Answer:
a) v = 0.9167 m / s, b) A = 0.350 m, c) v = 0.9167 m / s, d) A = 0.250 m
Explanation:
a) to find the velocity of the wave let us use the relation
v = λ f
the wavelength is the length that is needed for a complete wave, in this case x = 5.50 m corresponds to a wavelength
λ = x
λ = x
the period is the time for the wave to repeat itself, in this case t = 3.00 s corresponds to half a period
T / 2 = t
T = 2t
period and frequency are related
f = 1 / T
f = 1 / 2t
we substitute
v = x / 2t
v = 5.50 / 2 3
v = 0.9167 m / s
b) the amplitude is the distance from a maximum to zero
2A = y
A = y / 2
A = 0.700 / 2
A = 0.350 m
c) The horizontal speed of the traveling wave (waves) is independent of the vertical oscillation of the particles, therefore the speed is the same
v = 0.9167 m / s
d) the amplitude is
A = 0.500 / 2
A = 0.250 m
19,600 Newtons (about 4,400 pounds).
On Earth only.
Different in other places.
Answer:
his is a multi-part question. Once an answer is submitted, you will be unable to return to this part. Air enters a nozzle steadily at 2.21 kg/m3 and 40 m/s and leaves at 0.762 kg/m3 and 192 m/s. The inlet area of the nozzle is 90 cm2.
determine (a) the mass flow rate through the nozzle, and (b) the exit area of the nozzle.
a)0.7956kg/s
b)5.437 × 10⁻³m²
Explanation:
The concepts related to the change of mass flow for both entry and exit is applied
The general formula is defined by

Where,

values are divided by inlet(1) and outlet(2) by


PART A) Applying the flow equation

PART B) For the exit area we need to arrange the equation in function of Area, that is

Answer:
Mass doesn't change.
Weight is measured based on gravitational pull.
Explanation: