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barxatty [35]
3 years ago
8

An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

uch work is required to bring an additional electron from infinity to the origin? Express your answer with the appropriate units. Part B If, instead of the second electron coming in from infinity, it is initially at x = + 20.00 m on the axis and is given an initial velocity of 100.0 m/s toward the origin, does it reach the origin? Part C How close to the origin does it come?
Physics
1 answer:
r-ruslan [8.4K]3 years ago
4 0

PART A)

Electrostatic potential at the position of origin is given by

V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}

here we have

q_1 = 1.6 \times 10^{-19} C

q_2 = -1.6 \times 10^{-19} C

r_1 = r_2 = 1 m

now we have

V = \frac{Ke}{r} - \frac{Ke}{r}

V = 0

Now work done to move another charge from infinite to origin is given by

W = q(V_f - V_i)

here we will have

W = e(0 - 0) = 0

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}

U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}

U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})

U = 1.15\times 10^{-30}

Now we know

KE = \frac{1}{2}mv^2

KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2

KE = 4.55 \times 10^{-27} kg

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

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A charged particle is moving perpendicular to a magnetic field in a circle with a radius r. An identical charge particle enters
emmasim [6.3K]

Answer:

<em>The second particle will move through the field with a radius greater that the radius of the first particle</em>

Explanation:

For a charged particle, the force on the particle is given as

F = \frac{mv^{2} }{r}

also recall that work is force times the distance traveled

work = F x d

so, the work on the particle = F x d,

where the distance traveled by the particle in one revolution = 2\pi r

Work on a particle = 2πrF = 2\pi mv^{2}

This work is proportional to the energy of the particle.

And the work is also proportional to the radius of travel of the particles.

Since the second particle has a bigger speed v, when compared to the speed of the first particle, then, the the second particle has more energy, and thus will move through the field with a radius greater that the radius of the first particle.

3 0
3 years ago
How do you calculate acceleration
Eduardwww [97]

Answer: acceleration is equal to the change in velocity per unit time in seconds.

a= ∆v / t = vf - vi / t

Explanation: change in velocity or ∆v can be expressed as (vf - vi)

7 0
3 years ago
A 900 kg car moves around a 500 m radius curve at 25.0 m/s. What is the centripetal force on
Gnesinka [82]

Answer:

9375 N

Explanation:

From the question,

Centripetal force (F) = mv²/r.................. Equation 1

Where m = mass of the car, v = velocity of the car, r = radius of the curve.

Given: m = 900 kg, r = 600 m, v = 25 m/s

Substitute these values into equation 1

F = (900×25²)/600

F = 9375 N.

Hence the centripetal force on the car is 9375 N

3 0
3 years ago
A camcorder has a power rating of 15 watts. If the output voltage from its battery is 5 volts, what current does it use?
MrMuchimi

Formula

W = E * I

Givens

E = 5 volts

W = 15 watts

I = ?

Solution

W = E * I

15 = 5 * I

15/5 = I

I = 3 amps.   Answer

5 0
3 years ago
A bullet with a mass 2.25g is fired up into the air with a velocity of 187.5 m/s. What is the maximum height of the bullet
Ksivusya [100]

Answer:

1793.7m

Explanation:

From the principle of conservation of energy; the kinetic energy substended by the object equals the potential energy sustain by the object when it gets to its maximum position.

Now the kinetic energy; is

K.E = 1/2 × m × v2

Where m is mass

v is velocity

Hence.

K.E = 1/2 × 2.25 × (187.5)^2

Now this should be same with the potential energy which is given as;

P.E = m× g× h

Where m is mass of object

g is acceleration of free fall due to gravity = 9.8m/S2

h is maximum height substain by the object.

Hence P.E = 2.25 × 9.8 × h

From the foregoing analysis of energy conversation it implies;

1/2 × 2.25 × (187.5)^2 =2.25 × 9.8 × h

=> 1/2 × (187.5)^2 = 9.8 × h

=>1/2 × (187.5)^2 / 9.8 = h

=> 1793.69m = h

h= 1793.69m

h =1793.7m to 1 decimal place

3 0
3 years ago
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