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bazaltina [42]
3 years ago
9

Along the line connecting the two charges, at what distance from the charge q1 is the total electric field from the two charges

zero? Express your answer in terms of some or all of the variables s, q1, q2 and k =14πϵ0. If your answer is difficult to enter, consider simplifying it, as it can be made relatively simple with some work.
Physics
1 answer:
Nostrana [21]3 years ago
8 0

Answer:

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

Explanation:

Here two charges are placed at distance "d" apart

now the net value of electric field at some position between two charges will be ZERO

so we will have

electric field due to charge 1 = electric field due to charge 2

E_1 = E_2

Let the position where net field is zero will lie at distance "r" from q1

\frac{kq_1}{r^2} = \frac{kq_2}{(d-r)^2}

now we will have

\frac{(d - r)^2}{r^2} = \frac{q_2}{q_1}

now square root both sides

\frac{d}{r} - 1 = \sqrt{\frac{q_2}{q_1}}

now we have

\frac{d}{r} = \sqrt{\frac{q_2}{q_1}} + 1

so we have

r = d (\frac{\sqrt {q_1}}{\sqrt{q_1} + \sqrt{q_2}})

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lozanna [386]

Answer:velocity=93.12m/s

Explanation:

We shall use conservation of energy to solve this problem

we have

Potential energy of King Kong at top of building = His kinetic energy at the bottom

Potential energy of an object = m\times g\times h = 20000 kg

Where m is mass of an object(20000 kg)

g is accleration due to gravity = 9.81m/s^{2}

h is the height above the surface of earth = 1450 feet = 441.96 meters

Applying values we get

(P.E)_{TOP}=(20000\times 9.81\times 441.96) Joules\\\\(P.E)_{TOP}= 86712.552KiloJoules......................(i)

Now Kinetic energy is given by

(K.E)_{Bottom}=\frac{1}{2}mv^{2}.......................(ii)

Equating i and ii we get

86712.552 =\frac{1}{2}mv^{2}

v=\sqrt{\frac{2E}{m}}

Applying values we get

v=\sqrt{\frac{2\times 86712.552\times 10^{3}}{20\times 10^{3}}}\\\\v= 93.12m/s

3 0
3 years ago
Salmon often jump waterfalls to reach their
natta225 [31]

Answer:

5.0 m/s

Explanation:

The horizontal motion of the salmon is uniform, so the horizontal component of the salmon's velocity is constant and it is

v_x = u cos \theta

where u is the initial speed and \theta=37.7^{\circ}. The horizontal distance travelled by the salmon is

d=v_x t = (ucos \theta)t

where d = 1.95 m and t is the time needed to reach the final point.

Re-arranging for t,

t=\frac{d}{v_x}=\frac{d}{u cos \theta} (1)

Along the vertical direction, the equation of motion is

y=h+u_y t -\frac{1}{2}gt^2

where:

y = 0.311 m is the final height reached by the salmon

h = 0 is the initial height

u_y = u sin \theta is the vertical component of the initial velocity of the salmon

g=9.81 m/s^2 is the acceleration of gravity

t is the time

Substituting t as found in eq.(1), we get the equation

y=(u sin \theta) \frac{d}{u cos \theta}- \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}

and we can solve this formula for u, the initial speed of the salmon:

y=d tan \theta - \frac{1}{2}g\frac{d^2}{u^2 cos^2 \theta}\\\\u=\sqrt{\frac{gd^2}{2(dtan \theta -y)cos^2 \theta}}=\sqrt{\frac{(9.81)(1.95)^2}{2((1.95)(tan 37.7^{\circ}) -0.311)cos^2 37.7^{\circ}}}=5.0 m/s

5 0
3 years ago
The speed of sound in aluminum is 5200 m/s. Can you hear a sound with a
romanna [79]

Answer:

v = wavelength * frequency

frequency = 5200 m/s / .2 m = 26000 / sec

20,000 / sec is optimistic for the upper frequency of human hearing

So 26,000 is above the hearing range for human ears

5 0
2 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
What are the equivalent celsius and kelvin temperatures of -128 f?
Gre4nikov [31]

Answer:

128 Kelvin = 128 - 273.15 = -145.15 Celsius. Temperature conversion chart Sample temperature conversions 103.55 Kelvin to degrees Fahrenheit 39.82 degrees Fahrenheit to Kelvin

Explanation:

hope this helps have a good day

6 0
2 years ago
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