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Sergeeva-Olga [200]
4 years ago
5

If the force between two fixed Charged spheres is F and you triple the charge on one sphere and also triple the distance between

the spheres, what fraction/multiple of original force of attraction is now between the two spheres?
Physics
1 answer:
mezya [45]4 years ago
4 0
Since the charge has a direct proportionality, the increase will be a factor of three. But, since you triple the distance, which has an inversely squared proportion, it will be 3/9 F or 1/3 F.

*This is all based on the fact that this is true:
F_s = \frac{Kq_1q_2}{r^2}

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A centrifugal pump rotates at n˙ = 740 rpm. Suppose the pump has some swirl at the inlet (α1 = 10°) and exits at an angle of 35°
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Answer:

Net head = 380cm

bhp = 17.710kW

Explanation:

Angular velocity of centrifugal pump:

w=\frac{2\pi n}{60}=\frac{2\pi (750)}{60}=78.54\frac{rad}{s}

Normal velocity component at outlet of pump:

V_{2,n}=\frac{V}{2\pi r_{2}b_{2}}=\frac{0.573}{2\pi (0.24)(0.162)}}=2.346\frac{m}{s}

Tangential velocity component at exit of the pump:

V_{2,t}=v_{2,n}tan\alpha _{2}=(2.346)tan(35)=1.643\frac{m}{s}

Normal velocity component at inlet of pump:

V_{1,n}=\frac{V}{2\pi r_{1} b_{1}}=\frac{0.573}{2\pi (0.12)(0.18)}=4.22\frac{m}{s}

Tangential velocity component at inlet of the pump:

V_{1,t}=v_{1,n}tan\alpha _{1}=(4.22)tan(0)=0\frac{m}{s}

Equivalent head in centimetre of water column:

H_{water}=H(\frac{rho_{air}}{rho_{water}})\\\\H_{water}=(\frac{w}{g} )(r_{2}V_{2,t}-r_{1}V_{1,t})(\frac{rho_{air}}{rho_{water}}) \\\\H_{water}=(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0)})(\frac{1.2}{998})=38m=380cm

Break horse power:

bhp=rho_{water} gHV=rho_{water} g[(\frac{w}{g})(r_{2}V_{2,t}-r_{1}V_{1,t})]V\\\\bhp=(998)(9.81)[(\frac{78.54}{9.81})((0.24)(1.643)-(0.12)(0))](0.573)=17710W=17.710kW

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