Answer:
0.5 m
Explanation:
Givens:
ym1 = 2.5 mm
ym2 = 4.5 mm
Ф_1=π / 4
Ф_2=π / 2
We have 2 ways to solve this problem. The first one given that the 2 waves have the frequency then we know that the resultant wave amplitude is
Ym = (ym1 + ym2)cos(Ф_2/2)
By substitution we have
Ym= (0.025 + 0.045)cos(π/4) = 0.496 m
The second one is it treat them as Phasors where the phase between them is Ф_2=π / 2 Therefore
Ym^2=(ym1^2+ym2^2)
So we have Ym=√0.025^2+0.045^2
= 0.5 m
<span>Acceleration is the change in velocity divided by time taken. It has both magnitude and direction. In this problem, the change in velocity would first have to be calculated. Velocity is distance divided by time. Therefore, the velocity here would be 300 m divided by 22.4 seconds. This gives a velocity of 13.3928 m/s. Since acceleration is velocity divided by time, it would be 13.3928 divided by 22.4, giving a final solution of 0.598 m/s^2.</span>
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
Explanation:
The work done equals the change in energy.
W = ΔKE
W = 0 − ½mv²
W = -½ (0.270 kg) (-7.50 m/s)²
W = -7.59 J
Work is force times displacement.
W = Fd
-7.59 J = F (-0.150 m)
F = 50.6 N
