Answer:
r = 16 Km
Explanation:
given
m_n= 1.67 x 10^-27 Kg
M_star = 3.88 x 10^30 Kg
A= M_star/m_n
A= 3.88*10^30/1.67 x 10^-27
A=2.28 *10^57 neutrons A = The number of neutrons
we use the number of neutrons as a mass number because the star has only neutrons. = 1.2 x 10-15 m
r = r_o*A^1/3
r = 1.2*10^-15*2.28 *10^57^1/3
r = 16 Km
Answer:
20 m/s
30 m/s
Explanation:
Given:
v₀ = -10 m/s
a = -9.8 m/s²
When t = 1 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (1 s)
v = -19.8 m/s
When t = 2 s:
v = v₀ + at
v = (-10 m/s) + (-9.8 m/s²) (2 s)
v = -29.6 m/s
Rounded to one significant figures, the speed of the ball at 1 s and 2 s is 20 m/s and 30 m/s, respectively.
Answer:
Option (C) is the answer
Explanation:
may be it is possible if that we stand so far
It stands for Unified Computing System
I hope I helped :3