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3 years ago

9

According to the Law of Reflection, the angle of incidence the angle of reflection. O A. is greater than B. is less than C. equa

ls D. is opposite from

1 answer:

Amanda [17]3 years ago

8 0

Answer:

C. Equals

Explanation:

Law of reflection Equals the angle of incidence

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In Thomson’s experiment, why was the glowing beam repelled by a negatively charged plate?

Svetllana [295]

The glowing beam was repelled by a negatively charged plate because they were negatively charged

<h3>What are the nature of charges?</h3>

The nature of charges refers to the properties of charges.

There are two types of charges:

negative charges
positive charges

The law of electricity states that opposite charges attract whereas like charges repel.

Therefor, in Thomson’s experiment, the glowing beam was repelled by a negatively charged plate because they were negatively charged

In conclusion, like charges repel while opposite charges attract.

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1 year ago

14. Which is the relationship between photon energy and frequency?

yawa3891 [41]

Answer:

The amount of energy is directly proportional to the photon's electromagnetic frequency and thus, equivalently, is inversely proportional to the wavelength. The higher the photon's frequency, the higher its energy.

i hope this helps.

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2 years ago

What is the last planet discovered?

Bumek [7]

Pluto is the last planet discovered in our solar system.

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3 years ago

a car moves 6 km east, then 4km north. what was the distance covered? What was the total displacement?

Aliun [14]

Answer:

distance= 10 km

displacement= 7.21 km

Explanation:

distance= scalar (only magnitude)

displacement= vector (magnitude & direction)

distance= 6 km+ 4 km

= 10 km

displacement= shortest difference btwn 2 pts=

sqrt( 6^2+4^2)

sqrt(52)

7.21 km

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2 years ago

The volume of water in the Pacific Ocean is about 7.0 × 10 8 km 3 . The density of seawater is about 1030 kg/m3. (a) Determine t

Novay_Z [31]

To solve the problem it is necessary to consider the concepts related to Potential Energy and Kinetic Energy.

Potential Energy because of a planet would be given by the equation,

$PE=\frac{GMm}{r}$

Where,

G = Gravitational Universal Constant

M = Mass of Ocean

M = Mass of Moon

r = Radius

From the data given we can calculate the mass of the ocean water through the relationship of density and volume, then,

$m = \rho V$

$m = (1030Kg/m^3)(7*10^8m^3)$

$m = 7.210*10^{11}Kg$

It is necessary to define the two radii, when the ocean is far from the moon and when it is facing.

When it is far away, it will be the total diameter from the center of the earth to the center of the moon.

$r_1 = 3.84*10^8 + 6.4*10^6 = 3.904*10^8m$

When it's near, it will be the distance from the center of the earth to the center of the moon minus the radius,

$r_2 = 3.84*10^8-6.4*10^6 - 3.776*10^8m$

PART A) Potential energy when the ocean is at its furthest point to the moon,

$PE_1 = \frac{GMm}{r_1}$

$PE_1 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.904*10^8}$

$PE_1 = 9.05*10^{15}J$

PART B) Potential energy when the ocean is at its closest point to the moon

$PE_2 = \frac{GMm}{r_2}$

$PE_2 = \frac{(6.61*10^{-11})*(7.21*10^{11})*(7.35*10^{22})}{3.776*10^8}$

$PE_2 = 9.361*10^{15}J$

PART C) The maximum speed. This can be calculated through the conservation of energy, where,

$\Delta KE = \Delta PE$

$\frac{1}{2}mv^2 = PE_2-PE_1$

$v=\sqrt{2(PE_2-PE_1)/m}$

$v = \sqrt{\frac{2*(9.361*10^{15}-9.05*10^{15})}{7.210*10^{11}}}$

$v = 29.4m/s$

8 0

3 years ago