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Hoochie [10]
2 years ago
10

Which phenomenon supports the particle model of light?

Physics
2 answers:
Korolek [52]2 years ago
7 0

Answer:

The photoelectric effect

kondaur [170]2 years ago
3 0

Answer:

Explanation:

your answer is c Destructive interference

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A uniform electric field exists in the region between two oppositely charged plane parallel plates. a proton is released from re
dedylja [7]

According to Newton's second law

E.e = a * mp  ..... (1)

where

E is the magnitude of the electric field; e = 1.6 * 10^-19 is the elementary charge; mp = 1.67*10^-27 kg is the proton mass; a is the acceleration.

So, the distance

l = at^2/2 .......(2)

The proton accelerated

a = 2l / t^2 ...........(3)

From equations (1) and (3)

E= 32.51 V/m

Electric field

The physical field that surrounds electrically charged particles and exerts a force on all other charged particles in the field, either attracting or repelling them, is known as an electric field (also known as an E-field).  It can also refer to a system of charged particles' physical field. Electric charges and time-varying electric currents are the building blocks of electric fields. The electromagnetic field, one of the four fundamental interactions (also known as forces) of nature, manifests itself in both electric and magnetic fields.

To learn more about an electric field refer here:

brainly.com/question/15800304

#SPJ4

5 0
1 year ago
Give an example of a force applied to an object that does not change the object's velocity. Why does the object's velocity not c
Xelga [282]

Answer:

A chair at rest on the floor has two forces acting on it its own weight that pulls it downward and the floor pushing upward on the chair, both of these forces are acting on it but the net force is 0, so the chair remains at rest and its velocity stays at 0.

8 0
2 years ago
A 0.150 kg stone rests on a frictionless, horizontal surface. A bullet of mass 9.50 g, traveling horizontally at 380 m/s, strike
Anvisha [2.4K]

Answer:

(a)Magnitude=28.81 m/s

Direction=33.3 degree below the horizontal

(b) No, it is not perfectly elastic collision

Explanation:

We are given that

Mass of stone, M=0.150 kg

Mass of bullet, m=9.50 g=9.50\times 10^{3} kg

Initial speed of bullet, u=380 m/s

Initial speed of stone, U=0

Final speed of bullet, v=250m/s

a. We have to find the magnitude and direction of the velocity of the stone after it is struck.

Using conservation of momentum

mu+ MU=mv+ MV

Substitute the values

9.5\times 10^{-3}\times 380 i+0.150(0)=9.5\times 10^{-3} (250)j+0.150V

3.61i=2.375j+0.150V

3.61 i-2.375j=0.150V

V=\frac{1}{0.150}(3.61 i-2.375j)

V=24.07i-15.83j

Magnitude of velocity of stone

=\sqrt{(24.07)^2+(-15.83)^2}

|V|=28.81 m/s

Hence, the magnitude and direction of the velocity of the stone after it is struck, |V|=28.81 m/s

Direction

\theta=tan^{-1}(\frac{y}{x})

=tan^{-1}(\frac{-15.83}{24.07})

\theta=tan^{-1}(-0.657)

=33.3 degree below the horizontal

(b)

Initial kinetic energy

K_i=\frac{1}{2}mu^2+0=\frac{1}{2}(9.5\times 10^{-3})(380)^2

K_i=685.9 J

Final kinetic energy

K_f=\frac{1}{2}mv^2+\frac{1}{2}MV^2

=\frac{1}{2}(9.5\times 10^{-3})(250)^2+\frac{1}{2}(0.150)(28.81)^2

K_f=359.12 J

Initial kinetic energy is not equal to final kinetic energy. Hence, the collision is not perfectly elastic collision.

5 0
2 years ago
Type of energy transformed into chemical energy by plants is called type of energy transformed into chemical energy by plants is
yuradex [85]
I think the correct answer is light energy. It is light energy that is transformed into chemical energy by plants by the process called photosynthesis. In this process, plants<span> take in water, carbon dioxide, and sunlight and </span>turn<span> them </span>into<span> glucose and oxygen.</span>
6 0
3 years ago
Assume the radius of an atom, which can be represented as a hard sphere, is r = 1.95 Å. The atom is placed in a (a) simple cubic
Nuetrik [128]

Answer:

(a) A = 3.90 \AA

(b) A = 4.50 \AA

(c) A = 5.51 \AA

(d) A = 9.02 \AA

Solution:

As per the question:

Radius of atom, r = 1.95 \AA = 1.95\times 10^{- 10} m

Now,

(a) For a simple cubic lattice, lattice constant A:

A = 2r

A = 2\times 1.95 = 3.90 \AA

(b) For body centered cubic lattice:

A = \frac{4}{\sqrt{3}}r

A = \frac{4}{\sqrt{3}}\times 1.95 = 4.50 \AA

(c) For face centered cubic lattice:

A = 2{\sqrt{2}}r

A = 2{\sqrt{2}}\times 1.95 = 5.51 \AA

(d) For diamond lattice:

A = 2\times \frac{4}{\sqrt{3}}r

A = 2\times \frac{4}{\sqrt{3}}\times 1.95 = 9.02 \AA

6 0
2 years ago
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