To develop this problem it is necessary to apply the concepts given in the balance of forces for the tangential force and the centripetal force. An easy way to detail this problem is through a free body diagram that describes the behavior of the body and the forces to which it is subject.
PART A) Normal Force.


Here,
Normal reaction of the ring is N and velocity of the ring is v




PART B) Acceleration





Negative symbol indicates deceleration.
<em>NOTE: For the problem, the graph in which the turning radius and the angle of suspension was specified was not supplied. A graphic that matches the description given by the problem is attached.</em>
1. Humidity cannot be used to predict rain.
2. I'm pretty sure it's weather but I'm not 100% sure. Maybe like 89% sure.
3. Tempurature doesn't affect humidity.
4. Not sure but I think its the 3rd one
Answer:
like horror? or action haha
Explanation:
Answer:
Explanation:
If a number of less than 1, then the number has a decimal point like
0.085, 0.008 e.t.c.
The zeros before the none zero digit are insignificant. The significant figure is 8 and 5.
But if there a zero between the none zero e.g. 0.0087056
Here the zero between 7 and 5 is significant, then the significant numbers are 8,7,0,5,6
But if the zero is not in between the none zero digit, then the zero is insignificant
E.g 0.05800
The last two zero is insignificant, the significant number is 5 and 8
So, If a positive numbers less than 1, the zeros between the decimal point and a non-zero number are NOT significant.