Answer:
A star with 15 solar masses is too big to be a main-sequence star.
Hi There! :)
An equilibrium constant is not changed by a change in pressurea. True
b. False
False! :P
Answer:
Explanation:
reading of scale = reaction force of surface R
centripetal force = R - mg = m v² / R , m is mass , v is velocity and R is radius of the circular path .
R = mg + m v² / R
given ,
m v² / R = .80 mg
v² = .80 x g x R
= .8 x 9.8 x 9 = 70.56
v = 8.4 m /s
That's what scientists and other technical people call the object's "<em>volume</em>".
Answer:
9.36*10^11 m
Explanation
Orbital velocity v=√{(G*M)/R},
G = gravitational constant =6.67*10^-11 m³ kg⁻¹ s⁻²,
M = mass of the star
R =distance from the planet to the star.
v=ωR, with ω as the angular velocity and R the radius
ωR=√{(G*M)/R},
ω=2π/T,
T = orbital period of the planet
To get R we write the formula by making R the subject of the equation
(2π/T)*R=√{(G*M)/R}
{(2π/T)*R}²=[√{(G*M)/R}]²,
(4π²/T²)*R²=(G*M)/R,
(4π²/T²)*R³=G*M,
R³=(G*M*T²)/4π²,
R=∛{(G*M*T²)/4π²},
Substitute values
R=9.36*10^11 m
