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3 years ago

The de Broglie wave is produced only by sub atomic particle and photon. O True O False

Physics

1 answer:

Alina [70]3 years ago

8 0

Answer:

True

Explanation:

Matter can be in the form of a particle or a wave. This is known as the dual nature of matter. This concept was proposed by Louis de Broglie and was named after him. This phenomenon has been observed for all the elementary particles.

The de Broglie wavelength is given by

$\lambda=\frac{h}{p}=\frac{h}{mv}$

Where

h = Planck's constant

p = Particles momentum

m = Mass of particle

v = Velocity of particle

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Two large thin metal plates are parallel and close to each other. On their inner faces, the plates have excess surface charge of

wariber [46]

Answer:

For left = 0 N/C

For right = 0 N/C

At middle = $-7.6836 * 10^{-11} \vec{i}$ N/C

Explanation:

Given data :-

б = $6.8 * 10^{-22}$ C/ m²

Considering the two thin metal plates to be non conducting sheets of charges.

Electric field is given by

$E = \frac{\sigma }{2\varepsilon }$

1) To the left of the plate

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

2) To the right of them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(\vec{i})$ = 0 N/C.

3) Between them.

$\vec{E}= (\frac{\sigma }{2\varepsilon })(-\vec{i})+ (\frac{\sigma }{2\varepsilon })(-\vec{i})$ = $(\frac{\sigma }{\varepsilon })(-\vec{i})$ = $-\frac{6.8 * 10^{-22} }{8.85 * 10 ^{-12} } \vec{i}$ = $-7.6836 * 10^{-11} \vec{i}$ N/C

5 0

3 years ago

4. A 1,000-kilogram satellite completes a uniform circular orbit of radius 8.0 x 10 meters as

lesantik [10]

Answer:

Zero

Explanation:

The work done by a force on an object is given by:

$W=Fd cos \theta$

where

F is the magnitude of the force

d is the displacement of the object

$\theta$ is the angle between the direction of the force and the displacement of the object

In this situation, the force is the force of gravity acting on the satellite. This force always points towards the centre of the trajectory, so it is always perpendicular to the direction of motion of the satellite (since the orbit is circular), so $\theta=90^{\circ}$ and $cos \theta =0$ . Therefore, the work done by gravity is also zero.