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
For t1:
t1 = square root of 2h1 / g = square root of 2 * 0.5 / 9.8 = 0.319 sec
For t2:
t2 = sqaure root of 2h2 / g = square root of 2 * 1.0 / 9.8 = 0.451 sec
Wherein:
t = time(s) for the vertical movement
h= height
g = gravity (using the standard 9.8 m/sec measurement)
d1 = 1*0.319 = 0.319 m
d2 = 0.5 * 0.451 = 0.225 m
Where:
d = hor. distance
ratio = d1:d2
= 0.319 : 0.225
=3.19 : 2.25
The answer is 3.19 : 2.25
Answer:
t = 12,105.96 sec
Explanation:
Given data:
weight of spacecraft is 2000 kg
circular orbit distance to saturn = 180 km
specific impulse = 300 sec
saturn orbit around the sun R_2 = 1.43 *10^9 km
earth orbit around the sun R_1= 149.6 * 10^ 6 km
time required for the mission is given as t
![t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%5Cmu_sun%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28R_1%20%2B%20R_2%29%5D%5E%7B3%2F2%7D)
where
is gravitational parameter of sun = 1.32712 x 10^20 m^3 s^2.![t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}](https://tex.z-dn.net/?f=t%20%3D%20%5Cfrac%7B2%5Cpi%7D%7B%5Csqrt%7B%201.32712%20x%2010%5E%7B20%7D%7D%7D%20%5B%5Cfrac%7B1%7D%7B2%7D%28149.6%20%2A%2010%5E%206%20%2B1.43%20%2A10%5E9%20%29%5D%5E%7B3%2F2%7D)
t = 12,105.96 sec
Answer:
Explained
Explanation:
Metals are good conductors of electricity because they contain free electrons in their atoms. The outer shell of atom's of metal have free electrons. These free electrons are responsible of electrical conductivity of metals. These electron are not bounded by the attraction forces of the nucleous. They are free to wonder in lattice of positive ion and thus allow electrical conductivity.
The net speed due west is = distance traveled in west / time taken = 120/0.5 = 240 km/h.
so airspeed due west is = net speed - speed of plane = 240-220= 20 km/h.
airspeed due south is = distance traveled in west / time taken= 20/0.5= 40 km/h.
the magnitude of the wind velocity = √[(airspeed due south )² + (airspeed due west)²] = √ ( 40^2 + 20^2 ) = 44.72 km/h
the angle of airspeed south of west is tan⁻¹ ( airspeed due south / airspeed due west )= tan⁻¹(40/20)=63.43 degrees.
if wind velocity is 40 km/h due south, her velocity should have 20 km/h component in north.
so component west = sqrt ( 220^2 - 40^2 ) = 216.33 km/h.
the angle north of west is arctan( 40/216.33 ) = 10.47 degrees.