How does the frequency of infrared electromagnetic waves compare with the frequency of radio and microwaves?

Railroad tracks appear to converge in the distance. This provides a cue for depth perception known as _____. Please type the cor

evablogger [386]

Answer:

linear perspective

Explanation:

Railroad tracks appear to converge due to our field of view. When an object is far away it seems smaller that means there is angle that subtends from our line of sight this results in the illusion of railroad tracks converging.

Railroad tracks converging activates the linear perspective of our vision. Linear perspective is a technique used to convey a sense of depth in two dimensional representation of an object. The lines at an angle give us a sense of depth otherwise straight lines will not be able to convey depth.

6 0

3 years ago

Now assume that the mass of object 1 is 2m, while the mass of object 2 remains m. If the collision is elastic, what are the fina

Artemon [7]

Answer:

(v₁, v₂) = [(v/3), (4v/3)]

Or

(v₁, v₂) = (v, 0)

Explanation:

In elastic collisions, the momentum and kinetic energy is usually conserved.

The momentum before collision = momentum after collision

And

Kinetic energy before collision = Kinetic energy after collision

Momentum of object 1 before collision = (2m)v = 2mv

Momentum of object 2 before collision = (m)(0) = 0

Momentum of object 1 after collision = (2m)(v₁) = 2mv₁

Momentum of object 2 after collision = (m)(v₂) = mv₂

So, we have

2mv = 2mv₁ + mv₂

2v = 2v₁ + v₂

v₂ = 2v - 2v₁ (eqn 1)

Kinetic energy of object 1 before collision = (1/2)(2m)(v²) = mv²

Kinetic energy of object 2 before collision = (1/2)(m)(0²) = 0

Kinetic energy of object 1 after collision = (1/2)(2m)(v₁²) = mv₁²

Kinetic energy of object 2 after collision = (1/2)(m)(v₁²) = (mv₂²/2)

So, we have,

mv² = mv₁² + (mv₂²/2)

v² = v₁² + (v₂²/2)

2v² = 2v₁² + v₂² (eqn 2)

Substitute (v₂ = 2v - 2v₁) from (eqn 1) into (eqn 2)

2v² = 2v₁² + (2v - 2v₁)²

2v² = 2v₁² + 4v² - 8vv₁ + 4v₁²

6v₁² - 8vv₁ + 2v² = 0

6v₁² - 6vv₁ - 2vv₁ + 2v² = 0

6v₁(v₁ - v) - 2v(v₁ - v) = 0

(6v₁ - 2v)(v₁ - v) = 0

6v₁ = 2v or v₁ = v

v₁ = (v/3) or v₁ = v

If v₁ = (v/3)

From (eqn 1)

v₂ = 2v - 2v₁

v₂ = 2v - 2(v/3)

v₂ = 2v - (2v/3)

v₂ = (4v/3)

If v₁ = v,

From eqn 1,

v₂ = 2v - 2v₁

v₂ = 2v - 2v = 0

(v₁, v₂) = [(v/3), (4v/3)]

Or

(v₁, v₂) = (v, 0)

8 0

3 years ago

How fast should a moving clock travel if it is to be observed by a stationary observer as running at one-half its normal rate?A)

Elan Coil [88]

Answer:

Option (D) is correct.

Explanation:

Let the speed is v.

$\Delta t = \gamma \Delta t'\\\\\Delta t = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\times \frac{\Delta t}{2}\\\\\sqrt{1-\frac{v^2}{c^2}} =\frac{1}{2}\\\\1-\frac{v^2}{c^2}=\frac{1}{4}\\\\\frac{3}{4}c^2 = v^2\\\\v = 0.87 c$

Option (D) is correct.

4 0

3 years ago

Once carbon begins burning in the core of a high-mass star, the outer layers begin to fall inward, driving up the fusion rates a

sammy [17]

Answer:

Fusion rates and star evolution increase rapidly because of the conversion of hydrogen molecules to helium.

Explanation:

At the early stage of a star mass, hydrogen atoms are produced. After a few billion years, these hydrogen atoms formed at the core of the star mass begin to fuse on the external side of the core. Hydrogen atoms get rapidly fused, resulting to a contraction of the star core, and as more hydrogen atoms get fused, helium atoms are formed.

This action results into rapid into a rapid and violent burning action of hydrogen atoms around the core.

6 0

3 years ago

A crate is dragged up a ramp at constant speed. The work done on the system can be accounted for by:

Alex

Answer:

The work done on the system can be accounted for by;

Both $E_g$ and $E_{int}$

Explanation:

The speed of the crate = Constant

Therefore, the acceleration of the crate = 0 m/s²

The net force applied to the crate, $F_{NET}$ = 0

Therefore, the force of with which the crate is pulled = The force resisting the upward motion of the crate

However, we have;

The force resisting the upward motion of the crate = The weight of the crate + The frictional resistance of the ramp due to the surface contact between the ramp and the crate

The work done on the system = The energy to balance the resisting force = The weight of the crate × The height the crate is raised + The heat generated as internal energy to the system

The weight of the crate × The height the crate is raised = Gravitational Potential Energy = $E_g$

The heat generated as internal energy to the system = $E_{int}$

Therefore;

The work done on the system = $E_g$ + $E_{int}$ .

6 0

3 years ago