Helppp please

It takes light from the closest star to Earth (other than the Sun) about four years to reach Earth. If intelligent life were on

mafiozo [28]

If the intelligent aliens on that distant planet could receive our signal, and then reply to us INSTANTLY with no delay, then our Scientists on Earth would hear the response <em>eight years after</em> they sent their original greeting.

Light and radio waves both travel through space at the same speed. If our scientists sent a radio transmission to that distant planet, it would take four years for the radio message to get there and be heard. Any response would take four years to travel back to us. If the aliens didn't dilly-dally and sent their response immediately, the round trip would total up to eight years.

4 0

3 years ago

A small dog is trained to jump straight up a distance of 1.2 m. How much kinetic energy does the 7.2-kg dog need to jump this hi

LUCKY_DIMON [66]

Answer:84.672 joules.

Explanation:

1) Data:

m = 7.2 kg
h = 1.2 m
g = 9.8 m / s²

2) Physical principle

Using the law of mechanical energy conservation principle, you have that the kinetic energy of the dog, when it jumps, must be equal to the final gravitational potential energy.

3) Calculations:

The gravitational potential energy, PE, is equal to m × g × h

So, PE = m × g × h = 7.2 kg × 9.8 m/s² × 1.2 m = 84.672 joules.

And that is the kinetic energy that the dog needs.

8 0

4 years ago

What is the magnitude of the total acceleration of point A after 2 seconds? The bar starts from rest and has a constant angular

monitta

Answer:

a_total = 2 √ (α² + w⁴) , a_total = 2,236 m

Explanation:

The total acceleration of a body, if we use the Pythagorean theorem is

a_total² = a_T²2 + $a_{c}$ ²

where

the centripetal acceleration is

a_{c} = v² / r = w r²

tangential acceleration

a_T = dv / dt

angular and linear acceleration are related

a_T = α r

we substitute in the first equation

a_total = √ [(α r)² + (w r² )²]

a_total = 2 √ (α² + w⁴)

Let's find the angular velocity for t = 2 s if we start from rest wo = 0

w = w₀ + α t

w = 0 + 1.0 2

w = 2.0rad / s

we substitute

a_total = r √(1² + 2²) = r √5

a_total = r 2,236

In order to finish the calculation we need the radius to point A, suppose that this point is at a distance of r = 1 m

a_total = 2,236 m

7 0

4 years ago

Light rays in a material with index of refrection 1.29 1.29 can undergo total internal reflection when they strike the interface

deff fn [24]

Answer:

The second material's index of refraction is 1.17.

Explanation:

Given that,

Refractive index of the material, n = 1.29

Critical angle is 65.9 degrees.

We need to find the second material's index of refraction. We know that at critical angle of incidence, angle of refraction is equal to 90 degrees. Using Snell's law as:

$n_1\sin \theta_c=n_2\sin (90)\\\\n_2=n_1\sin \theta_c\\\\n_2=1.29\times \sin (65.9)\\\\n_2=1.17$

So, the second material's index of refraction is 1.17.

7 0

3 years ago

A small rock is launched straight upward from the surface of a planet with no atmosphere. The initial speed of the rock is twice

Scorpion4ik [409]

If gravitational effects from other objects are negligible, the speed of the rock at a very great distance from the planet will approach a value of $\sqrt{3} v_{e}$