Answer:
finding Cepheid variable and measuring their periods.
Explanation:
This method is called finding Cepheid variable and measuring their periods.
Cepheid variable is actually a type of star that has a radial pulsation having a varying brightness and diameter. This change in brightness is very well defined having a period and amplitude.
A potent clear link between the luminosity and pulsation period of a Cepheid variable developed Cepheids as an important determinants of cosmic criteria for scaling galactic and extra galactic distances. Henrietta Swan Leavitt revealed this robust feature of conventional Cepheid in 1908 after observing thousands of variable stars in the Magellanic Clouds. This in fact turn, by making comparisons its established luminosity to its measured brightness, allows one to evaluate the distance to the star.
The time lapse between when the bat emits the sound and when it hears the echo is 0.05 s.
From the question given above, the following data were obtained:
Velocity of sound (v) = 343 m/s
Distance (x) = 8.42 m
Time (t) =?
We can obtain obtained the time as illustrated below:
v = 2x / t
343 = 2 × 8.42 / t
343 = 16.84 / t
Cross multiply
343 × t = 16.84
Divide both side by 343
t = 16.84/343
t = 0.05 s
Thus, the time between when the bat emits the sound and when it hears the echo is 0.05 s.
<h3>
How does a bat know how far away something is?</h3>
A bat emits a sound wave and carefully listens to the echoes that return to it. The returning information is processed by the bat's brain in the same way that we processed our shouting sound with a stopwatch and calculator. The bat's brain determines the distance of an object by measuring how long it takes for a noise to return.
Learn more about time elapses between when the bat emits the sound :
<u>brainly.com/question/16931690</u>
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Correction question:
A bat emits a sonar sound wave (343 m/s) that bounces off a mosquito 8.42 m away. How much time elapses between when the bat emits the sound and when it hears the echo? (Unit = s)
Each hour 430 quintillion Joules of energy from the sun hits the Earth.
In a year it is very hard to determine because of the night and different light levels.
Answer:
The surface gravity g of the planet is 1/4 of the surface gravity on earth.
Explanation:
Surface gravity is given by the following formula:
So the gravity of both the earth and the planet is written in terms of their own radius, so we get:
The problem tells us the radius of the planet is twice that of the radius on earth, so:
If we substituted that into the gravity of the planet equation we would end up with the following formula:
Which yields:
So we can now compare the two gravities:
When simplifying the ratio we end up with:
So the gravity acceleration on the surface of the planet is 1/4 of that on the surface of Earth.
Answer:
Momentum is always conserved, and kinetic energy may be conserved.
Explanation:
For an object moving on a horizontal, frictionless surface which makes a glancing collision with another object initially at rest on the surface, the type of collision experienced by this objects can either be elastic or an inelastic collision depending on whether the object sticks together after collision or separates and move with a common velocity after collision.
If the body separates and move with a common velocity after collision, the collision is elastic but if they sticks together after collision, the collision is inelastic.
Either ways the momentum of the bodies are always conserved since they will always move with a common velocity after collision but their kinetic energy may or may not be conserved after collision, it all depends whether they separates or stick together after collision and since we are not told in question whether or not they separate, we can conclude that their kinetic energy "may" be conserved.
