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:
It is given that,
Speed of a wave, v = 251 m/s
Wavelength of the wave, λ = 5.1 cm = 0.051 m
(1) The frequency of the wave is given by :
(2) Angular frequency of the wave is given by :
(3) The period of oscillation is given by T as :
T = 0.000203 seconds
or
T = 0.203 milliseconds
Hence, this is the required solution.
Explanation :
The interaction between two objects is termed as the collision. The collision can be of two types i.e. elastic collision and inelastic collision.
In this case, two identical carts travel at the same speed toward each other, and then a collision occurs. In an inelastic collision, the momentum before and after the collision remains the same but its kinetic energy gets lost.
After the collision, both the object sticks over each other and moves with one velocity.
Out of the given graph, the graph that shows a perfectly inelastic collision is attached. It shows that after the collision both the carts move with the same velocity.
Answer: The same current flows through bth cables
Explanation:
Lets have a look to the next two equations
The Ohm´s V = I*R (1)
where:
V is voltage (potencial dfference) in volts
I is the electric current in ampers
R is the electric resistance
When a voltage is applied as the electrc load is not specified ( we have to assume is the same) the current will be the same
And in the other hand the resistance R =ρL/s
Where ρ is the resistivity of the conductor L the length and s square section of the conductor
If we assume that the smaller diameter cable is able to conduct the current then nothing happens. The point is that the capacity of conduction of current depend on the section of the cable (the area)
Tables exist where to find the capacity of each cable according to its diameter.
W = 1/2k*x^2.
k = spring constant = 2500 n/m.
x = distance = 4 cm = 0.04m (convert to same units).
W = 1/2(2500)(0.04)^2 = 2J.
