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3 years ago

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A WLAN transmitter that emites a 100 mW signal is connected to a cable with a 3dB loss. If the cable is connected to an antenna

with 7 dBi of gain, what is the EIRP at the antenna element?

1 answer:

irakobra [83]3 years ago

6 0

Answer: 24 dBm

Explanation:

in the attachment

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A ___ is when too much water moves into an area

Marina86 [1]

A. Flood.
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3 years ago

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An airplane is moving at 350 km/hr. If a bomb is

Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

$y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2}$ (1)

$x=V_{ox}t$ (2)

$V_{f}=V_{oy}-gt$ (3)

Where:

$y=0 m$ is the bomb's final jeight

$y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m$ is the bomb'e initial height

$V_{oy}=0 m/s$ is the bomb's initial vertical velocity, since the airplane was moving horizontally

$t$ is the time

$g=9.8 m/s^{2}$ is the acceleration due gravity

$x$ is the bomb's range

$V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s$ is the bomb's initial horizontal velocity

$V_{f}$ is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

$y_{o}=\frac{1}{2}gt^{2}$ (4)

Isolating $t$ :

$t=\sqrt{\frac{2 y_{o}}{g}}$ (5)

$t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}}$ (6)

$t=17.49 s$ (7)

<h3>a) Final velocity</h3>

Since $V_{oy}=0 m/s$ , equation (3) is written as:

$V_{f}=-gt$ (8)

$V_{f}=-(97.22)(17.49 s)$ (9)

$V_{f}=-171.402 m/s$ (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

$x=(97.22 m/s)(17.49 s)$ (11)

$x=1700.99 m$ (12)

5 0

2 years ago

Science help please!

OLga [1]

Answer:

104 N

Explanation:

m = 1300 kg

a = 0.08m/s^2

F = 1300*0.08

F = 104 N

Newtons is the unit of force.

3 0

3 years ago

A discus thrower turns with angular acceleration of 50 rad/s2, moving the discus in a circle of radius 0.80m. Find the radial an

anyanavicka [17]

Answer:

The value of tangential acceleration $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

The value of radial acceleration $\alpha_{r} = 80 \frac{m}{s^{2} }$

Explanation:

Angular acceleration = 50 $\frac{rad}{s^{2} }$

Radius of the disk = 0.8 m

Angular velocity = 10 $\frac{rad}{s}$

We know that tangential acceleration is given by the formula $\alpha_{t} =$ $r \alpha$

Where r = radius of the disk

$\alpha$ = angular acceleration

⇒ $\alpha_{t} =$ 0.8 × 50

⇒ $\alpha_{t} =$ 40 $\frac{m}{s^{2} }$

This is the value of tangential acceleration.

Radial acceleration is given by

$\alpha_{r} = \frac{V^{2} }{r}$

Where V = velocity of the disk = r $\omega$

⇒ V = 0.8 × 10

⇒ V = 8 $\frac{m}{s}$

Radial acceleration

$\alpha_{r} = \frac{8^{2} }{0.8}$

$\alpha_{r} = 80 \frac{m}{s^{2} }$

This is the value of radial acceleration.

7 0

3 years ago

MY LAST ONE<,Brainliest for best answer!

grigory [225]

_Award brainliest if helped!
Mechanical Advantage = Force by Hammer / Force by Nail = 160/40 = 4

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3 years ago

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