Respond with TRUE if the symbol of the valve shown belows

A sinusoidal wave of frequency 420 Hz has a speed of 310 m/s. (a) How far apart are two points that differ in phase by π/8 rad?

Olin [163]

Answer:

a) Two points that differ in phase by π/8 rad are 0.0461 m apart.

b) The phase difference between two displacements at a certain point at times 1.6 ms apart is 4π/3.

Explanation:

f = 420 Hz, v = 310 m/s, λ = wavelength = ?

v = fλ

λ = v/f = 310/420 = 0.738 m

T = periodic time of the wave = 1/420 = 0.00238 s = 0.0024 s = 2.4 ms

a) Two points that differ in phase by π/8 rad

In terms of the wavelength of the wave, this is equivalent to [(π/8)/2π] fraction of a wavelength,

[(π/8)/2π] = 1/16 of a wavelength = (1/16) × 0.738 = 0.0461 m

b) two displacements at times 1.6 ms apart.

In terms of periodic time, 1.6ms is (1.6/2.4) fraction of the periodic time.

1.6/2.4 = 2/3.

This means those two points are 2/3 fraction of a periodic time away from each other.

1 complete wave = 2π rad

Points 2/3 fraction of a wave from each other will have a phase difference of 2/3 × 2π = 4π/3.

8 0

3 years ago

Tech A says that 18 AWG wire is larger than 12 AWG wire. Tech B says that the larger the diameter of the conductor, the more ele

shutvik [7]

Answer:

Both of them are wrong

Explanation:

The two technicians have given the wrong information about the wires.

This is because firstly, a higher rating of AWG means it is smaller in diameter. Thus, the diameter of a 18 AWG wire is smaller than that of a 12 AWG wire and that makes the assertion of the technician wrong.

Also, the higher the resistance, the smaller the cross sectional area meaning the smaller the diameter. A wire with bigger cross sectional area will have a smaller resistance

So this practically makes the second technician wrong too

8 0

3 years ago

Tech A says that a cylinder leakage test is performed on a cylinder with low compression to determine the severity of the leak a

Karolina [17]

Tech A djjdjdndnndndbdbx

4 0

2 years ago

. The flexure strength test was performed on a concrete beam having a cross section of 0.15m by 0.15m and a span of 0.45m. If th

ivann1987 [24]

Answer:

σ =5.39Mpa

Explanation:

step one:

The flexure strength is defined as the tendency with which unreinforced concrete yield to bending forces

Flexural strength test Flexural strength is calculated using the equation:

σ = FL/ (bd^2 )----------1

Where

σ = Flexural strength of concrete in Mpa

F= Failure load (in N).

L= Effective span of the beam

b= Breadth of the beam

step two:

Given data

F=40.45 kN= 40450N

b=0.15m

d=0.15m

L=0.45m

step three:

substituting into the expression we have

σ = 40450*0.45/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.15^2 )

σ =18202.5/ (0.15*0.0225 )

σ =18202.5/0.003375

σ =5393333.3

σ =5393333.3/1000000

σ =5.39Mpa

Therefore the flexure strength of the concrete is 5.39Mpa

5 0

2 years ago

Water flows in a tube that has a diameter of D= 0.1 m. Determine the Reynolds number if the average velocity is 10 diameters per

Cloud [144]

Answer:

a) $Re_{D} = 111896.745$ , b) $Re_{D} = 1.119\times 10^{-7}$

Explanation:

a) The Reynolds number for the water flowing in a circular tube is:

$Re_{D} = \frac{\rho\cdot v\cdot D}{\mu}$

Let assume that density and dynamic viscosity at 25 °C are $997\,\frac{kg}{m^{3}}$ $0.891\times 10^{-3}\,\frac{kg}{m\cdot s}$ , respectively. Then:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (1\,\frac{m}{s} )\cdot (0.1\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 111896.745$

b) The result is:

$Re_{D}=\frac{(997\,\frac{kg}{m^{3}} )\cdot (10^{-6}\,\frac{m}{s} )\cdot (10^{-7}\,m)}{0.891\times 10^{-3}\,\frac{kg}{m\cdot s} }$

$Re_{D} = 1.119\times 10^{-7}$

6 0

3 years ago