(C)
Explanation:
The circle has a radius r = 0.5 m, which means that its circumference C is
![C = 2\pi r = 2\pi(0.5\:\text{m}) = 3.14\:\text{m}](https://tex.z-dn.net/?f=C%20%3D%202%5Cpi%20r%20%3D%202%5Cpi%280.5%5C%3A%5Ctext%7Bm%7D%29%20%3D%203.14%5C%3A%5Ctext%7Bm%7D)
One revolution means that the stopper travels a distance equal to the circumference of the circle so the velocity of the stopper is
![v = \dfrac{C}{t} =\dfrac{3.14\:\text{m}}{0.2\:\text{s}} = 15.7\:\text{m/s} \approx 16\:\text{m/s}](https://tex.z-dn.net/?f=v%20%3D%20%5Cdfrac%7BC%7D%7Bt%7D%20%3D%5Cdfrac%7B3.14%5C%3A%5Ctext%7Bm%7D%7D%7B0.2%5C%3A%5Ctext%7Bs%7D%7D%20%3D%2015.7%5C%3A%5Ctext%7Bm%2Fs%7D%20%5Capprox%2016%5C%3A%5Ctext%7Bm%2Fs%7D)
Answer:
1.07 nT
Explanation:
We know that E/B = c where E = electric field amplitude = 320 mV/m = 0.32 V/m, B = magnetic field amplitude and c = speed of light = 3 × 10⁸ m/s.
So, B = E/c
Substituting E and c into B, we have
B = E/c
= 0.32 V/m ÷ 3 × 10⁸ m/s
= 0.1067 × 10⁻⁸ T
= 1.067 × 10⁻⁹ T
= 1.067 nT
≅ 1.07 nT
The greatest height the ball will attain is 3.27 m
<h3>Data obtained from the question</h3>
- Initial velocity (u) = 8 m/s
- Final velocity (v) = 0 m/s (at maximum height)
- Acceleration due to gravity (g) = 9.8 m/s²
The maximum height to which the ball can attain can be obtained as follow:
v² = u² – 2gh (since the ball is going against gravity)
0² = 8² – (2 × 9.8 × h)
0 = 64 – 19.6h
Collect like terms
0 – 64 = –19.6h
–64 = –19.6h
Divide both side by –19.6
h = –64 / –19.6h
h = 3.27 m
Thus, the greatest height the ball can attain is 3.27 m
Learn more about motion under gravity:
brainly.com/question/13914606
Simply, apply the formula
![E = \frac{1}{2}mv^{2}](https://tex.z-dn.net/?f=%20E%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E%7B2%7D%20)
and insert the values of m = mass, v = velocity and E = Energy.
The result will be
![72J = \frac{1}{2}m(12m/s)^{2}](https://tex.z-dn.net/?f=72J%20%3D%20%20%5Cfrac%7B1%7D%7B2%7Dm%2812m%2Fs%29%5E%7B2%7D%20%20)
, m = 1 kg
Answer:
The inlet velocity is 21.9 m/s.
The mass flow rate at reach exit is 1.7 kg/s.
Explanation:
Given that,
Mass flow rate = 2 kg/s
Diameter of inlet pipe = 5.2 cm
Fifteen percent of the flow leaves through location (2) and the remainder leaves at (3)
The mass flow rate is
![m_{2}=0.15\times2](https://tex.z-dn.net/?f=m_%7B2%7D%3D0.15%5Ctimes2)
We need to calculate the mass flow rate at reach exit
Using formula of mass
![m_{3}=m_{1}-m_{2}](https://tex.z-dn.net/?f=m_%7B3%7D%3Dm_%7B1%7D-m_%7B2%7D)
![m_{3}=2-0.15\times2](https://tex.z-dn.net/?f=m_%7B3%7D%3D2-0.15%5Ctimes2)
![m_{3}=1.7\ kg/s](https://tex.z-dn.net/?f=m_%7B3%7D%3D1.7%5C%20kg%2Fs)
We need to calculate the inlet velocity
Using formula of velocity
![v=\dfrac{m}{\rho A}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bm%7D%7B%5Crho%20A%7D)
Put the value into the formula
![v=\dfrac{2}{42.868\times\dfrac{\pi}{4}\times(5.2\times10^{-2})^2}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B2%7D%7B42.868%5Ctimes%5Cdfrac%7B%5Cpi%7D%7B4%7D%5Ctimes%285.2%5Ctimes10%5E%7B-2%7D%29%5E2%7D)
![v=21.9\ m/s](https://tex.z-dn.net/?f=v%3D21.9%5C%20m%2Fs)
Hence, The inlet velocity is 21.9 m/s.
The mass flow rate at reach exit is 1.7 kg/s.