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3 years ago

7

Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s

tring makes with the vertical. (Assume 0 is small so sin 0~tan 0~0.) [Hint: consider the vertical and horizontal components of the forces acting on each particle.]

1 answer:

ozzi3 years ago

7 0

Answer:

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe ..... (1)

T Cosθ = mg .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

$tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}$

$tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}$

$tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}$

$tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}$

As θ is very small, so tanθ and Sinθ is equal to θ.

$\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}$

$\theta =\left (\frac{kq^{2}}{4L^{2}\times mg} \right )^{\frac{1}{3}}$

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A tube 1.20 m long is closed at one end. A stretched wire is placed near the open end. The wire is 0.350 m long and has a mass o

Ksju [112]

Answer:

71.4583 Hz

67.9064 N

Explanation:

L = Length of tube = 1.2 m

l = Length of wire = 0.35 m

m = Mass of wire = 9.5 g

v = Speed of sound in air = 343 m/s

The fundamental frequency of the tube (closed at one end) is given by

$f=\frac{v}{4L}\\\Rightarrow f=\frac{343}{4\times 1.2}\\\Rightarrow f=71.4583\ Hz$

The fundamental frequency of the wire and tube is equal so he fundamental frequency of the wire is 71.4583 Hz

The linear density of the wire is

$\mu=\frac{m}{l}\\\Rightarrow \mu=\frac{9.5\times 10^{-3}}{0.35}\\\Rightarrow \mu=0.02714\ kg/m$

The fundamental frequency of the wire is given by

$f=\frac{1}{2l}\sqrt{\frac{T}{\mu}}\\\Rightarrow f^2=\frac{1}{4l^2}\frac{T}{\mu}\\\Rightarrow T=f^2\mu 4l^2\\\Rightarrow T=71.4583^2\times 0.02714\times 4\times 0.35^2\\\Rightarrow T=67.9064\ N$

The tension in the wire is 67.9064 N

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Answer:

a) v= 1732.05m/s

b) d=250000m

c) v= 1414.214m/s

Explanation:

Notation

M= mass of the asteroid

m= mass of the particle moving upward

R= radius

v= escape speed

G= Universal constant

h= distance above the the surface

Part a

For this part we can use the principle of conservation of energy. for the begin the initial potential energy for the asteroid would be $U_i =-\frac{GMm}{R}$ .

The initial kinetic energy would be $\frac{1}{2}mv^2$ . The assumption here is that the particle escapes only if is infinetely far from the asteroid. And other assumption required is that the final potential and kinetic energy are both zero. Applying these we have:

$-\frac{GMm}{R}+\frac{1}{2}mv^2=0$ (1)

Dividing both sides by m and replacing $\frac{GM}{R}$ by $a_g R$

And the equation (1) becomes:

$-a_g R+\frac{1}{2} v^2=0$ (2)

If we solve for v we got this:

$v=\sqrt{2 a_g R}=\sqrt{2x3\frac{m}{s^2}x500000m}=1732.05m/s$

Part b

When we consider a particule at this surface at the starting point we have that:

$U_i=-\frac{GMm}{R}$

$K_i=\frac{1}{2}mv^2$

Considering that the particle is at a distance h above the surface and then stops we have that:

$U_f=-\frac{GMm}{R+h}$

$K_f=0$

And the balance of energy would be:

$-\frac{GMm}{R}+\frac{1}{2}mv^2 =-\frac{GMm}{R+h}$

Dividing again both sides by m and replacing $\frac{GM}{R}$ by $a_g R^2$ we got:

$-a_g R+\frac{1}{2}v^2 =-\frac{a_g R^2}{R+h}$

If we solve for h we can follow the following steps:

$R+h=-\frac{a_g R^2}{-a_g R+\frac{1}{2}v^2}$

And subtracting R on both sides and multiplying by 2 in the fraction part and reordering terms:

$h=\frac{2a_g R^2}{2a_g R-v^2}-R$

Replacing:

$h=\frac{2x3\frac{m}{s^2}(500000m)^2}{2(3\frac{m}{s^2})(500000m)-(1000m/s)^2}- 500000m=250000m$

Part c

For this part we assume that the particle is a distance h above the surface at the begin and start with 0 velocity so then:

$U_i=-\frac{GMm}{R+h}$

$K_i=0$

And after the particle reach the asteroid we have this:

$U_f=-\frac{GMm}{R}$

$K_f=\frac{1}{2}mv^2$

So the balance of energy would be:

$-\frac{GMm}{R+h}=-\frac{GMm}{R}+\frac{1}{2}mv^2$

Replacing again $a_g R^2$ instead of GM and dividing both sides by m we have:

$-\frac{a_g R^2}{R+h}=-a_g R+\frac{1}{2}v^2$

And solving for v:

$a_g R-\frac{a_g R^2}{R+h}=\frac{1}{2}v^2$

Multiplying both sides by two and taking square root:

$v=\sqrt{2a_g R-\frac{2a_g R^2}{R+h}}$

Replacing

$v=\sqrt{2(3\frac{m}{s^2})(500000m)-\frac{2(3\frac{m}{s^2}(500000m)^2}{500000+1000000m}}=1414.214m/s$

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