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3 years ago

6

Give a physical description of radio waves and compare and contrast it with the frequency, wavelength, energy and wave speed wit

h those of other regions of the EMS.

1 answer:

Ad libitum [116K]3 years ago

8 0

There is nothing physical about radio waves. They are pulses of energy caused by periodic variations in electric and magnetic fields. These are caused by electrons “jumping” between energy levels. Radio waves have the lowest energy, longest wavelengths and lowest frequencies. These values change gradually as we go through microwaves, infra red, visible light, ultra violet, x-rats abd gamma rays which have the highest energy, highest frequency and shortest wavelength.

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What is the force per unit area at this point acting normal to the surface with unit nor- Side View √√ mal vector n = (1/ 2)ex +

Mumz [18]

Complete Question:

Given $\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$ at a point. What is the force per unit area at this point acting normal to the surface with $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$ ? Are there any shear stresses acting on this surface?

Answer:

Force per unit area, $\sigma_n = 28 MPa$

There are shear stresses acting on the surface since $\tau \neq 0$

Explanation:

$\sigma = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right]$

equation of the normal, $\b n = (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z$

$\b n = \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

Traction vector on n, $T_n = \sigma \b n$

$T_n = \left[\begin{array}{ccc}10&12&13\\12&11&15\\13&15&20\end{array}\right] \left[\begin{array}{ccc}\frac{1}{\sqrt{2} }\\0\\\frac{1}{\sqrt{2} }\end{array}\right]$

$T_n = \left[\begin{array}{ccc}\frac{23}{\sqrt{2} }\\0\\\frac{27}{\sqrt{33} }\end{array}\right]$

$T_n = \frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z$

To get the Force per unit area acting normal to the surface, find the dot product of the traction vector and the normal.

$\sigma_n = T_n . \b n$

$\sigma \b n = (\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z) . ((1/ \sqrt{2} ) \b e_x + 0 \b e_y +(1/ \sqrt{2}) \b e_z)\\\\\sigma \b n = 28 MPa$

If the shear stress, $\tau$ , is calculated and it is not equal to zero, this means there are shear stresses.

$\tau = T_n - \sigma_n \b n$

$\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - 28( (1/ \sqrt{2} ) \b e_x + (1/ \sqrt{2}) \b e_z)\\\\\tau = [\frac{23}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{33}{\sqrt{2} } \b e_z] - [ (28/ \sqrt{2} ) \b e_x + (28/ \sqrt{2}) \b e_z]\\\\\tau = \frac{-5}{\sqrt{2} } \b e_x + \frac{27}{\sqrt{2} } \b e_y + \frac{5}{\sqrt{2} } \b e_z$

$\tau = \sqrt{(-5/\sqrt{2})^2 + (27/\sqrt{2})^2 + (5/\sqrt{2})^2} \\\\ \tau = 19.74 MPa$

Since $\tau \neq 0$ , there are shear stresses acting on the surface.

3 0

2 years ago

Please help Math Phys

Lorico [155]

Answer:

269 m

45 m/s

-58.6 m/s

Explanation:

Part 1

First, find the time it takes for the package to land. Take the upward direction to be positive.

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: t

Δy = v₀ t + ½ at²

(-175 m) = (0 m/s) t + ½ (-9.8 m/s²) t²

t = 5.98 s

Next, find the horizontal distance traveled in that time:

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: Δx

Δx = v₀ t + ½ at²

Δx = (45 m/s) (5.98 s) + ½ (0 m/s²) (5.98 s)²

Δx = 269 m

Part 2

Given (in the x direction):

v₀ = 45 m/s

a = 0 m/s²

t = 5.98 s

Find: v

v = at + v₀

v = (0 m/s²) (5.98 s) + (45 m/s

v = 45 m/s

Part 3

Given (in the y direction):

Δy = -175 m

v₀ = 0 m/s

a = -9.8 m/s²

Find: v

v² = v₀² + 2aΔy

v² = (0 m/s)² + 2 (-9.8 m/s²) (-175 m)

v = -58.6 m/s

6 0

2 years ago

Define polarization in reference in the simple cell

vampirchik [111]

Answer:polarization pertains to the act or process of producing a positive electrical charge and a negative electrical charge such that between a nerve cell internal electrical charge..................

Explanation:

7 0

3 years ago

The uniform slender bar AB has a mass of 6.4 kg and swings in a vertical plane about the pivot at A. If θ˙ = 2.7 rad/s when θ =

dolphi86 [110]

Answer:

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Explanation:

Given data,

The mass of the bar AB, m = 6.4 kg

The angular velocity of the bar, θ˙ = 2.7 rad/s

The angle of the bar at A, θ = 24°

Let the length of the bar be, L = l

The angular moment at point A is,

∑ Mₐ = Iα

Where, Mₐ - the moment about A

α - angular acceleration

I - moment of inertia of the rod AB

$-mg(\frac{lcos\theta}{2})=\frac{1}{3}(ml^{2})\alpha$

$\alpha=\frac{-3gcos\theta}{2l}$

Let G be the center of gravity of the bar AB

The position vector at A with respect to the origin at G is,

$\vec{r_{G}}=[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

The acceleration at the center of the bar

$\vec{a_{G}}=\vec{a_{a}}+\vec{\alpha}X\vec{r_{G}}-\omega^{2}\vec{r_{G}}$

Since the point A is fixed, acceleration is 0

The acceleration with respect to the coordinate axes is,

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=0+(\frac{-3gcos\theta}{2l})\hat{k}\times[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]-\omega^{2}[\frac{lcos\theta}{2}\hat{i}-\frac{lcos\theta}{2}\hat{j}]$

$(\vec{a_{G}})_{x}\hat{i}+(\vec{a_{G}})_{y}\hat{j}=[-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}\hat{i}+(\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4})\hat{j}]$

Comparing the coefficients of i

$=-\frac{cos\theta(2l\omega^{2}+3gsin\theta)}{4}$

Comparing coefficients of j

$(\vec{a_{G}})_{y}=\frac{2l\omega^{2}sin\theta-3gcos^{2}\theta}{4}$

Net force on x direction

$F_{x}=(\vec{a_{G}})_{x}$

substituting the values

$F_{x}$ =1.5(14.58L+11.96)

Similarly net force on y direction

$F_{y}=(\vec{a_{G}})_{y}+mg$

= 3.2(2.97L - 157.03) + 62.72

Where L is the length of the bar AB

Therefore the net force,

$F=\sqrt{F_{x}^{2}+F_{y}^{2}}$

F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]

Substituting the value of L gives the force at pin A

8 0

2 years ago

A siren on the top of the police car emits sound at frequency of 823 Hz. The car is approaching a bystander on a sidewalk at spe

statuscvo [17]

Answer:

The bystander will hear a frequency of 931.18 Hz.

Explanation:

let V = 39.5 m/s be velocity of the police car and Fs be the frequency a siren of the police car emits. let v = 0 m/s be the velocity of the bystander and Vs = 340 m/s is the velocity of sound in air.

then, we know that the bystander has to hear a higher frequency than the one emitted and the frequency that the bystander hears is given by the doppler relation:

Fo = [(Vs)/(Vs - V)]×Fs

= [(Vs)/(Vs - V)]×Fs

= [(340)/(340 - 39.5)]×(823)

= 931.18 Hz

Therefore, the bystander will hear a frequency of 931.18 Hz.

3 0

2 years ago