Answer:
According to the travellers, Alpha Centauri is <em>c) very slightly less than 4 light-years</em>
<em></em>
Explanation:
For a stationary observer, Alpha Centauri is 4 light-years away but for an observer who is travelling close to the speed of light, Alpha Centauri is <em>very slightly less than 4 light-years. </em>The following expression explains why:
v = d / t
where
- v is the speed of the spaceship
- d is the distance
- t is the time
Therefore,
d = v × t
d = (0.999 c)(4 light-years)
d = 3.996 light-years
This distance is<em> very slightly less than 4 light-years. </em>
Answer:
The air-water interface is an example of<em> </em>boundary. The <u><em>transmitted</em></u><em> </em> portion of the initial wave energy is way smaller than the <u><em>reflected</em></u><em> </em> portion. This makes the <u><em>boundary</em></u> wave hard to hear.
When both the source of the sound and your ears are located underwater, the sound is louder because the sound waves can <u><em>travel directly to your ear</em></u>.
Explanation:
The air-to-water sound wave transmission is inhibited because more of reflection than transmission of the wave occurs at the boundary. In the end, only about 30% of the sound wave eventually reaches underwater. For sound generated underwater, all the wave energy is transmitted directly to the observer. Sound wave travel faster in water than in air because, the molecules of water are more densely packed together, and hence can easily transmit their vibration to their neighboring molecules, when compared to air.
Answer:
μ = 0.408
Explanation:
given,
speed of the automobile (u)= 20 m/s
distance = 50 m
final velocity (v) = 0 m/s
kinetic friction = ?
we know that,
v² = u² + 2 a s
0 = 20² + 2 × a × 50

a = 4 m/s²
We know
F = ma = μN
ma = μ mg
a = μ g


μ = 0.408
hence, Kinetic friction require to stop the automobile before it hit barrier is 0.408
Answer:0
Explanation:
Given
circumference of circle is 2 m
Tension in the string 


In this case Force applied i.e. Tension is Perpendicular to the Displacement therefore angle between Tension and displacement is 



Kepler's third law states that, for a planet orbiting around the Sun, the ratio between the cube of the radius of the orbit and the square of the orbital period is a constant:

(1)
where
r is the radius of the orbit
T is the period
G is the gravitational constant
M is the mass of the Sun
Let's convert the radius of the orbit (the distance between the Sun and Neptune) from AU to meters. We know that 1 AU corresponds to 150 million km, so

so the radius of the orbit is

And if we re-arrange the equation (1), we can find the orbital period of Neptune:

We can convert this value into years, to have a more meaningful number. To do that we must divide by 60 (number of seconds in 1 minute) by 60 (number of minutes in 1 hour) by 24 (number of hours in 1 day) by 365 (number of days in 1 year), and we get