) = 1.767 meters above the trampoline.
With no other forces other than gravity acting on him, it takes him
the same time to come down from the peak as it took to rise to it.
(0.6 sec up) + (0.6 sec down) = 1.2 seconds until he hits rubber again.
Answer:
In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. r' = r/2.
Explanation :
The electric force between two point charges is directly proportional to the product of charges and inversely proportional to the square of the distance between charges. It is given by :
r is the separation between charges
If F'= 2F
In order to make this force twice as strong, F' = 2 F, the distance would have to be changed to half i.e. . Hence, this is the required solution.
Answer:
6.75m
Explanation:
To calculate the distance in this question, we can use the formula:
S = ut + 1/2at^2
Where; S = distance
u = initial velocity = 0m/s
t = 3s
a = 1.5m/s^2
Hence:
S = (0 × 3) + 1/2 (1.5 × 3 × 3)
S = 0 + 1/2 (13.5)
S = 13.5/2
S = 6.75
Therefore, the car will travel 6.75m in 3seconds.
Answer:
.......................
Explanation:
A net force = unbalanced force. If however, the forces are balanced (in equilibrium) and there is no net force, the object will not accelerate and the velocity will remain constant.
Answer:
a) 
, b) 
, c) 
Explanation:
a) The linear acceleration of the watermelon seed is:
![a_{r} = \left[\left(33\,\frac{rev}{min} \right)\cdot \left(2\pi\,\frac{rad}{rev} \right)\cdot \left(\frac{1}{60}\,\frac{min}{s} \right)\right]^{2}\cdot (0.023\,m)](https://tex.z-dn.net/?f=a_%7Br%7D%20%3D%20%5Cleft%5B%5Cleft%2833%5C%2C%5Cfrac%7Brev%7D%7Bmin%7D%20%5Cright%29%5Ccdot%20%5Cleft%282%5Cpi%5C%2C%5Cfrac%7Brad%7D%7Brev%7D%20%5Cright%29%5Ccdot%20%5Cleft%28%5Cfrac%7B1%7D%7B60%7D%5C%2C%5Cfrac%7Bmin%7D%7Bs%7D%20%5Cright%29%5Cright%5D%5E%7B2%7D%5Ccdot%20%280.023%5C%2Cm%29)
b) The watermelon seed is experimenting a centrifugal acceleration. The coefficient of static friction between the seed and the turntable is calculated by the Newton's Laws:
c) Angular acceleration experimented by the turntable is:
The tangential acceleration experimented by the watermelon seed is:
The linear acceleration experimented by the watermelon seed is:
The minimum coefficient of static friction is:
