# Electricity and magentism questions.

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This is for Jhansi, Utter Pradesh, India OTA ID#103642 ONLY no others please....

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1.After bungee jumping off a bridge, a person begins to swing as a pendulum, the rope being 77m long. The person's speed is 14m/s as it passes through it lowest point. A). Find the height to which it rises above this position before stopping and B). What angle does the pendulum then make with the vertical?

2.a mine explodes at sea, and there is an interval of 5.0sec between the arrival of the sound through the water and its arrival through the air at a nearby ship. How far away is the mine from the ship?

3. The Earth has a net charge that produces an electric field of approximately 150N/C downward at its surface. IF you think of the charge as being all concentrated at the center of the earth as a point charge, what is the sign and the magnitude of the charge, and what acceleration will the field produce on a free electron near the earth's surface?

4. Two charged parallel plates, at equal distance above and below the x-axis create a downward electric field of strength 1000N/C. The plates are 4cm long. An electron is shot along the x-axis from one end of the plates towards the other, and is deflected upward by the field. At the end of the plates there is a screen perpendicular to the x-axis. If the speed of the electron is 3*10^6m/s, how far above the x-axis does it strike the screen? What is its vertical velocity? At what angle does it exit?

5.In Rutherford's famous scattering experiments that led to the planetary model of the atom, alpha particles(having charges of +2e and masses of 6.6*10^-27kg) were fired toward a fixed gold nucleus with charge +79e. An alpha particle, initially very far from the gold nucleus, is fired at 2.0*10^7m/s directly toward the gold nucleus. How close to the gold nucleus does the alpha particled get before turning around?

6.An electron gun consists of two vertical parallel plates with an opening in one of them. The gun accelerates electrons using a potential difference between the plates shooting them form the (--) plate twards the (++) plate and through the opening. Calculate the final speed of a free electron accelerated from rest through a potential difference of 100V.

HERE IS WHAT I GOT FROM ANOTHER OTA.... As you can see I need 1 and 4

could answer only four questions within the time limit of two hours. Sorry.

2)

The speed of sound in sea water will depend on the temperature,salinity and such parameters. We need the velocity of sound in sea water to do this problem. Let it be V (capital letter V)

Speed of sound in air v = 330 m/s

Let the sound arrives through water in x seconds, then it will arrive through air in x+5 seconds.

Since the distance D is the same, we can write, (Distance=vel x time)

D = 330 * (x +5) through air and,

D = V * x through water, take V=1500 m/s and we can equate these two equations

V x = 330 (x+5)

From this, (after substitutng V=1500m/s) we get x as,

x = 1650/1160 Second = 1.42 sec

Now, we have D = 330 * (x + 5) = 330 * (1.42+5)

= 2118.6 meters (Answer)

3)

Since the field direction is downward the surface, the earth is a negatively charged sphere. (answer)

Since the electric field strength is given as 150N/C (150 Newton on 1 coulomb charge on the surface),

we get, the force on an electron of charge e = 1.6 x 10^-19 Coulomb will be,

F = 150 x 1.6 x 10^-19 = 240 x 10^-19 Newton

ie, F = k Q1 q2/R^2 = 240 x 10^-19 where K=9 x 10^9

Let q2 = 1.6 x 10^-19, the charge of the electron and R = 6371x10^3 m

therefore, Q1 = 240 x 10^-19 x (6371x10^3)^2/(1.6 x 10^-19)x9x10^9

Q2 = 67649.401 Coulomb (answer)

Also, F = ma with m = 9.1x10^-31Kg, the electronic mass, we get

a = F/m = 240x10^-19/9.1x10^-31 = 26.37 x 10^12 m/s^2 (answer)

5) The alpha particles will come to a point at which the electrostatic repulsion from the nucleus just balances the kinetic energy of the alpha particles

Ke of alpha particles (1/2) mv^2 = 0.5 x 6.6 x 10^-27 x (2x10^7)^2

= 13.2x10^-13 Joule

Electrostatic repulsive force F = K q1 Q2/R^2

q1= 2 x 1.6 x 10^-19 C = 3.2x10^-19 C

Q2= 79x 1.6 x 10^-19 C = 126.4 x 10^-19 and K = 9 x 10^9

The point of nearest approach

is when (1/2) mv^2 = K q1 Q2/R^2

Substitute the values to get R,

R =[ 9 x 10^9 x 3.2 x 10^-19 x 126.4 x 10^-19/13.2x10^-13 ]^0.5

= 16.6 x 10^-8 m (Answer)

6) The Kinetic Energy acquired by an electron accelerated through a potential difference of V volts is given by the formula

K E = eV = 100 eV

But we have, 1 eV = 1.6 x 10^-19 C and electronic mass=9.1 x 10^-31 kg

KE = (1/2)mv^2 = 100 x 1.6 x 10^-19 Joules

Velocity v = (2KE/m)^(1/2) = (320/9.1)^(1/2) x 10^6 m/s

= 5.93 x 10^6 m/s (Answer

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1.) A.)

<br>In oscillations the K.E. be maximum at mean position and zero at extreme and P.E. maxm at extremum and least at mean point. In the case of simple pendulam the lowest point is equivalent to mean position.

<br>by energy law:

<br>The K.E at lowest point = P.E. at heighest point(

<br>(1/2)* m*v^2 = m*g*hmax

<br>=> hmax = v^2/2g = 14 * 14/(2*9.8) = 10 m Ans

<br>

<br>B.)

<br> Look at the attached figure:

<br>cos(theta) = 67/77 => theta = 29.52 degree Ans

<br>

<br>

<br>2.)actually

<br>v_w*t = v_a*(t+5)

<br>=>1500*t = 330 * (t +5)

<br>=> (1500-330)*t = 330*5 = 1650

<br>=> t = 1650/1170 = 1.41 sec (though he has calculated 1650/1160)

<br>and ...

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