Answer:
a) Time = 2.67 s
b) Height = 35.0 m
Explanation:
a) The time of flight can be found using the following equation:
(1)
Where:
: is the final position in the horizontal direction = 80 m
: is the initial position in the horizontal direction = 0
: is the initial velocity in the horizontal direction = 30 m/s
a: is the acceleration in the horizontal direction = 0 (the stone is only accelerated by gravity)
t: is the time =?
By entering the above values into equation (1) and solving for "t", we can find the time of flight of the stone:
b) The height of the hill is given by:
Where:
: is the final position in the vertical direction = 0
: is the initial position in the vertical direction =?
: is the initial velocity in the vertical direction =0 (the stone is thrown horizontally)
g: is the acceleration due to gravity = 9.81 m/s²
Hence, the height of the hill is:
I hope it helps you!
The heat lost by the water will be equivalent to the energy gained by the alcohol. Thus:
maCaΔT = -mwCwΔT
400 x 2.64 x (T - 10) = 400 x 4.186 x (88 - T)
T = 57.8 °C
The habitable zone is the range of distances from a star where a planet’s temperature allows liquid water oceans, critical for life on Earth. The earliest definition of the zone was based on simple thermal equilibrium, but current calculations of the habitable zone include many other factors, including the greenhouse effect of a planet’s atmosphere. This makes the boundaries of a habitable zone "fuzzy."
Astronomers announced in August 2016 that they may have found such a planet orbiting Proxima Centauri. The newfound world, known as Proxima b, is about 1.3 times more massive than Earth, which suggests that the exoplanet is a rocky world, researchers said. The planet is also in the star's habitable zone, just 4.7 million miles (7.5 million kilometers) from its host star. It completes one orbit every 11.2 Earth-days. As a result, it's likely that the exoplanet is tidally locked, meaning it always shows the same face to its host star, just as the moon shows only one face (the near side) to Earth.
The young sun would have had a very strong magnetic field, whose lines of force reached out into the disk of swirling gas from which the planets would form. These field lines connected with the charged particles in the gas, and acted like anchors, slowing down the spin of the forming sun and spinning up the gas that would eventually turn into the planets. Most stars like the sun rotate slowly, so astronomers inferred that the same “magnetic braking” occurred for them, meaning that planet formation must have occurred for them. The implication: Planets must be common around sun-like
A Canadian team discovered a Jupiter-size planet around Gamma Cephei in 1988, but because its orbit was much smaller than Jupiter’s, the scientists did not claim a definitive planet detection. “We weren’t expecting planets like that. It was different enough from a planet in our own solar system that they were cautious," Matthews said.
Most of the first exoplanet discoveries were huge Jupiter-size (or larger) gas giants orbiting close to their parent stars. That's because astronomers were relying on the radial velocity technique, which measures how much a star “wobbles” when a planet or planets orbit it. These large planets close in produce a correspondingly big effect on their parent star, causing an easier-to-detect wobble.
Before the era of exoplanet discoveries, instruments could only measure stellar motions down to a kilometer per second, too imprecise to detect a wobble due to a planet. Now, some instruments can measure velocities as low as a centimeter per second, according to Matthews. “Partly due to better instrumentation, but also because astronomers are now more experienced in teasing subtle signals out of the data.”
Today, there are more than 1,000 confirmed exoplanets discovered by a single telescope: the Kepler space telescope, which reached orbit in 2009 and hunted for habitable planets for four years. Kepler uses a technique called the “transit” method, measuring how much a star's light dims when a planet passes in front of it.
Kepler has revealed a cornucopia of different types of planets. Besides gas giants and terrestrial planets, it has helped define a whole new class known as “super-Earths”: planets that are between the size of Earth and Neptune. Some of these are in the habitable zones of their stars, but astrobiologists are going back to the drawing board to consider how life might develop on such worlds.
In 2014, Kepler astronomers (including Matthews’ former student Jason Rowe) unveiled a “verification by multiplicity” method that should increase the rate at which astronomers promote candidate planets to confirmed planets. The technique is based on orbital stability — many transits of a star occurring with short periods can only be due to planets in small orbits, since multiply eclipsing stars that might mimic would gravitationally eject each other from the system in just a few million years.
While the Kepler (and French CoRoT) planet-hunting satellites have ended their original missions, scientists are still mining the data for discoveries, and there are more to come. MOST is still operating, and the NASA TESS (Transiting Exoplanet Survey Satellite), Swiss CHEOPS (Characterizing ExOPlanets Satellite) and ESA’s PLATO missions will soon pick up the transit search from space. From the ground, the HARPS spectrograph on the European Southern Observatory's La Silla 3.6-meter telescope in Chile is leading the Doppler wobble search, but there are many other telescopes in the hunt.
With almost 2,000 to choose from, it’s hard to narrow down a few. Small solid planets in the habitable zone are automatically standouts, but Matthews singled out five other exoplanets that have expanded our perspective on how planets form and
Answer:
λ = 5.656 x 10⁻⁷ m = 565.6 nm
Explanation:
Using the formula of fringe spacing from the Young's Double Slit experiment, which is given as follows:
where,
λ = wavelength = ?
Δx = fringe spacing = 1.6 cm = 0.016 m
L = Distance between slits and screen = 4.95 m
d = slit separation = 0.175 mm = 0.000175 m
Therefore,
<u>λ = 5.656 x 10⁻⁷ m = 565.6 nm</u>
