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3 years ago

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Two charged particles, with charges q1 = q and q2 = 4q are located at a distance d = 2cm apart on the x axis A third charged par

ticle, with charge q3 = q is placed on the x axis such that the magnitude of the force that charge 1 exerts on charge 3 is equal to the force that charge 2 exerts on charge 3. Find the position of charge 3 when q = 10n C

1 answer:

weqwewe [10]3 years ago

3 0

Answer:

0.667 cm

Explanation:

|F_1on3| = |F_2on3|

k×q1×q3/x^2 = k×q2×q3/(d-x)^2

q1/x^2 = q2/(d-x)^2

(d-x)^2/x^2 = q2/q1

(d-x)/x = sqrt(q2/q1)

(2 - x)/x = sqrt(4×q/q)

(2 - x)/x = 2

2 - x = 2x

2 = 3x

x = 2/3

x= 0.667 cm

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Answer:

The deceleration is $a = -0.7273 \ m/s^2$

Explanation:

From the question we are told that

The distance of the car from the crossing is $d = 360 \ m$

The speed is $u = 16 \ m/s$

The reaction time of the engineer is $t = 0.53 \ s$

Generally the distance covered during the reaction time is

$d_r = u * t$

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Generally distance of the car from the crossing after the engineer reacts is

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=> $D = 360 - 8.48$

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Generally from kinematic equation

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So

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3 years ago

A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar

denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

Explanation:

Given that,

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$\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}$

Put the value into the formula

$\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}$

$\beta=8.86\times10^{9}$

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Using formula of tunnel barrier

$T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}$

Put the value into the formula

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}$

$T=5.45\times10^{-4}$

(b). We need to calculate the tunnel probability for width 1.00 nm

$T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}$

$T=7.74\times10^{-8}$

Hence, The tunnel probability for 0.5 nm and 1.00 nm are $5.45\times10^{-4}$ and $7.74\times10^{-8}$ respectively.

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A 2.0-cm-diameter, 15-cm-long solenoid is tightly wound with one layer of wire. A 2.5 A current through the wire generates a 3.0

sweet [91]

Answer:

d = 1.047 mm

Explanation:

given,

diameter of the wire = 2.0-cm

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$B = \dfrac{\mu_0 N I}{L}$

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$\ d = \dfrac{4\pi \times 10^{-7}\times 0.15\times 2.5}{0.45 \times 10^{-3}}$

d = 1.047 x 10⁻³ m

d = 1.047 mm

diameter of the wire is d = 1.047 mm

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Arlecino [84]

Answer:

The average force exerted by the water on the ground is 17.53 N.

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the average force exerted by the water on the ground;

F = mg

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F = 17.53 N

Therefore, the average force exerted by the water on the ground is 17.53 N.

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3 years ago

A vector is given in terms of its magnitude, v=8.7m/s and its direction, 256 degrees standard position. a) what is this vector i

Wittaler [7]

Answer:

Explanation:

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