Explanation:
It is given that,
A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.
Mass of the Sun, 
Radius of Mercury's orbit, 
Radius of discovered planet, 
Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :
T = 4135214.625 s
or
T = 47.86 days
So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.
<h2>
Answer: Pulsars</h2>
A <u>pulsar</u> is a neutron star that emits very intense electromagnetic radiation at short and periodic intervals ( rotating really fast) due to its intense magnetic field that induces this emission.
Nevertheless, it is important to note that all pulsars are neutron stars, but not all neutron stars are pulsars.
Let's clarify:
A neutron star, is the name given to the remains of a supernova. In itself it is the result of the gravitational collapse of a massive supergiant star after exhausting the fuel in its core.
Neutron stars have a small size for their very high density and they rotate at a huge speed.
However, the way to know that a pulsar is a neutron star is because of its high rotating speed.
Answer:
A) d = 11.8m
B) d = 4.293 m
Explanation:
A) We are told that the angle of incidence;θ_i = 70°.
Now, if refraction doesn't occur, the angle of the light continues to be 70° in the water relative to the normal. Thus;
tan 70° = d/4.3m
Where d is the distance from point B at which the laser beam would strike the lakebottom.
So,d = 4.3*tan70
d = 11.8m
B) Since the light is moving from air (n1=1.00) to water (n2=1.33), we can use Snell's law to find the angle of refraction(θ_r)
So,
n1*sinθ_i = n2*sinθ_r
Thus; sinθ_r = (n1*sinθ_i)/n2
sinθ_r = (1 * sin70)/1.33
sinθ_r = 0.7065
θ_r = sin^(-1)0.7065
θ_r = 44.95°
Thus; xonsidering refraction, distance from point B at which the laser beam strikes the lake-bottom is calculated from;
d = 4.3 tan44.95
d = 4.293 m
No, it couldn't be.
On that scale, Neptune would be almost 1,740 MILES from the sun.
ON THAT SCALE !
Answer:
No. of moles, n = 25.022 moles
Given:
Volume of gas in tank, V = 29.1 l
Temperate of gas, T = 
= 273 + 35.8 = 308.8 K
Pressure of gas, P = 21.8 atm
Solution:
Making use of the ideal gas equation which given as:
PV = nRT
where
R = Rydberg's constant = 0.0821 L-atm/mol-K
Re-arranging the above formula for 'n' and putting the values in the above formula:
n = 25.022
