Answer:
Distance, d = 192 meters
Explanation:
We have,
Initial velocity of an object is 10 m/s
Acceleration of the object is 3.5 m/s²
Time, t = 8 s
We need to find the distance travelled by the object during that time. Second equation of motion gives the distance travelled by the object. It is given by :
So, the distance travelled by the object is 192 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
Answer:51.44 units
Explanation:
Given
x component of vector is
y component of vector is
so position vector is
Magnitude of vector is
|r|=51.44 units
Direction
vector is in 2nd quadrant thus
The answer would 0. The reasoning of this is because freezing point in celsius is always 0 degrees but in fahrenheit the freezing point is 32 degrees.