Answer:
Magnitude of vector A = 0.904
Explanation:
Vector A , which is directed along an x axis, that is
Vector B , which has a magnitude of 5.5 m
The sum is a third vector that is directed along the y axis, with a magnitude that is 6.0 times that of vector A
Comparing we will get
Substituting in
So we have
Magnitude of vector A = 0.904
Answer:
The maximum radius the asteroid can have for her to be able to leave it entirely simply by jumping straight up is approximately 1782.45 meters
Explanation:
Whereby the height the astronaut can jump on Earth = 0.500 m, we have the following kinematic equation;
v² = u² - 2·g·h
Where;
v = The final velocity
u = The initial velocity
g = The acceleration due to gravity ≈ 9.8 m/s²
h = The height she jumps
At the maximum height, = 0.500 m, she jumps, v = 0, therefore, we have;
0² = u² - 2·g·
u² = 2 × 9.8 × 0.5 = 9.8
u = √9.8 ≈ 3.13
u = 3.13 m/s
Her initial jumping velocity ≈ 3.13 m/s
Escape velocity,
Where;
M = The mass of the asteroid
G = The Universal gravitational constant = 6.67408 × 10⁻¹¹ m³/(kg·s²)
r = The radius of the asteroid
The average density of the Earth = 5515 kg/m³
The mass of the asteroid, M = Density × Volume = 5515 kg/m³× 4/3 × π × r³
The escape velocity, she has, ≈ 3.13 m/s is therefore;
Therefore, the maximum radius of the asteroid can have for her jumping velocity to be equal to the escape velocity for her to be able to leave it entirely simply by jumping straight up = r ≈ 1782.45 meters.
Answer:
a) - 72.5°c
b) pressure = 3625.13 Pa
c) density = 0.063 kg/m^3
d) it is a subsonic aircraft
Explanation:
a) Determine Temperature
Temperature at 19.5 km ( 19500 m )
T = -131 + ( 0.003 * altitude in meters )
= -131 + ( 0.003 * 19500 ) = - 72.5°c
b) Determine pressure and density at 19.5 km altitude
Given :
Po (atmospheric pressure at sea level ) = 101kpa
R ( gas constant of air ) = 0.287 KJ/Kgk
T = -72.5°c ≈ 200.5 k
pressure = 3625.13 Pa
hence density = 0.063 kg/m^3
attached below is the remaining part of the solution
C) determine if the aircraft is subsonic or super sonic
Velocity ( v ) = =
= 283.8 m/s
hence it is a subsonic aircraft
