Answer:
the fundamental frequency produced by the open pipe is 120.78 Hz
Explanation:
Given;
length of the open pipe, L = 1.42 m
speed of sound in air, v = 343 m/s
The length of the open pipe for the fundamental frequency is equivalent to half of wavelength;
The fundamental frequency produced by the open pipe is calculated as;
Therefore, the fundamental frequency produced by the open pipe is 120.78 Hz
Answer:
see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas
Explanation:
We will propose this exercise using Archimedes' principle, which establishes that the thrust on a body is equal to the volume of the desalted liquid.
B = ρ g V
The weight of a submerged body is the net force between the weight and the thrust
F_net = W - B
we can write the weight as a function of the density
ρ_body = m / V
m = ρ_body V
W = mg
W = ρ _body g V
we substitute
F_net= ( ρ_body - ρ _fluid) g V
In general this force is directed downwards, we can call this value the apparent weight of the body. This is the weight of the submerged body.
W_aparente = ( ρ_body - ρ _fluid) g V
If some air bubbles formed in this body, the net force of these bubbles is
F_net ’= #_bubbles ( ρ_fluido - ρ_air) g V’
this force is directed upwards
whereby the measured force is
F = W_aparente - F_air
As we can see from this analysis, the apparent weight of the body is lower due to the push created by the air brujuleas
Answer:
F = k q1 q2 / r^2
r^2 = k q1 q2 / F = 9E9 * 4E-5 * 10.8E-5 / 4
r^2 = 9 * 4 * 10.8 / 4 * E-1 = 9.72 m^2
r = 3.12 m
Answer:
D. 0.1
Explanation:
Using transformer equation,
N2/N1 = I1/I2................... Equation 1
Where N2 = secondary coil, N1 = primary coil, I1 = input current, I2 = output current.
make I2 the subject of the equation
I2 = I1/(N2/N1)............ Equation 2
From equation 2 above, For the output current of the secondary coil to be 10 times the input current, N2/N1 = 0.1
Hence the right option is D. 0.1
