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baherus [9]
3 years ago
10

Estimate the force a person must exert on a massless string attached to a 0.15 kg ball to make the ball revolve in horizontal ci

rcle of radius 0.6 m. The ball makes 2 revolutions per second.
Physics
1 answer:
Mrrafil [7]3 years ago
6 0

Answer:

F = 14.2 N  

Explanation:

From the question we are told that:

Mass m=0.15kg

Radius r=0.6

Angular Velocity \omega=2rev/s

                            \omega= =2x2 \pi rad/s=>4 \pi rad/s

Generally the equation for Force applied is mathematically given by

 F =mrw2

 F=0.15*0.6* (4*x3.14^)2

 F = 14.2 N    

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r-ruslan [8.4K]

Answer:

nπ + π/2 for any integer n

Explanation:

Since destructive interference occurs every odd multiple of half wavelength, that is π/2, 3π/2, 5π/2 where the interference is half wavelength and in general, (n + 1/2)π where n is an integer. So, nπ + π/2 for any integer n

4 0
3 years ago
What is entropy and how is it related to string theory?
balu736 [363]

Answer:

Explanation:

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5 0
3 years ago
A +27 nCnC point charge is placed at the origin, and a +6 nCnC charge is placed on the xx axis at x=1mx=1m. At what position on
svet-max [94.6K]

Answer:

The position on the x axis is 0.32 m.

Explanation:

Given that,

Point charge = 27 nC

Charge = 6 nC

Distance = 1

We need to calculate the distance

Using formula of electric field

\dfrac{kq_{1}}{x^2}=\dfrac{kq_{2}}{(r-x)^2}

Put the value into the formula

\dfrac{27\times10^{-9}}{x^2}=\dfrac{6\times10^{-9}}{(1-x)^2}

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\dfrac{(1-x)^2}{x^2}=\dfrac{27}{6}

\dfrac{1-x}{x}=\sqrt{\dfrac{27}{6}}

\dfrac{1}{x}=\sqrt{\dfrac{27}{6}}+1

x=0.32\ m

Hence, The position on the x axis is 0.32 m.

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fomenos
The answer is c. 
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Hope this helped :)
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