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goldenfox [79]
3 years ago
10

Two long straight parallel lines of charge, #1 and #2, carry positive charge per unit lengths of λ1 and λ2, respectively. λ1 &gt

; λ2. The locus of points where the electric field is zero in this case is
Physics
1 answer:
Stella [2.4K]3 years ago
3 0

Answer:

Somewhere between the two wires, but closer to the wire carrying λ₂

Explanation:

Electric Field for a point at distance x from an electric charge Q is Ef = K*Q/x².

Electric Fied due to an electric charge is a vector and its direction is  such that  if we place a positive charge in the point it will be rejected ( equal sign charge repulse each other and different attract each other)

According to that previous explanation, it is no possible two have Ef=0 out of the two wires region, since above the upper wire and below the lower wire we have to add the two electric fields (both have the same direction). Therefore we only have possibilities of Ef = 0 inside the two wires, where the repulsion produced over a positive charge due to the two wires  are opposite

In the particular case in which λ₁ and λ₂ are equals then all the points exactly in the middle of d (distance between the two wires ) will have Ef =0.

As we can see at the beginning of the step by step explanation Electric field is proportional to the electric charge, or for a bigger charge, bigger Ef (keeping constant distance). In our case λ₁ >λ₂ then E₁ (Electric field produced by a wire carrying λ₁ will be bigger than (Electric field produced by wire carrying λ₂ at the middle way between the wires.

But for points closer to wire with λ₂  ( where E₂ is bigger than E₁ ) we will surely find an appropriate distance  to get equals E and then Ef = 0

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Un péndulo de 4 m de longitud tiene una frecuencia de 5Hz. Calcular la longitud de otro péndulo que en el mismo lugar tiene una
Murrr4er [49]

Conociendo la longitud y frecuencia de un péndulo, queremos encontrar la longitud de otro pendulo de tal forma que tenga otra frecuencia.

Veremos que la longitud del nuevo péndulo debe ser 6.25m

Sabemos que un péndulo de 4m de longitud tiene una frecuencia de 5Hz.

La frecuencia de un péndulo está dada por:

f = \frac{1}{2*\pi} *\frac{g}{l}

Donde g es la aceleración gravitatoria y l es la longitud del péndulo, remplazando los datos que tenemos en esa ecuación obtenemos:

5 Hz = \frac{1}{2*3.14} *\sqrt{\frac{g}{4m} } \\\\(5Hz*2*3.14)^2*4m = g = 3,943.8 m/s^2

Ahora debemos encontra la longitud de tal forma que la frecuencia sea 4Hz, entonces debemos resolver:

4Hz = \frac{1}{2*3.14} *\sqrt{ \frac{3943.8m/s^2}{l} }\\\\4hz*2*3.14 = \sqrt{ \frac{3943.8m/s^2}{l} }\\\\l =\frac{3943.8m/s^2}{ (4hz*2*3.14)^2}  = 6.25m

La longitud del nuevo péndulo deve ser 6.25m

Sí quieres aprender más, puedes leer:

brainly.com/question/23483504

8 0
2 years ago
You get pulled over for running a red light. You explain to the police officer that the light appeared green to you due to the D
Ludmilka [50]

Answer:

$ 3085713685.71

Explanation:

\lambda_0 = Actual wavelength = 700 nm

\lambda = Changed wavelength = 500 nm

Let the wavelength of red color be 700 nm and green be 500 nm

Change in wavelength is

\Delta \lambda=700-500\\\Rightarrow \theta\lambda=200\ nm

We have the relation

\dfrac{\Delta\lambda}{\lambda_0}=\dfrac{v}{c}\\\Rightarrow v=\dfrac{\Delta\lambda}{\lambda_0}c\\\Rightarrow v=\dfrac{200}{700}\times 3\times 10^8\\\Rightarrow v=85714285.7143\ m/s

The speed of the vehicle is 85714285.7143 m/s

\dfrac{85714285.7143\times 3600}{1000}=308571428.571\ km/h

By how much was the car speeding

308571428.571-60=308571368.571\ km/h

The number of 10 km/h in the above speed

308571368.571\times\dfrac{1}{10}=30857136.8571

Cost of the ticket

30857136.8571\times 100=\$ 3085713685.71

The cost of the ticket is $ 3085713685.71

8 0
3 years ago
Read 2 more answers
Air at 38oC and 97% relative humidity is to be cooled to 14oC and fed into a plant area at a rate of 510 m3/min. Calculate the r
bekas [8.4K]

Answer:

mass flow rate at  water condenses is 36.72 kg/min

Explanation:

given data

temperature t1 = 38°C

temperature t2 = 14°C

humidity ∅= 97 % = 0.97

rate v = 510 m³/min

to find out

mass flow rate at  water condenses

solution

by gas equation we find here mass flow rate  that is

pv = mRT

put here value and p is 0.066626 bar at 38°C and find m

m = 0.06626 × 10^{5} × 510  / 287×311

m = 37.85 kg/min

so at water condenses mass flow rate is express as

∅ = M / m

Mass flow rate M = ∅ × m

M = 0.97 × 37.85

mass flow rate = 36.72 kg/min

so mass flow rate at  water condenses is 36.72 kg/min

8 0
3 years ago
Is the momentum of a 0.1 kg Mass moving with a velocity of 5 miles per second West
dem82 [27]
<span>Momentum equals Mass x Velocity
Mass equals 0.1kg
Velocity equals 5m/s

So the momentum has to = 0.1 x 5 = 0.5kgm/s

I hope this helped
</span>
6 0
3 years ago
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Answer:

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5 0
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