Answer:
![a=12.97\ m/s^2](https://tex.z-dn.net/?f=a%3D12.97%5C%20m%2Fs%5E2)
Explanation:
Given that,
The length of a string, l = 0.87 m
Speed of the ball, v = 3.36 m/s
We need to find the acceleration of the ball. The acceleration acting on the ball is centripetal acceleration. It is given by :
![a=\dfrac{v^2}{r}\\\\a=\dfrac{(3.36)^2}{0.87}\\\\=12.97\ m/s^2](https://tex.z-dn.net/?f=a%3D%5Cdfrac%7Bv%5E2%7D%7Br%7D%5C%5C%5C%5Ca%3D%5Cdfrac%7B%283.36%29%5E2%7D%7B0.87%7D%5C%5C%5C%5C%3D12.97%5C%20m%2Fs%5E2)
So, the acceleration of the ball is
.
Answer:
physical change
Explanation:
Physical change are changes affecting the form of a chemical substance, but not its chemical composition. Examples of physical change: melting, freezing, shape, size, color...
Answer:
probably liquid cause it can be contained easily
The speed of air at the intake area is
(m/s).
<h3 /><h3>Continuity equation</h3>
The continuity equation is used to determine the flow rate at different sections of a pipe or fluid conduit.
The continuity equation is given as;
![A_1v_1 = A_2 v_2](https://tex.z-dn.net/?f=A_1v_1%20%3D%20A_2%20v_2)
- Let the intake area = A₁
- Let the velocity of air in intake area = v₁
- Let the area of the test section = A₂
- Velocity of air in test section, v₂ = 10 m/s
The speed of air at the intake area is calculated as follows;
![A_1 v_1 = A_2 v_2\\\\ v_1 = \frac{A_2 v_2}{A_1} \\\\ v_1 = \frac{10A_2}{A_1}](https://tex.z-dn.net/?f=A_1%20v_1%20%3D%20A_2%20v_2%5C%5C%5C%5C%0Av_1%20%3D%20%5Cfrac%7BA_2%20v_2%7D%7BA_1%7D%20%5C%5C%5C%5C%0Av_1%20%3D%20%5Cfrac%7B10A_2%7D%7BA_1%7D%20)
Learn more about continuity equation here: brainly.com/question/14619396
Answer:
v = 17.9 m/s
Explanation:
As we know that the normal force measured by the sensor before the ride is started is given as
![F_n = mg = 990 N](https://tex.z-dn.net/?f=F_n%20%3D%20mg%20%3D%20990%20N)
now when the rider has reached at the top position of the loop then the normal force is given as
![F_n' = 360 N](https://tex.z-dn.net/?f=F_n%27%20%3D%20360%20N)
now at the top position we have
![F_n' + mg = ma](https://tex.z-dn.net/?f=F_n%27%20%2B%20mg%20%3D%20ma)
![F_n' + mg = \frac{mv^2}{R}](https://tex.z-dn.net/?f=F_n%27%20%2B%20mg%20%3D%20%5Cfrac%7Bmv%5E2%7D%7BR%7D)
so we have
![990 + 360 = \frac{990 v^2}{9.8 \times 24}](https://tex.z-dn.net/?f=990%20%2B%20360%20%3D%20%5Cfrac%7B990%20v%5E2%7D%7B9.8%20%5Ctimes%2024%7D)
![v = 17.9 m/s](https://tex.z-dn.net/?f=v%20%3D%2017.9%20m%2Fs)