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2 years ago

14

Find the distance from a point charge q=100nC where the field intensity is equal to E=6kN/C. please include description of the r

esult

1 answer:

madam [21]2 years ago

4 0

Answer: 0.3872m

Explanation:

q= 100nC --> 100x10^-9 C

k= 9x10^9 Nm^2/C^2

E= 6kN/C --> 600 N/C

r=?

$E= K\frac{q}{r^{2} }$ --> $r=\sqrt{\frac{kq}{E} }$ Despejas "r"

Resuelves

<h3> $r=\sqrt{\frac{(9x10^9 Nm^2/C^2)(100x10x^{-9} C)}{6000N/C} }$ (la x es por, no es una variable)</h3><h3>r= 0.3872983346m</h3>

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Gala2k [10]

Answer:

I think, (remember think) it might be 2.0 m/s

Explanation:

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2 years ago

Your car's speedometer works in much the same way as its odometer, except that it converts the angular speed of the wheels to a

brilliants [131]

Answer:

Speed will be 30810 rpm

Explanation:

We have given diameter of the tire d = 24 inch

So radius $r=\frac{d}{2}=\frac{24}{2}=12imch$

We have given linear velocity v = 35 mph

We know that linear velocity is given by $v=\omega r$

$35=\omega \times 12$

$\omega =\frac{35}{12}\times \frac{63360}{60}=3080rad/min$

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3 years ago

A steel marble with 0.05 kg of mass starts from rest and rolls down a ramp. It travels 0.25 m in 1.2 seconds. Find the force act

emmasim [6.3K]

After finding acceleration, it is found that 0.02 N of force is acting on the marble

<h3>What is Force ?</h3>

Force can simply be defined as a pull or push. It is the product of mass and acceleration of the object. It is a vector quantity and it is measured in Newton.

Given that a steel marble with 0.05 kg of mass starts from rest and rolls down a ramp. It travels 0.25 m in 1.2 seconds.

The parameters to consider are;

m = 0.05 Kg

u = 0

s = 0.25 m

t = 1.2 s

a = ?

F = ?

Before we find the force acting on the marble, let us first find the acceleration by using the formula: s = ut + 1/2at²

Substitute all the parameters into the formula

0.25 = 0 + 1/2 × a × 1.2²

0.25 = 1/2 × a × 1.44

0.25 = 0.72a

a = 0.25/0.72

a = 0.35 m/s²

The force acting on the marble will be ;

F = ma

F = 0.05 × 0.35

F = 0.017

F = 0.02 N

Therefore, the force acting on the marble is 0.02 N

Learn more about Force here: brainly.com/question/388851

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1 year ago

A photon of wavelength 2.78 pm scatters at an angle of 147° from an initially stationary, unbound electron. What is the de Brogl

Elena-2011 [213]

Answer:

2.07 pm

Explanation:

The problem given here is the very well known Compton effect which is expressed as

$\lambda^{'}-\lambda=\frac{h}{m_e c}(1-cos\theta)$

here, $\lambda$ is the initial photon wavelength, $\lambda^{'}$ is the scattered photon wavelength, h is he Planck's constant, $m_e$ is the free electron mass, c is the velocity of light, $\theta$ is the angle of scattering.

Given that, the scattering angle is, $\theta=147^{\circ}$

Putting the respective values, we get

$\lambda^{'}-\lambda=\frac{6.626\times 10^{-34} }{9.11\times 10^{-31}\times 3\times 10^{8} } (1-cos147^\circ ) m\\\lambda^{'}-\lambda=2.42\times 10^{-12} (1-cos147^\circ ) m.\\\lambda^{'}-\lambda=2.42(1-cos147^\circ ) p.m.\\\lambda^{'}-\lambda=4.45 p.m.$

Here, the photon's incident wavelength is $\lamda=2.78pm$

Therefore,

$\lambda^{'}=2.78+4.45=7.23 pm$

From the conservation of momentum,

$\vec{P_\lambda}=\vec{P_{\lambda^{'}}}+\vec{P_e}$

where, $\vec{P_\lambda}$ is the initial photon momentum, $\vec{P_{\lambda^{'}}}$ is the final photon momentum and $\vec{P_e}$ is the scattered electron momentum.

Expanding the vector sum, we get

$P^2_{e}=P^2_{\lambda}+P^2_{\lambda^{'}}-2P_\lambda P_{\lambda^{'}}cos\theta$

Now expressing the momentum in terms of De-Broglie wavelength

$P=h/\lambda,$

and putting it in the above equation we get,

$\lambda_{e}=\frac{\lambda \lambda^{'}}{\sqrt{\lambda^{2}+\lambda^{2}_{'}-2\lambda \lambda^{'} cos\theta}}$

Therefore,

$\lambda_{e}=\frac{2.78\times 7.23}{\sqrt{2.78^{2}+7.23^{2}-2\times 2.78\times 7.23\times cos147^\circ }} pm\\\lambda_{e}=\frac{20.0994}{9.68} = 2.07 pm$

This is the de Broglie wavelength of the electron after scattering.

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3 years ago

Deep in the interiors of the giant planets, water is still a liquid even though the temperatures are tens of thousands of degree

Nataliya [291]

Answer:

High pressure inside the giant planet

Explanation:

As we move in the interior of the giant planet, the pressure and temperature in the interior of the planet increases. Since, the giant planets have hardly any solid surface and thus they are mostly constituted of atmosphere.

Also, the gravitational forces keep even the lightest of the matter bound in it contributing to the large mass of the planet.

If we look at the order of the magnitude of the temperature of these giant planets than nothing should be able to stay in liquid form but as the depth of the planet increases with the increase in temperature, pressure also increases which keeps the particle of the matter in compressed form.

Thus even at such high order of magnitude water is still found in liquid state in the interior of the planet.

7 0

2 years ago