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2 years ago

( PLEASE HURRY, WILL GIVE BRAINLIEST IF CORRECT )

Physics

2 answers:

Korolek [52]2 years ago

5 0

Answer:

Explanation:

Ugo [173]2 years ago

5 0

Answer:

The answer is D: Light from the sun is blocked from Earth's surface by the moon.

Explanation:

First contact—when the Moon's limb (edge) is exactly tangential to the Sun's limb.

Second contact—starting with Baily's Beads (caused by light shining through valleys on the Moon's surface) and the diamond ring effect. ...

Totality—the Moon obscures the entire disk of the Sun and only the solar corona is visible.

Hope this helps please mark brainliest :)

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Before the car launches, which objects have stored energy?

Sloan [31]

Answer:

maybeeeeeeeee launcherrr im not sureee hehe

3 0

2 years ago

Light is incident on the left face of an isosceles prism; with an apex angle of 49o, such that the light exiting the right face

Sunny_sXe [5.5K]

Answer:

$\mu = 1.645$

Explanation:

By Snell's law we know at the left surface

$\theta_i = 19^o$

$\theta_r = ?$

$\mu_1 = 1$

$\mu_2 = \mu$

now we have

$1 sin19 = \mu sin\theta_r$

$0.33 = \mu sin\theta_r$

now on the other surface we know that

angle of incidence = $\theta_r'$

$\theta_e = 90$

so again we have

$\mu sin\theta_r' = 1 sin90$

so we have

$\theta_r = sin^{-1}\frac{0.33}{\mu}$

$\theta_r' = sin^{-1}\frac{1}{\mu}$

also we know that

$\theta_r + \theta_r' = 49$

$sin^{-1}\frac{0.33}{\mu} + sin^{-1}\frac{1}{\mu} = 49$

By solving above equation we have

$\mu = 1.645$

3 0

2 years ago

Read 2 more answers

A converging lens of focal length 20 cm is placed in contact with, and to the left of, a diverging lens of focal length 30 cm. I

scZoUnD [109]

Answer:

Magnification will be equal to 3

Explanation:

We have given focal length of the converging lens $F_1=20cm$

Focal length of the diverging lens $F_2=30cm$

Object is placed 40 cm to the length of the converging lens d = 40 cm

Combination of the focal length will be equal to

$\frac{1}{F}=\frac{1}{F_1}+\frac{1}{F_2}$

$\frac{1}{F}=\frac{1}{20}+\frac{1}{-30}=\frac{1}{20}-\frac{1}{30}=\frac{1}{60}$

F = 60 cm

So combination of the focal length will be 60 cm

Magnification is given by

$M=\frac{F}{F-d}=\frac{60}{60-40}=3$

So magnification will be equal to 3

3 0

2 years ago

What is E of a hydrogen atom in the 3p state?

notsponge [240]

Answer:

E=-1.51 eV.

$L=\hbar\sqrt{2}$

Explanation:

The nth level energy of a hydrogen atom is defined by the formula,

$E_{n}=-\frac{13.6}{n^{2} }$

Given in the question, the hydrogen atom is in the 3p state.

Then energy of n=3 state is,

$E_{n}=-\frac{13.6}{(3)^{2} }\\E_{n}=-1.51eV$

Therefore, energy of the hydrogen atom in the 3p state is -1.51 eV.

Now, the value of L can be calculated as,

$L=\hbar\sqrt{l(l+1)}$

For 3p state, l=1

$L=\hbar\sqrt{1(1+1)}\\L=\hbar\sqrt{2}$

Therefore, the value of L of a hydrogen atom in 3p state is $L=\hbar\sqrt{2}$ .

4 0

3 years ago

Assume (unrealistically) that a TV station acts as a point source broadcasting isotropically at 3.2 MW. What is the intensity of

Dahasolnce [82]

Answer:

$I=1.5\times10^{-28}W/m^2$

Explanation:

The intensity is related to the power and surface area by $I=\frac{P}{A}=\frac{P}{4\pi r^2}$ . We need to calculate the surface area of a sphere of radius r=4.3ly.

Since 4.3ly is the distance light travels in 4.3 years at 299792458m/s, we can obtain it in meters by doing:

$r=vt=(299792458m/s)(4.3\times365\times24\times60\times60s)=4.1\times10^{16}m$

So we have:

$I=\frac{P}{4\pi r^2}=\frac{3.2\times10^6W}{4\pi (4.1\times10^{16}m)^2}=1.5\times10^{-28}W/m^2$

4 0

2 years ago