The centripetal force acting on the ball will be 23.26 N.The direction of the centripetal force is always in the path of the center of the course.
<h3>What is centripetal force?</h3>
The force needed to move a body in a curved way is understood as centripetal force. This is a force that can be sensed from both the fixed frame and the spinning body's frame of concern.
The given data in the problem is;
m is the mass of A ball = 0.25 kg
r is the radius of circle= 1.6 m rope
v is the tangential speed = 12.2 m/s
is the centripetal force acting on the ball
The centripetal force is found as;
![\rm F_C = \frac{mv^2}{r} \\\\ F_C = \frac{0.25 \times (12.2)^2}{1.6} \\\\ F_C=23.26\ N](https://tex.z-dn.net/?f=%5Crm%20F_C%20%3D%20%5Cfrac%7Bmv%5E2%7D%7Br%7D%20%20%5C%5C%5C%5C%20F_C%20%3D%20%5Cfrac%7B0.25%20%5Ctimes%20%2812.2%29%5E2%7D%7B1.6%7D%20%20%5C%5C%5C%5C%20F_C%3D23.26%5C%20N)
Hence the centripetal force acting on the ball will be 23.26 N.
To learn more about the centripetal force refer to the link;
brainly.com/question/10596517
Answer:
1. ![t_2 = 2t_1](https://tex.z-dn.net/?f=t_2%20%3D%202t_1)
2. ![t_2 = t_1\sqrt{2}](https://tex.z-dn.net/?f=t_2%20%3D%20t_1%5Csqrt%7B2%7D)
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.
![a_1 = 2a_2](https://tex.z-dn.net/?f=a_1%20%3D%202a_2)
Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time
![t = v / a_1](https://tex.z-dn.net/?f=t%20%3D%20v%20%2F%20a_1)
Now that acceleration is halved:
![t = \frac{v}{2a_2}](https://tex.z-dn.net/?f=%20t%20%3D%20%5Cfrac%7Bv%7D%7B2a_2%7D)
![\frac{v}{a_2} = 2t](https://tex.z-dn.net/?f=%20%5Cfrac%7Bv%7D%7Ba_2%7D%20%3D%202t)
You would need to push for twice amount of time ![t_2 = 2t_1](https://tex.z-dn.net/?f=t_2%20%3D%202t_1)
2. The distance traveled by the puck is as the following equation:
![d = at^2](https://tex.z-dn.net/?f=d%20%3D%20at%5E2)
So if the acceleration is halved while maintaining the same d:
![\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}](https://tex.z-dn.net/?f=%5Cfrac%7Bd_1%7D%7Bd_2%7D%20%3D%20%5Cfrac%7Ba_1t_1%5E2%2F2%7D%7Ba_2t_2%5E2%2F2%7D)
As
, then
. Also ![a_1 = 2a_2](https://tex.z-dn.net/?f=a_1%20%3D%202a_2)
![1 = \frac{2a_2t_1^2}{a_2t_2^2}](https://tex.z-dn.net/?f=1%20%3D%20%5Cfrac%7B2a_2t_1%5E2%7D%7Ba_2t_2%5E2%7D)
![t_2^2 = 2t_1^2](https://tex.z-dn.net/?f=%20t_2%5E2%20%3D%202t_1%5E2)
![t_2 = t_1\sqrt{2}\approx 1.14t_1](https://tex.z-dn.net/?f=t_2%20%3D%20t_1%5Csqrt%7B2%7D%5Capprox%201.14t_1)
So t increased by 1.14
Answer:
The movement of an object depends on the reference frame, so it is important to predicate it.
Explanation:
The definition of a scale of 1: 166 will mean that the scale of 1 in the model will be equivalent to 166 times the measurement in the real model, therefore we will have that the height would be 166 times smaller than the 179m given:
![h = \frac{179m}{166} = 1.078m = 107.8cm](https://tex.z-dn.net/?f=h%20%3D%20%5Cfrac%7B179m%7D%7B166%7D%20%3D%201.078m%20%3D%20107.8cm)
The same for the diameter,
![\phi = \frac{48m}{166}= 0.2891m = 28.91cm](https://tex.z-dn.net/?f=%5Cphi%20%3D%20%5Cfrac%7B48m%7D%7B166%7D%3D%200.2891m%20%3D%2028.91cm)
The volume of a cylinder is given as
![V = (\pi r^2)(h)](https://tex.z-dn.net/?f=V%20%3D%20%28%5Cpi%20r%5E2%29%28h%29)
![V = (\pi (\frac{d}{2})^2)(h)](https://tex.z-dn.net/?f=V%20%3D%20%28%5Cpi%20%28%5Cfrac%7Bd%7D%7B2%7D%29%5E2%29%28h%29)
![V = (\pi (\frac{ 28.91}{2})^2)(107.8)](https://tex.z-dn.net/?f=V%20%3D%20%28%5Cpi%20%28%5Cfrac%7B%2028.91%7D%7B2%7D%29%5E2%29%28107.8%29)
![V = 70762.8cm^3](https://tex.z-dn.net/?f=V%20%3D%2070762.8cm%5E3)
Therefore the volume would be ![V = 70762.8cm^3](https://tex.z-dn.net/?f=V%20%3D%2070762.8cm%5E3)