All categories

3 years ago

When light (or another wave) bounces back from a surface ?

1 answer:

3 0

Light bounces back due to the phenomenon called reflection. When light incidents on a plane shiny surface like mirror, it travels again in the same medium(medium from which light was incident) with no loss in energy(energy before and after reflection will remain same iff incident surface completely reflects light)

You might be interested in

Transform boundaries are classified under which type of fault?

cluponka [151]

Answer:

Strike-slip fault

Explanation:

Transform boundaries play the role of connecting the other plate boundary segments.

When the plates are rubbed against each other, they result in enormous amount of stresses which leads to the breaking of the part of a rock causing earthquakes. Places of occurrence of these breaks are termed as faults.

Strike slip faults results from compression which takes place horizontally, but but in this the rock displacement releases energy and takes place in a horizontal direction which is parallel to the force of compression.

8 0

3 years ago

At a certain instant, the earth, the moon, and a stationary 1160 kg spacecraft lie at the vertices of an equilateral triangle wh

Afina-wow [57]

Answer:

$W = 1.22 \times 10^9 J$

Explanation:

Initial potential energy of the given spacecraft is given as

$U = -\frac{GM_e m}{r} - \frac{GM_m m}{r}$

so we have

$U = - \frac{Gm}{r}(M_e + M_m)$

so we have

$M_e = 5.98 \times 10^{24} kg$

$M_m = 7.35 \times 10^{22} kg$

$m = 1160 kg$

$r = 3.84 \times 10^8 m$

$U = - \frac{(6.67 \times 10^{-11})(1160)}{3.84 \times 10^8}(5.98 \times 10^{24} + 7.35 \times 10^{22})$

$U = -1.22 \times 10^9 J$

now total work done to move it to infinite is given

W = 0 - U

$W = 1.22 \times 10^9 J$

6 0

3 years ago

A spring scale has a spring with a force constant of 250 N/m and a weighing pan with a mass of 0.075 kg. During one weighing, th

mr Goodwill [35]

elasticity stretches and can also return to it's normal size ..

8 0

2 years ago

The electric field strength in the space between two closely spaced parallel disks is 1.0 105 N/C. This field is the result of t

balu736 [363]

Answer:

$D=2.996\times 10^{-2} m$

Explanation:

*Assume the parallel disks have equal diameters.

Given the electric strength as $1.0\times 10^5 N/C.$ transferring $3.9\times 10^9$ electrons, the disk's Area can be calculated using the formula:

$E=\frac{\eta}{\epsilon_o}=\frac{Q}{A\epsilon_o}\\\\A=\frac{Q}{E\epsilon_o}\\\\=\frac{(3.9\times 10^9)\times (1.6\times10^{-19})}{(1.0\times 10^5 )\times (8.85\times10^{-12})}\\\\A=7.0508\times 10^{-4} \ m^2$

#We now calculate the disks diameter:

$A=\pi(D/2)^2\\\\2\sqrt{\frac{A}{\pi}}=D\\\\=2\sqrt{7.0508\times 10^{-4}/\pi}\\\\D=2.996\times 10^{-2} \ m$

Hence, the diameter of the disks is $D=2.996\times 10^{-2} m$

8 0

3 years ago

The formation of the solar system in order from start to finish

KatRina [158]

<h2><u>FORMAT</u><u>ION</u></h2>

$\\$

Our solar system formed about 4.5 billion years ago from a dense cloud of interstellar gas and dust. The cloud collapsed, possibly due to the shockwave of a nearby exploding star, called a supernova. When this dust cloud collapsed, it formed a solar nebula—a spinning, swirling disk of material

$\\$

Hope It Helps!

6 0

3 years ago