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2 years ago

Background information about reflection and refraction of light

Physics

1 answer:

KIM [24]2 years ago

4 0

Answer:

reflection :

angle of reflection and angle of incidence is equal
Incident ray and reflect d ray plus normal lies on the same plan on a same point
light reflect uniformly if it is incident on a plan surface

refraction :

light will refrect if light goes from rare medium to a dense medium or vice versa
after the light is refracted, it retunes to its original direction
angle of incidence is not equal to the angle of refraction

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A comet is in an elliptical orbit around the sun. its closest approach to the sun is a distance of 4.5 1010 m (inside the orbit

barxatty [35]

r1 = 5*10^10 m , r2 = 6*10^12 m

v1 = 9*10^4 m/s

From conservation of energy

K1 +U1 = K2 +U2

0.5mv1^2 - GMm/r1 = 0.5mv2^2 - GMm/r2

0.5v1^2 - GM/r1 = 0.5v2^2 - GM/r2

M is mass of sun = 1.98*10^30 kg

G = 6.67*10^-11 N.m^2/kg^2

0.5*(9*10^4)^2 - (6.67*10^-11*1.98*10^30/(5*10^10)) = 0.5v2^2 - (6.67*10^-11*1.98*10^30/(6*10^12))

v2 = 5.35*10^4 m/s

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3 years ago

Draw a schematic diagram of a circuit consisting of a battery of five 2 V cells, a 5 ohm resistor, a 10 ohm resistor, a 15 ohm r

MakcuM [25]

Answer:

Current = 0.33 A

Explanation:

For diagram refer the attachment.

It is given that five cells of 2V are connected in series, so total voltage of the battery:

$\dashrightarrow \: \: \sf V = 2 \times 5 = 10 V$

Three resistor of 5 $\Omega$ , 10 $\Omega$ , 15 $\Omega$ are connected in Series, so the net resistance:

$\dashrightarrow \: \: \sf R_{n} = R_{1} + R_{2} + R_{3}$

$\dashrightarrow \: \: \sf R = 5 + 10 + 15$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{R = 30 \: \Omega}}}}$

According to ohm's law:

$\dashrightarrow \sf\: \: V = IR$

$\dashrightarrow \sf \: \: I = \dfrac{V}{R}$

On substituting resultant voltage (V) as 10 V and resultant resistant, as 30 ${\pmb{\sf{\Omega}}}$ we get:

$\dashrightarrow \sf \: \: I = \dfrac{10V}{30\Omega}$

${ \pink{\dashrightarrow \sf \: \: { \underbrace{I = 0.33 A}}}}$

$\therefore$ The electric current passing through the above circuit when the key is closed will be <u>0.33 A</u>

4 0

1 year ago

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find the mass of a ball on a roof 30 meters high, if the ball's gravitational potential energy is 58.8 joules

Sidana [21]

We are given the gravitational potential energy and the height of the ball and is asked in the problem to determine the mass of the ball. the formula to be followed is PE = mgh where g is the gravitational acceleration equal to 9.81 m/s^2. substituting, 58.8 J = m*9.8 m/s^2 * 30 m; m = 0.2 kg.

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2 years ago

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Two metal spheres are suspended from strings. The

nikdorinn [45]

Answer:

B. A repulsive force of 8.0*10^3 N.

Explanation:

As we know by Coulomb's law that the electrostatic force between two charges is given as

$F = \frac{kq_1q_2}{r^2}$