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vladimir1956 [14]
2 years ago
9

The electric potential at the dot in the figure is 3160 V. What is charge q?

Physics
1 answer:
PSYCHO15rus [73]2 years ago
5 0

Hi there!

Recall the equation for electric potential of a point charge:

V = \frac{kQ}{r}

V = Electric potential (V)
k = Coulomb's Constant(Nm²/C²)

Q = Charge (C)
r = distance (m)

We can begin by solving for the given electric potentials. Remember, charge must be accounted for. Electric potential is also a SCALAR quantity.

Upper right charge's potential:

V = \frac{(8.99*10^9)(-5 * 10^{-9})}{0.04} = -1123.75 V

Lower left charge's potential:

V = \frac{(8.99*10^9)(5*10^{-9})}{0.02} = 2247.5 V

Add the two, and subtract from the total EP at the point:

3160 + 1123.75 - 2247.5 = 2036.25


The remaining charge must have a potential of 2036.25 V, so:

2036.25 = \frac{(8.99*10^9)(Q)}{\sqrt{0.02^2 + 0.04^2}}\\\\2036.25 = \frac{(8.99*10^9)Q}{0.0447} \\\\Q = 0.000000010127 = \boxed{10.13nC}


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Answer:

Minimum uncertainty in velocity of a proton,\Delta v\ge 3.15\times 10^7\ m/s      

Explanation:

It is given that,

A proton is confined to a space 1 fm wide, \Delta x=10^{-15}\ m

We need to find the minimum uncertainty in its velocity. We know that the Heisenberg Uncertainty principle gives the uncertainty between position and the momentum such that,

\Delta p.\Delta x\ge \dfrac{h}{4\pi}

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m \Delta v.\Delta x\ge \dfrac{h}{4\pi}

\Delta v\ge \dfrac{h}{4\pi m\Delta x}

\Delta v\ge \dfrac{6.63\times 10^{-34}}{4\pi \times 1.67\times 10^{-27}\times 10^{-15}}

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So, the minimum uncertainty in its velocity is greater than 3.15\times 10^7\ m/s. Hence, this is the required solution.

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PLEASE HELP
Anvisha [2.4K]

The speed of the rock at 20 m is 34.3 m/s

Explanation:

We can solve this problem by using the law of conservation of energy: the mechanical energy of the rock, sum of its potential energy + its kinetic energy) must be conserved in absence of air resistance. So we can write:

U_i +K_i = U_f + K_f

where :

U_i is the initial potential energy

K_i is the initial kinetic energy

U_f is the final potential energy

K_f is the final kinetic energy

The equation can also be rewritten as  follows:

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 100 kg is the mass of the rock

g=9.8 m/s^2 is the acceleration of gravity

h_i = 80 is the initial height

u = 0 is the initial speed  (the rock starts at rest)

h_f = 20 m is the final height of the rock

v is the final speed when h = 20 m

And solving for v, we find:

v=\sqrt{2g(h_i-h_f)}=\sqrt{2(9.8)(80-20)}=34.3 m/s

Learn more about kinetic energy and potential energy here:

brainly.com/question/6536722

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A hydrogen discharge lamp emits light with two prominent wavelengths: 656 nm (red) and 486 nm (blue). The light enters a flint-g
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The angle between the red and blue light is 1.7°.

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Angle = 37°

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n_{r}=1.572

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We need to calculate the angle for red wavelength

Using Snell's law,

n_{r}\sin\theta_{i}=n_{a}\sin\theta_{r}

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Put the value into the formula

1.587\sin37=1\times\sin\theta_{b}

\theta_{b}=\sin^{-1}(\dfrac{1.587\sin37}{1})

\theta_{b}=72.7^{\circ}

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Using formula of angle

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Put the value into the formula

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\Delta \theta=1.7^{\circ}

Hence, The angle between the red and blue light is 1.7°.

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