Plzzz helpp mee guysss!!! plzzzz

Ran 300 meter in 40 seconds, what is the speed?

nignag [31]

Answer:

7.5

Explanation:

Speed= distance/ time

Speed= 300m/40sec

Speed= 7.5m/s

5 0

3 years ago

7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

3 years ago

Leonard is stressed and feeling overwhelmed by the college selection process. What should he do to combat this problem?

exis [7]

He should ask other people for advice

6 0

3 years ago

Your Friend Wants To Travel Abroad For University Education.Write a Letter Expressing Your Views about his or her intentions and

Elena L [17]

Answer:

ALT TO FRIEND

HI FIREND

COPY BOB IN ALT ACC

CANT GET BAN

───▄▀▀▀▄▄▄▄▄▄▄▀▀▀▄───

───█▒▒░░░░░░░░░▒▒█───

────█░░█░░░░░█░░█──── this is Taco

─▄▄──█░░░▀█▀░░░█──▄▄─

█░░█─▀▄░░░░░░░▄▀─█░░█

░░░░░▐▀█▀▌░░░░▀█▄░░░

░░░░░▐█▄█▌░░░░░░▀█▄░░

░░░░░░▀▄▀░░░▄▄▄▄▄▀▀░░

░░░░▄▄▄██▀▀▀▀░░░░░░░

░░░█▀▄▄▄█░▀▀░░

░░░▌░▄▄▄▐▌▀▀▀░░ This is Bob

▄░▐░░░▄▄░█░▀▀ ░░ they have made an alliance

▀█▌░░░▄░▀█▀░▀ ░░ Copy and paste them so we can take over brainly.

░░░░░░░▀███▀█░▄░░

░░░░░░▐▌▀▄▀▄▀▐▄░░

░░░░░░▐▀░░░░░░▐▌░

░░░░░░█░░░░░░░░█░

Explanation:

7 0

3 years ago

Una fuerza F de 200 lb actúa a lo largo de AB, sobre la rampa mostrada. fuerza de F respecto del eje OC. Calcule el momento de f

ruslelena [56]

Answer:

Moc = -613.25 [lb*in]

Explanation:

Este problema se puede resolver mediante la mecánica vectorial, es decir se realizara un analisis de vectores.

Primero se calculara el momento de la fuerza F_AB con respecto al punto O, debemos recordar que el momento con respecto a un punto se define como el producto cruz de la distancia por la fuerza.

$M_{o}=r_{A/O} * F_{AB}$ (producto cruz)

Necesitamos identificar los puntos:

O (0,0,0) [in]

A (12,0,0) [in]

B (0, 24,8) [in]

C (12,24,0) [in]

$r_{A/O}=(12,0,0) - (0,0,0)\\r_{A/O} = 12 i + 0j+0k [in]\\AB = (0,24,8) - (12,0,0)\\AB = -12i+24j+8k [in]\\[LAB]=\frac{-12i+24j+8k}{\sqrt{(12)^{2} +(24)^{2} +(8)^{2} } }\\ LAB=-\frac{3}{7} i+\frac{6}{7}j+\frac{2}{7}k$

El ultimo vector calculado corresponde al vector unitario (magnitud = 1) de AB. El vector fuerza corresponderá al producto del vector unitario por la magnitud de la fuerza = 200 [lb].

$F_{AB}=-\frac{600}{7} i +\frac{1200}{7}j+\frac{400}{7} k [Lb]$

De esta manera realizando el producto cruz tenemos

$M_{O}=r_{A/O} * F_{AB}$

$M_{O}=0i-685.7j+2057.1k [Lb*in]$

Para calcular el momento con respecto a la diagonal OC, necesitamos el vector unitario de esta diagonal.

$OC = (12,24,0)-(0,0,0)\\OC= 12i+24j+0k[Lb]\\LOC = \frac{12i+24j+0k}{\sqrt{(12)^{2} +(24)^{2} +(0)^{2} } } \\LOC=\frac{12}{\sqrt{720}}i+\frac{24}{\sqrt{720}}j +0k$

El vector con respecto al eje OC, es igual al producto punto del momento en el punto O por el vector unitario LOC

$M_{OC}=L_{OC}*M_{O}\\M_{OC}=(\frac{12}{\sqrt{720}}i +\frac{24}{\sqrt{720}} j+0k )* (0i-685.7j+2057.1k)\\M_{OC}= -613.32[Lb*in]$

7 0

2 years ago