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3 years ago

5

What is the magnitude of the electric field at a point midway between a −5.0μC and a +5.8μC charge 8.4cm apart? Assume no other

charges are nearby.]

1 answer:

Alex73 [517]3 years ago

4 0

Answer:

Electric Field = E = 36.848 N/C

Explanation:

In accordance with Columb's law

E = k Q1 Q2 / r.r = 8.99 x 10^9 x 5.0 x 10^-6 x 5.8 x 10^-6 / 0.084 x 0.084

= 36948.6961 x 10^-3 = 36.848 N/C

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3 years ago

A thin double convex glass lens with an index of 1.56 while surrounded by air has a 10 cm focal length. If it is placed under wa

bearhunter [10]

Explanation:

Formula which holds true for a leans with radii $R_{1}$ and $R_{2}$ and index refraction n is given as follows.

$\frac{1}{f} = (n - 1) [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

Since, the lens is immersed in liquid with index of refraction $n_{1}$ . Therefore, focal length obeys the following.

$\frac{1}{f_{1}} = \frac{n - n_{1}}{n_{1}} [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

$\frac{1}{f(n - 1)} = [\frac{1}{R_{1}} - \frac{1}{R_{2}}]$

and, $\frac{n_{1}}{f(n - n_{1})} = \frac{1}{R_{1}} - \frac{1}{R_{2}}$

or, $f_{1} = \frac{fn_{1}(n - 1)}{(n - n_{1})}$

$f_{w} = \frac{10 \times 1.33 \times (1.56 - 1)}{(1.56 - 1.33)}$

= 32.4 cm

Using thin lens equation, we will find the focal length as follows.

$\frac{1}{f} = \frac{1}{s_{o}} + \frac{1}{s_{i}}$

Hence, image distance can be calculated as follows.

$\frac{1}{s_{i}} = \frac{1}{f} - \frac{1}{s_{o}} = \frac{s_{o} - f}{fs_{o}}$

$s_{i} = \frac{fs_{o}}{s_{o} - f}$

$s_{i} = \frac{32.4 \times 100}{100 - 32.4}$

= 47.9 cm

Therefore, we can conclude that the focal length of the lens in water is 47.9 cm.

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3 years ago

At resonance, what is impedance of a series RLC circuit? less than R It depends on many other considerations, such as the values

denis23 [38]

Answer:

at resonance impedence is equal to resistance and quality factor is dependent on R L AND C all

Explanation:

we know that for series RLC circuit impedance is given by

$Z=\sqrt{R^2+\left ( X_L-X_C \}right )^2$

but we know that at resonance $X_L=X_C$

putting $X_L=X_C$ in impedance formula , impedance will become

Z=R so at resonance impedance of series RLC is equal to resistance only

now quality factor of series resonance is given by

$Q=\frac{\omega L}{R}=\frac{1}{\omega CR}=\frac{1}{R}\sqrt{\frac{L}{C}}$ so from given expression it is clear that quality factor depends on R L and C

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3 years ago

Two long, parallel, current-carrying wires lie in an xy-plane. The first wire lies on the line y = 0.300 m and carries a current

Marrrta [24]

Answer:

Explanation:

Since the wires attract each other , the direction of current will be same in both the wires .

Let I be current in wire which is along x - axis

force of attraction per unit length between the two current carrying wire is given by

$\frac{\mu_0}{4\pi}$ x $\frac{2 I_1\times I_2}{d}$

where I₁ and I₂ are currents in the wires and d is distance between the two

Putting the given values

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I₂ = 16.76 A

Current in the wire along x axis is 16.76 A

To find point where magnetic field is zero due the these wires

The point will lie between the two wires as current is in the same direction.

Let at y = y , the neutral point lies

k 2 x $\frac{16.76}{y}$ = k 2 x $\frac{25.5}{.3-y}$

25.5y = 16.76 x .3 - 16.76y

42.26 y = 5.028

y = .119

= .12 m

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3 years ago

.A box falls to the ground from a delivery truck traveling at 30 m/s. After hitting the road, it slides 45 m to

Romashka-Z-Leto [24]

Answer:

t = 3 seconds

Explanation:

Given that,

Initial speed, u = 30 m/s

Final speed, v = 0

It slides 45 m to rest.it take the box to come to rest

We need to find how long it take the box to come to rest.

Let a be the acceleration and t is time.

$v^2-u^2=2ad\\\\a=\dfrac{v^2-u^2}{2d}\\\\a=\dfrac{(30)^2-u^2}{2(45)}\\\\=10\ m/s^2$

Now finding time.

$t=\dfrac{v-u}{a}\\\\t=\dfrac{30-0}{10}\\\\t=3\ s$

So, the required time is 3 seconds.

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3 years ago