I really wish I could be helping you. I don't know.
Answer:
the expected distance is 4.32 m
Explanation:
given data
half life time = 1.8 ×
s
speed = 0.8 c = 0.8 × 3 ×
to find out
expected distance over
solution
we know c is speed of light in air is 3 ×
m/s
we calculate expected distance by given formula that is
expected distance = half life time × speed .........1
put here all these value
expected distance = half life time × speed
expected distance = 1.8 ×
× 0.8 × 3 ×
expected distance = 4.32
so the expected distance is 4.32 m
F=ma
m=total mass = 2300kg+2500kg=4800
F=18000N
a=?
a=F/m
a=18000/4800
a=3.8m/s^2
Final answer
Answer:
V_infinty=98.772 m/s
Explanation:
complete question is:
The following problem assume an inviscid, incompressible flow. Also, standard sea level density and pressure are 1.23kg/m3(0.002377slug/ft3) and 1.01imes105N/m2(2116lb/ft2), respectively. A Pitot tube on an airplane flying at standard sea level reads 1.07imes105N/m2. What is the velocity of the airplane?
<u>solution:</u>
<u>given:</u>
<em>p_o=1.07*10^5 N/m^2</em>
<em>ρ_infinity=1.23 kg/m^2</em>
<em>p_infinity=1.01*10^5 N/m^2</em>
p_o=p_infinity+(1/2)*(ρ_infinity)*V_infinty^2
V_infinty^2=9756.097
V_infinty=98.772 m/s
Answer:
b) Document lessons learned.
Explanation:
First he should do documentation
then C
then D
then A