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2 years ago

The forward movement of orbital waves classifies them as ____ waves.

1 answer:

7 0

The forward movement of orbital waves classifies them as progressive waves.
Orbital waves are waves that occur at the interface of two materials, with different densities e.g water and air. Ocean waves for example are all characterized by a circular motion. They combine the characteristics of longitudinal and transverse waves. Orbital waves are classified as progressive wave as the wave fronts move through a medium.

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A 2650-lb car is traveling at sea level at a constant speed. its engine is running at 4500 rev/min and is producing 175 ft-lb of

Zepler [3.9K]

Check the attached file for the answer.

6 0

2 years ago

Read 2 more answers

3-m-high large tank is initially filled with water. The tank water surface is open to the atmosphere, and a sharp-edged 10-cm-di

irinina [24]

Answer:

The initial velocity of the water from the tank is 5.42 m/s

Explanation:

By applying Bernoulli equation between point 1 and 2

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

At the point 1

P₁=0 ( Gauge pressure)

V₁= 0 m/s

Z₁=3 m

At point 2

P₂=0 ( Gauge pressure)

Z₂= 0 m/s

$h_L=1.5\ m$

Now by putting the values

$\dfrac{P_1}{\rho g}+\dfrac{V_1^2}{2g}+Z_1=\dfrac{P_2}{\rho g}+\dfrac{V_2^2}{2g}+Z_2+h_L$

$Z_1-h_L=\dfrac{V_2^2}{2g}$

$3-1.5=\dfrac{V_2^2}{2\times 9.81}$

$V_2=\sqrt{2\times 1.5\times 9.81}\ m/s$

V₂= 5.42 m/s

The initial velocity of the water from the tank is 5.42 m/s

3 0

3 years ago

A material that has a fracture toughness of 33 MPa.m0.5 is to be made into a large panel that is 2000 mm long by 250 mm wide and

scoray [572]

Answer:

$F_{allow} = 208.15kN$

Explanation:

The word 'nun' for thickness, I will interpret in international units, that is, mm.

We will begin by defining the intensity factor for the steel through the relationship between the safety factor and the fracture resistance of the panel.

The equation is,

$K_{allow} =\frac{K_c}{N}$

We know that $K_c$ is 33Mpa*m^{0.5} and our Safety factor is 2,

$K_{allow} = \frac{33Mpa*m^{0.5}}{2} = 16.5MPa.m^{0.5}$

Now we will need to find the average width of both the crack and the panel, these values are found by multiplying the measured values given by 1/2

<em>For the crack;</em>

$\alpha = 0.5*L_c = 0.5*4mm = 2mm$

<em>For the panel</em>

$\gamma = 0.5*W = 0.5*250mm = 125mm$

To find now the goemetry factor we need to use this equation

$\beta = \sqrt{sec(\frac{\pi\alpha}{2\gamma})}\\\beta = \sqrt{sec(\frac{2\pi}{2*125mm})}\\\beta = 1$

That allow us to determine the allowable nominal stress,

$\sigma_{allow} = \frac{K_{allow}}{\beta \sqrt{\pi\alpha}}$

$\sigma_{allow} = \frac{16.5}{1*\sqrt{2*10^{-3} \pi}}$

\sigma_{allow} = 208.15Mpa

So to get the force we need only to apply the equation of Force, where

$F_{allow}=\sigma_{allow}*L_c*W$

$F_{allow} = 208.15*250*4$

$F_{allow} = 208.15kN$

That is the maximum tensile load before a catastrophic failure.

4 0

2 years ago

To practice Problem-Solving Strategy 29.1: Faraday's Law. A metal detector uses a changing magnetic field to detect metallic obj

ExtremeBDS [4]

Answer:

$1.138\times 10^{-3}V$

Explanation:

Apply Faraday's Newmann Lenz law to determine the induced emf in the loop:

$\epsilon=\frac{d\phi}{dt}$

where:

$d\Phi-$ variation of the magnetic flux

$dt-$ is the variation of time

#The magnetic flux through the coil is expressed as:

$\Phi=NBA \ Cos \theta$

Where:

N- number of circular loops

A-is the Area of each loop( $A=\pi r^2=\pi \times 5^2=78.5398$ )

B-is the magnetic strength of the field.

$\theta=15\textdegree$ - is the angle between the direction of the magnetic field and the normal to the area of the coil.

$\epsilon=-\frac{d(78.5398\times 10^{-3}NB \ Cos \theta)}{dt}\\\\=-(78.5398\times 10^{-3}N\ Cos \theta)}{\frac{dB}{dt}$

$\frac{dB}{dt}-$ =0.0250T/s is given as rate at which the magnetic field increases.

#Substitute in the emf equation:

$=-(78.5398\times 10^{-3} m^2 \times 6\ Cos 15 \textdegree)\times 0.0250T/s\\\\=1.138\times 10^{-3}V$

Hence, the induced emf is $1.138\times 10^{-3}V$

4 0

2 years ago

If a scale on Earth reads 650 N, what is your mass?

OlgaM077 [116]

If the scale reads 650N, then the mass of whoever it is standing on the scale is

(weight) / (gravity) = (650N) / (9.8 m/s²) = 66.3 kilograms .

It's not MY mass, even if I'm the one standing on the scale.
If I stand on a scale and it reads 650 N, the scale is broken.

4 0

2 years ago