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alexdok [17]
3 years ago
10

What happens when light from the sun passes through any type of matter ?

Physics
2 answers:
fiasKO [112]3 years ago
8 0

Answer: Depending on the matter that the light pass through can give you different results.

The matter might be transparent, translucent, or opaque depending on how it interacts with light.

Explanation:

Transmission of light occurs when light passes through matter. As the light being transmitted, it may reflected, absorbed, scattered or pass straight through the matter.

Reflection of light occurs when light bounces back from a surface that it cannot pass through example mirror.

While in absorption, when light is absorbed, the temperature of the matter increases, thus it is said more energy has been added to the matter. Absorption is mostly high in black matters. Example when you wore a black shirt and work under the sun, you found out that you havr to deal with heat more than when you wore white or different color.

dsp733 years ago
5 0
Most of the light is absorbed.
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A speed boat travels from the dock to the first buoy a distance of 20 meters in 18 seconds it began the trip at a speed of 0 m/s
Lunna [17]

Answer:

1.11 m/s

Explanation:

The motion of the boat is an example of accelerated motion, since the velocity is not constant. However, we don't need to find the acceleration, because we are only interested in the average velocity of the boat, which is given by:

v=\frac{d}{t}

where d is the total distance covered and t the time taken. In this problem, the boat covered a distance of d = 20 m and it takes t = 18 s, therefore the average velocity is

v=\frac{20 m}{18 s}=1.11 m/s

6 0
3 years ago
Neutron stars are extremely dense objects that are formed from the remnants of supernova explosions. Many rotate very rapidly. S
Vlada [557]

Answer:

The required angular speed the neutron star is 10992.32 rad/s

Explanation:

Given the data in the question;

mass of the sun M_S = 1.99 × 10³⁰ kg

Mass of the neutron star

M_N = 2( M_S )

M_N = 2( 1.99 × 10³⁰ kg )

M_N = ( 3.98 × 10³⁰ kg )

Radius of neutron star R_N = 13.0 km = 13 × 10³ m

Now, let mass of a small object on the neutron star be m

angular speed be ω_N.

During rotational motion, the gravitational force on the object supplies the necessary centripetal force.

GmM_N = / R_N² = mR_Nω_N²

ω_N² = GM_N = / R_N³

ω_N = √(GM_N = / R_N³)

we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²

we substitute

ω_N = √( (  6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)

ω_N = √( 2.65466 × 10²⁰ / 2.197 × 10¹²

ω_N = √ 120831133.3636777

ω_N = 10992.32 rad/s

Therefore, The required angular speed the neutron star is 10992.32 rad/s

5 0
3 years ago
Which statement is true about a planet’s orbital motion?
lana66690 [7]

Answer:

Orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

Explanation:

The gravitational force is responsible for the orbital motion of the planet, satellite, artificial satellite, and other heavenly bodies in outer space.

When an object is applied with a velocity that is equal to the velocity of the orbit at that location, the body continues to move forward. And, this motion is balanced by the gravitational pull of the second object.

The orbiting body experience a centripetal force that is equal to the gravitational force of the second object towards the body.

The velocity of the orbit is given by the relation,

                                    V = \sqrt{\frac{GM}{R + h} }

Where

                   V - velocity of the orbit at a height h from the surface

                    R - Radius of the second object

                    G - Gravitational constant

                    h - height from the surface

The body will be in orbital motion when its kinetic motion is balanced by gravitational force.

                         1/2 mV^{2} = GMm/R

Hence, the orbital motion results when the object’s forward motion is balanced by a second object’s gravitational pull.

3 0
3 years ago
an Olympic runner leaps over a hurdle.if the runner initial vertical speed is 2.2m/s how much will the runners center of mass be
Snowcat [4.5K]
Supposing the runner is condensed to a point and moves upward at 2.2 m/s.
It takes a time = 2.2/g = 2.2/9.8 = 0.22 seconds to increase to max height. 
Now looking at this condition in opposite - that is the runner is at max height and drops back to earth in 0.22 s (symmetry of this kind of motion). 
From what height does any object take 0.22 s to fall to earth (supposing there is no air friction)? 
d = 1/2gt²= (0.5)(9.8)(0.22)²= 0.24 m
7 0
3 years ago
Which of the following is the flow of electrons through a wire or a conductor
Hatshy [7]
Wires or silver and copper
8 0
3 years ago
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