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3 years ago

The bright, visible surface of the sun is called the

1 answer:

katrin2010 [14]3 years ago

5 0

<span>The bright, visible surface of the Sun is called corona. The outermost layer of the Sun's atmosphere is called chromosphere.</span>

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What is the momentum of a photon having the same total energy as an electron with a kinetic energy of 100 keV?

statuscvo [17]

Answer:

The momentum of the photon is 1.707 x 10⁻²² kg.m/s

Explanation:

Given;

kinetic of electron, K.E = 100 keV = 100,000 eV = 100,000 x 1.6 x 10⁻¹⁹ J = 1.6 x 10⁻¹⁴ J

Kinetic energy is given as;

K.E = ¹/₂mv²

where;

v is speed of the electron

$K.E = \frac{1}{2}mv^2\\\\mv^2 = 2K.E \\\\v^2 = \frac{2K.E}{m} \\\\v = \sqrt{\frac{2K.E}{m}} \\\\but \ momentum ,P = mv\\\\(v)m = (\sqrt{\frac{2K.E}{m}})m\\\\P_{photon} = (\sqrt{\frac{2K.E}{m_e}})m_e\\\\P_{photon} = (\sqrt{\frac{2\times 1.6\times 10^{-14}}{9.11\times10^{-31}}})(9.11\times 10^{-31})\\\\P_{photon} = 1.707 \times 10^{-22} \ kg.m/s$

Therefore, the momentum of the photon is 1.707 x 10⁻²² kg.m/s

6 0

3 years ago

A force of 100N is applied to move an object a horizontal distance of 20m to the right. The work done by this force on the objec

horsena [70]

WORKDONE = FORCE * DISPLACEMENT
W=F*S
HERE, THE FORCE = 100N AND DISTANCE = 20M
WORKDONE = 100*20
WORKDONE=2000
ITS S.I UNIT IS JOULE OR J
SO, 2000J

5 0

3 years ago

Nearsightedness can be corrected by using eyeglass lenses that are

Nat2105 [25]

Concaved lenses...........

3 0

3 years ago

Read 2 more answers

Force → F = ( − 8.0 N ) ˆ i + ( 6.0 N ) ˆ j acts on a particle with position vector → r = ( 3.0 m ) ˆ i + ( 4.0 m ) ˆ j . What a

andrey2020 [161]

Explanation:

It is given that,

Force, $F=(-8\ N)i+(6\ N)j$

Position vector, $r=(3i+4j)\ m$

(a) The torque on the particle about the origin is given by :

$\tau=F\times r\\\\\tau=(-8i+6j)\times (3i+4j)\\\\\tau=(-50k)\ N-m$

(b) To find the angle between r and F use dot product formula as :

$F{\cdot} r=|F||r|\ \cos\theta\\\\\cos\theta=\dfrac{F{\cdot} r}{|F| |r|}\\\\\cos\theta=\dfrac{(-8i+6j){\cdot} (3i+4j)}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=\dfrac{-24+24}{\sqrt{(-8)^2+6^2} \times \sqrt{3^2+4^2} }\\\\\cos\theta=0\\\\\theta=90^{\circ}$

Hence, this is the required solution.

8 0

3 years ago

Read 2 more answers

John says that the value of the function cos[ω(t + T) + ϕ], obtained one period T after time t, is greater than cos(ωt + ϕ) by 2

Svetllana [295]

Answer:

No one is right

Explanation:

John Case:

The function $cos(\omega t +\phi)$ is defined between -1 and 1, So it is not possible obtain a value $2\pi$ greater.

In addition, if you move the function cosine a T Value, and T is the Period, the function take the same value due to the cosine is a periodic function.

Larry case:

Is you have $f=1+cos(\omega t +\phi)$ , the domain of this is [0,2].

it is equivalent to adding 1 to the domain of the $f=1+cos(\omega t +\phi)$ , and its mean that the function $f=cos(\omega t +\phi)$ , in general, is not greater than $cos(\omega t +\phi)$ .

3 0

3 years ago