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2 years ago

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In the Bohr model of the hydrogen atom, the electron moves in a circular orbit of radius 5.3×10−11m with a speed of 2.2×106m/s.

Find the direction of the electric field that the electron produces at the location of the nucleus (treated as a point).

1 answer:

lisov135 [29]2 years ago

6 0

Answer:

The magnitude of the electric field is 512171146711.7046 N/C.

The direction is toward the electron as electron has negative charge.

Explanation:

e = Charge of electron = $1.6\times 10^{-19}\ C$

r = Radius of circular orbit = $5.3\times 10^{-11}\ m/s$

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

The Electric field is given by

$E=\frac{e}{4\pi\epsilon_0 r^2}\\\Rightarrow B=\frac{1.6\times 10^{-19}}{4\pi\times 8.85\times 10^{-12}\times (5.3\times 10^{-11})^2}\\\Rightarrow E=512171146711.7046\ N/C$

The magnitude of the electric field is 512171146711.7046 N/C.

The direction is toward the electron as electron has negative charge.

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Explanation:

Formula to calculate electric field because of the plate is as follows.

E = $\frac{\sigma}{2 \times \epsilon_{o}}$

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a = $\frac{1.6 \times 10^{-19} \times 1.18 \times 10^{5}}{1.67 \times 10^{-27}}$

= $1.13 \times 10^{13} m/s^2$

According to the third equation of motion,

$v^{2} = 2 \times a \times d$

or, d = $\frac{v^{2}}{2d}$

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= 0.254 m

Thus, we can conclude that the proton will travel 0.254 m before reaching its turning point.

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Answer:

$E=88200\ J$

Explanation:

Given:

mass of car, $m=1500\ kg$

distance of skidding after the application of brakes, $d=15\ m$

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<u>So, the energy dissipated during the skidding of car:</u>

<em>Frictional force:</em>

$f=\mu_k.N$

where N = normal reaction by ground on the car

$f=0.4\ties 1500\times 9.8$

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Correct option: A

An object remains at rest until a force acts on it.

As the water moves faster, it applies greater force on the sediment, which over comes the frictional forces between the bed and the sediment. So, when the river flows faster, more and larger sediment particles are carried away. When the flow slows down, the river couldn't apply enough force on the larger sediments which can overcome the frictional force between the sediment and the river bed. So, the net force on the heavier particles become zero. Hence, the heavier particles of the load will settle out.

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It is definitely letter D. <span>A1 and B1 are like poles, but there is not enough information to tell whether they are north poles or south poles.

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2 years ago

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A projectile is launched at an angle of 30° and lands 20 s later at the same height as it was launched. (a) What is the initial

Elina [12.6K]

Answer:

a)Initial speed of the projectile = 196.2 m/s

b)Maximum altitude = 490.5 m

c) Range of projectile = 3398.28 m

d) Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

Explanation:

Time of flight of a projectile is given by the expression,

$t=\frac{2usin\theta}{g}$

Here θ = 30° and t = 20 s

a) $t=\frac{2usin\theta}{g}\\\\20=\frac{2\times usin30}{9.81}\\\\u=196.2m/s$

Initial speed of the projectile = 196.2 m/s

b) Maximum altitude is given by

$H=\frac{u^2sin^2\theta}{2g}=\frac{196.2^2\times sin^230}{2\times 9.81}=490.5m$

Maximum altitude = 490.5 m

c) Range of projectile is given by

$R=\frac{u^2sin2\theta}{g}=\frac{196.2^2\times sin(2\times 30)}{9.81}=3398.28m$

Range of projectile = 3398.28 m

d) Horizontal velocity = ucosθ = 196.2 x cos 30 = 169.91 m/s

Vertical velocity = usinθ = 196.2 x sin 30 = 98.1 m/s

We have equation of motion s = ut + 0.5 at²

Horizontal motion

u = 169.91 m/s

a = 0 m/s²

t = 15 s

Substituting

s = 169.91 x 15 + 0.5 x 0 x 15² = 2548.71 m

Vertical motion

u = 98.1 m/s

a = -9.81 m/s²

t = 15 s

Substituting

s = 98.1 x 15 + 0.5 x -9.81 x 15² = 367.88 m

$\texttt{Total displacement =}\sqrt{2548.71^2+367.88^2}=2575.12m$

Displacement from the point of launch to the position on its trajectory at 15 s = 2575.12 m

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