Answer:
a) 19.4 m/s
b) 19 m/s
Explanation:
a) In the given question,
the potential energy at the initial point = Ui = 0
the potential energy at the final point = Uf = mgh
the kinetic energy at the initial point = Ki = 1/2 mv₀².
the kinetic energy at the final point = Kf = 0
work done by air= Ea= fh = 0.262 N
Now, using the law of conservation of energy
initial energy= final energy
Ki +Ui = Kf + Uf +Ea
1/2 mv₀² + 0 = 0 + mgh + fh
1/2 mv₀² = mgh + fh
h = v₀²/ 2g (1 +f/w)
calculate m
m= w/g = 5.29 /9.8
= 0.54 kg
h = 20 ²/ (2 x9.80) x (1 0.265/5.29)
h = 19.4 m.
b) 1/2 mv² + 2fh = 1/2 mv₀²
Vg = 19 m/s
Answer:
Use of telemetry and radar astronomy
Explanation:
An astronomical Unit (AU) is a unit of measuring distances in outer space, which is based on the approximate distance between the earth and the Sun.
After several years of trying to approximate the distance between the Sun and the Earth using several methods based on geometry and some other calculations, advancements in technology made available the presence of special motoring equipment, which can be placed in outer space to remotely monitor and measure the position of the sun.
The use of direct radar measurements to the sun (radar astronomy) have also made the determination of the AU more accurate.
A standard radar pulse of known speed is sent to the Sun, and the time with which it takes to return is measured, once this is recorded, the distance between the Earth and the Sun can be calculated using
distance = speed X time.
However, most of these means have to be corrected for parallax errors
Answer:
heliocentric is when the earth and the other planets revolves around the sun.
Explanation:
I'm not sure about which one but I do know what heliocentric is.
Answer:
B. The number of electrons emitted from the metal per second increases.
Explanation:
Light consists of photons . Energy of each photon depends upon frequency of light . The increase in intensity increases the number of photons . It does not increase energy of photons .
So if a high intensity light falls on a photosensitive plate , each photon ejects one electron . So number of electrons increases if we increase intensity of photon. It does not increase kinetic energy of ejected electrons . Work function depends upon the nature of plate.
