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Sever21 [200]
4 years ago
7

g A beam of light incident on a smooth flat mirror at an angle of 25o relative to the surface normal of the mirror. What is the

angle of the reflected beam of light made with the surface normal?

Physics
1 answer:
WINSTONCH [101]4 years ago
5 0

To solve this problem we will resort to the concept of angle of incidence and refraction.

Since it is a reflection on a mirror, the angle provided for refraction will be equal to that of the incidence, that is, 25 °

The angle of reflation is always perpendicular to the surface so it is necessary to find the angle with respect to it.

\theta = 90-25

\theta = 65\°

Therefore the angle of the reflected beam of light made with the surface normal is 65°

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when charge 2 is 3.0 m away from charge 1, the strength of the electric force on charge 2 by charge 1 is 0.80 n. if instead, cha
Nikolay [14]

The strength of the electrostatic force is inversely proportional to the square of the distance

between the charges. If the distance is doubled, the force is 1/4.

The new force is 0.80/4 = 0.20 N.

<h3>What is electrostatic force? </h3>

Electrostatic force is the attractive or repulsive force that exists between two charged particles. Also known as Coulomb interaction or Coulomb force. For example, the electrostatic forces between the protons and electrons of an atom are responsible for the stability of the atom.

The force acting along a line joining two point charges is directly proportional to the product of the two charges and inversely proportional to the square of the distance between them.

F=K |\frac{q_{1} q_{2}}{r^{2} }   |

In the above formula, k is arbitrary and can be chosen to be any positive value. Since k is a constant, I chose to give the value of k as follows:

Therefore, with q₁ and q₂ values ​​of 1 and r = 1 (two charges with 1 Coulomb charge each at a distance of 1 m), we get F = 9 \times 10^9 N. In the above equation, ε₀ is given as the permittivity of free space and its value in SI units is 8.854\times10^{-12} C^{2} N^{-1}  m^{-2}.

To learn more about electrostatic force , visit:

brainly.com/question/14870624

#SPJ4

5 0
2 years ago
What condition is necessary for an object to make a good reference point?
777dan777 [17]
The object is fixed relative to the motion you are trying to describe.
8 0
4 years ago
A 1.00 kg block of ice, at -25.0°C, is warmed by 35 kJ of energy. What is the final temperature of the ice?
ahrayia [7]

Answer:

-8.4°C

Explanation:

From the principle of heat capacity.

The heat sustain by an object is given as;

H = m× c× (T2-T1)

Where H is heat transferred

m is mass of substance

T2-T1 is the temperature change from starting to final temperature T2.

c- is the specific heat capacity of ice .

Note : specific heat capacity is an intrinsic capacity of a substance which is the energy substained on a unit mass of a substance on a unit temperature change.

Hence ; 35= 1× c× ( T2-(-25))

35= c× ( T2+25)

35 =2.108×( T2+25)

( T2+25)= 35/2.108= 16.60°{ approximated to 2 decimal place}

T2= 16.60-25= -8.40°C

C, specific heat capacity of ice is =2.108 kJ/kgK{you can google that}

6 0
3 years ago
How much energy does a 100-W light bulb use in 10 s?
telo118 [61]

Answer:c

Explanation:

8 0
4 years ago
We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top
notka56 [123]

Answer:

A,)FD= 114.1N

B)Torque=798.5Nm

Explanation:

We can model a pine tree in the forest as having a compact canopy at the top of a relatively bare trunk. Wind blowing on the top of the tree exerts a horizontal force, and thus a torque that can topple the tree if there is no opposing torque. Suppose a tree's canopy presents an area of 9.0 m^2 to the wind centered at a height of 7.0 m above the ground. (These are reasonable values for forest trees.)

If the wind blows at 6.5 m/s, what is the magnitude of the drag force of the wind on the canopy? Assume a drag coefficient of 0.50 and the density of air of 1.2 kg/m^3

B)What torque does this force exert on the tree, measured about the point where the trunk meets the ground?

A)The equation of Drag force equation can be expressed below,

FD =[ CD × A × ρ × (v^2/ 2)]

Where CD= Drag coefficient for cone-shape = 0.5

ρ = Density

Area of of the tree canopy = 9.0 m^2

density of air of = 1.2 kg/m^3

V= wind velocity= 6.5 m/s,

If we substitute those values to the equation, we have;

FD =[ CD × A × ρ × (v^2/ 2)]

F= [ 0.5 × 9.0 m^2 × 1.2 kg/m^3 ( 6.5 m/s/ 2)]

FD= 114.1N

B) the torque can be calculated using below formula below

Torque= (Force × distance)

= 114.1 × 7

= 798.5Nm

8 0
3 years ago
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