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3 years ago

14

1When you look through a magnifying glass, the objects you are looking at

1 answer:

Vadim26 [7]3 years ago

8 0

convex lenses.

eyeglasses with convex lenses increase refraction, and accordingly reduce the focal lenght

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A thin ring of radius 73 cm carries a positive charge of 610 nC uniformly distributed over it. A point charge q is placed at the

kow [346]

Answer:

q = - 93.334 nC

Explanation:

GIVEN DATA:

Radius of ring 73 cm

charge on ring 610 nC

ELECTRIC FIELD p FROM CENTRE IS AT 70 CM

E = 2000 N/C

Electric field due tor ring is guiven as

$E = \frac{KQx}{[x^2+ R^2]^{3/2}}$

$E = \frac{9\time 10^9 \times 610\times 10^[-9} 0.70}{(0.70^2 + 0.73^2)^{3/2}}$

E1 = 3714.672 N/C

electric field due to point charge q

$E =\frac[kq}{x^2}$

$E = \frac{9\times 10^9 \times q}{0.70^2}$

$E2 = 1.837\times 10^{10}\times q$

now the eelctric charge at point P is

E = E1 + E2 $2000 = 3714.672 + 1.837\times 10[10} \times q$

solving for q

q = - 93.334 nC

7 0

4 years ago

In some proton accelerators, proton beams are directed toward each other for head-on collisions. Suppose that in such an acceler

ipn [44]

Answer:

a) 0.9995c

b) 5641MeV

c) 91670 MeV

Explanation:

(a) The speed of approach is given by the formula:

$u=\frac{v_1+v_2}{1+\frac{v_1v_2}{c}}=\frac{2(0.9898c)}{1+\frac{(0.9898)^2c^2}{c^2}}=0.99995c$

(b) the kinetic energy is given by:

$E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}-1]$

by replacing c=3*10^8m/s, m_0=1.67*10^{-27}kg we obtain:

$E_k=5641MeV$

(c) in the rest frame of the other proton we have:

$E_k=m_0c^2[\frac{1}{\sqrt{1-\frac{u^2}{c^2}}}-1]$

by replacing we get

$E_k=91670MeV$

hope this helps!!

7 0

3 years ago

fiasKO [112]

checlknthfd sissExplanation: 2+3 ls g g fg

3 0

3 years ago

A 1 uF parallel-plate capacitor is charged to 12 V and then disconnected from the voltage supply. The plate separation distance

nlexa [21]

Answer:

22

Explanation:

6 0

3 years ago

The ball will oscillate along the z axis between z=dz=d and z=−dz=−d in simple harmonic motion. What will be the angular frequen

Eddi Din [679]

Answer:

$\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }$

Explanation:

Additional information:

<em>The ball has charge </em> $-q_0$ <em>, and the ring has positive charge </em> $+Q$ <em> distributed uniformly along its circumference. </em>

The electric field at distance $z$ along the z-axis due to the charged ring is

$E_z= \dfrac{kQz}{(z^2+a^2)^{3/2}}.$

Therefore, the force on the ball with charge $-q_0$ is

$F=-q_oE_z$

$F=- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}$

and according to Newton's second law

$F=ma=m\dfrac{d^2z}{dz^2}$

substituting $F$ we get:

$- \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=m\dfrac{d^2z}{dz^2}$

rearranging we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(z^2+a^2)^{3/2}}=0$

Now we use the approximation that

$z^2+a^2\approx a^2$ <em>(we use this approximation instead of the original </em> $d^2+a^2\approx a^2$ <em> since </em> $z$ <em>, our assumption still holds )</em>

and get

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{(a^2)^{3/2}}=0$

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Qz}{a^{3}}=0$

Now the last equation looks like a Simple Harmonic Equation

$m\dfrac{d^2z}{dz^2}+kz=0$

where

$\omega=\sqrt{ \dfrac{k}{m} }$

is the frequency of oscillation. Applying this to our equation we get:

$m\dfrac{d^2z}{dz^2}+ \dfrac{kq_0Q}{a^{3}}z=0\\\\m=m\\\\k= \dfrac{kq_0Q}{a^{3}}$

$\boxed{\omega = \sqrt{\dfrac{kq_0Q}{ma^3} }}$

5 0

4 years ago