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Lesechka [4]
2 years ago
15

6. (a) Suppose the earth is revolving round the sun in a circular orbit of radius one b astronomical unit (1.5% 10 km). Find the

mass of the sun. G = 6.67 x 10" N m² kg 2. (b) If the mass of sun is 2 x 10" kg, distance of earth from the sun is 1.5 x 10¹ m and period of revolution of the former around the latter is 365.3 days, find the value of G.​
Physics
1 answer:
kkurt [141]2 years ago
5 0

Answer:

tough

ques

Explanation:

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PLEASE HELP (WILL GIVE BRAINLIEST TO BEST ANSWER!!!!!)
Luden [163]

Answer:

1) a radio are uses by astronomy

2) 6 bilion waves

3) expert vertified

Explanation:

1) in contrast to an "ordinary" telescope, which receives visible light, a radio telescope "sees" radio waves emitted by radio sources, typically by means of a large parabolic ("dish") antenna, or arrays of them.and Radio telescopes are also the primary means to track space probes, and are used in the SETI project. so must been

radio are almostly

ceiver by astronomy

2) Radio waves have the longest wavelengths in the electromagnetic spectrum.

3)Expert Verified

Radio telescopes are telescopes that are specially designed for observation of long light wavelengths

CARRY ON ✨

3 0
3 years ago
Where would you find convergent and divergent plate boundaries relative to
Sergio [31]

Answer:

When two tectonic plates meet, we get a “plate boundary.” There are three major types of plate boundaries, each associated with the formation of a variety of geologic features.

Explanation:

8 0
3 years ago
NaOH + FeCl3* Na Cl + Fe 10H)3<br> balanced
zvonat [6]
3NaOH + FeCl3 → 3NaCl + Fe(OH)3
8 0
3 years ago
Radar uses radio waves of a wavelength of 2.4 \({\rm m}\) . The time interval for one radiation pulse is 100 times larger than t
blondinia [14]

Answer:

120 m

Explanation:

Given:

wavelength 'λ' = 2.4m

pulse width 'τ'= 100T ('T' is the time of one oscillation)

The below inequality express the range of distances to an object that radar can detect

τc/2 < x < Tc/2 ---->eq(1)

Where, τc/2 is the shortest distance

First we'll calculate Frequency 'f' in order to determine time of one oscillation 'T'

f = c/λ (c= speed of light i.e 3 x 10^{8} m/s)

f= 3 x 10^{8} / 2.4

f=1.25 x  10^{8} hz.

As, T= 1/f

time of one oscillation T= 1/1.25 x  10^{8}

T= 8 x 10^{-9} s

It was given that pulse width 'τ'= 100T

τ= 100 x 8 x 10^{-9} => 800 x 10^{-9} s

From eq(1), we can conclude that the shortest distance to an object that this radar can detect:

x_{min}= τc/2 =>  (800 x 10^{-9} x 3 x 10^{8})/2

x_{min}=120m

8 0
3 years ago
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

8 0
4 years ago
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