Answer:
W = 1.06 MJ
Explanation:
- We will use differential calculus to solve this problem.
- Make a differential volume of water in the tank with thickness dx. We see as we traverse up or down the differential volume of water the side length is always constant, hence, its always 8.
- As for the width of the part w we see that it varies as we move up and down the differential element. We will draw a rectangle whose base axis is x and vertical axis is y. we will find the equation of the slant line that comes out to be y = 0.5*x. And the width spans towards both of the sides its going to be 2*y = x.
- Now develop and expression of Force required:
F = p*V*g
F = 1000*(2*0.5*x*8*dx)*g
F = 78480*x*dx
- Now, the work done is given by:
W = F.s
- Where, s is the distance from top of hose to the differential volume:
s = (5 - x)
- We have the work as follows:
dW = 78400*x*(5-x)dx
- Now integrate the following express from 0 to 3 till the tank is empty:
W = 78400*(2.5*x^2 - (1/3)*x^3)
W = 78400*(2.5*3^2 - (1/3)*3^3)
W = 78400*13.5 = 1058400 J
Answer:
78 percent
Explanation:
I guess that's the right answer
Answer:
0.54454
104.00902 N
Explanation:
m = Mass of wheel = 100 kg
r = Radius = 0.52 m
t = Time taken = 6 seconds
= Final angular velocity
= Initial angular velocity
= Angular acceleration
Mass of inertia is given by
![I=\dfrac{mr^2}{2}\\\Rightarrow I=\dfrac{100\times 0.52^2}{2}\\\Rightarrow I=13.52\ kgm^2](https://tex.z-dn.net/?f=I%3D%5Cdfrac%7Bmr%5E2%7D%7B2%7D%5C%5C%5CRightarrow%20I%3D%5Cdfrac%7B100%5Ctimes%200.52%5E2%7D%7B2%7D%5C%5C%5CRightarrow%20I%3D13.52%5C%20kgm%5E2)
Angular acceleration is given by
![\alpha=\dfrac{\tau}{I}\\\Rightarrow \alpha=\dfrac{\mu fr}{I}\\\Rightarrow \alpha=\dfrac{\mu 50\times 0.52}{13.52}](https://tex.z-dn.net/?f=%5Calpha%3D%5Cdfrac%7B%5Ctau%7D%7BI%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B%5Cmu%20fr%7D%7BI%7D%5C%5C%5CRightarrow%20%5Calpha%3D%5Cdfrac%7B%5Cmu%2050%5Ctimes%200.52%7D%7B13.52%7D)
Equation of rotational motion
![\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=\omega_i+\dfrac{\mu (-50)\times 0.52}{13.52}t\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{\mu (-50)\times 0.52}{13.52}\times 6\\\Rightarrow 0=6.28318-11.53846\mu\\\Rightarrow \mu=\dfrac{6.28318}{11.53846}\\\Rightarrow \mu=0.54454](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Calpha%20t%5C%5C%5CRightarrow%20%5Comega_f%3D%5Comega_i%2B%5Cdfrac%7B%5Cmu%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7Dt%5C%5C%5CRightarrow%200%3D60%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%2B%5Cdfrac%7B%5Cmu%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7D%5Ctimes%206%5C%5C%5CRightarrow%200%3D6.28318-11.53846%5Cmu%5C%5C%5CRightarrow%20%5Cmu%3D%5Cdfrac%7B6.28318%7D%7B11.53846%7D%5C%5C%5CRightarrow%20%5Cmu%3D0.54454)
The coefficient of friction is 0.54454
At r = 0.25 m
![\omega_f=\omega_i+\dfrac{0.54454 (-50)\times 0.52}{13.52}6\\\Rightarrow 0=60\times \dfrac{2\pi}{60}+\dfrac{0.54454 f\times 0.25}{13.52}6\\\Rightarrow 2\pi=0.06041f\\\Rightarrow f=\dfrac{2\pi}{0.06041}\\\Rightarrow f=104.00902\ N](https://tex.z-dn.net/?f=%5Comega_f%3D%5Comega_i%2B%5Cdfrac%7B0.54454%20%28-50%29%5Ctimes%200.52%7D%7B13.52%7D6%5C%5C%5CRightarrow%200%3D60%5Ctimes%20%5Cdfrac%7B2%5Cpi%7D%7B60%7D%2B%5Cdfrac%7B0.54454%20f%5Ctimes%200.25%7D%7B13.52%7D6%5C%5C%5CRightarrow%202%5Cpi%3D0.06041f%5C%5C%5CRightarrow%20f%3D%5Cdfrac%7B2%5Cpi%7D%7B0.06041%7D%5C%5C%5CRightarrow%20f%3D104.00902%5C%20N)
The force needed to stop the wheel is 104.00902 N
Answer:
Capacitive reactance is 132.6 Ω.
Explanation:
It is given that,
Capacitance, ![C=20\ \mu F=20\times 10^{-6}\ F=2\times 10^{-5}\ F](https://tex.z-dn.net/?f=C%3D20%5C%20%5Cmu%20F%3D20%5Ctimes%2010%5E%7B-6%7D%5C%20F%3D2%5Ctimes%2010%5E%7B-5%7D%5C%20F)
Voltage source, V = 20 volt
Frequency of source, f = 60 Hz
We need to find the capacitive reactance. It is defined as the reactance for a capacitor. It is given by :
![X_C=\dfrac{1}{2\pi fC}](https://tex.z-dn.net/?f=X_C%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20fC%7D)
![X_C=\dfrac{1}{2\pi \times 60\ Hz\times 2\times 10^{-5}\ F}](https://tex.z-dn.net/?f=X_C%3D%5Cdfrac%7B1%7D%7B2%5Cpi%20%5Ctimes%2060%5C%20Hz%5Ctimes%202%5Ctimes%2010%5E%7B-5%7D%5C%20F%7D)
![X_C=132.6\ \Omega](https://tex.z-dn.net/?f=X_C%3D132.6%5C%20%5COmega)
So, the capacitive reactance of the capacitor is 132.6 Ω. Hence, this is the required solution.