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3 years ago

In which region of the electromagnetic spectrum is most of earth's outgoing terrestrial radiation?

1 answer:

6 0

The answer is Infrared. The infrared of the electromagnetic spectrum is most of earth's outgoing terrestrial radiation. <span>Earth is the hot body with temperature of 30 degrees on the average.</span>

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A spinning tossed baseball veers off course in the direction ofA) reduced air pressure on the ball.B) increased air pressure on

katovenus [111]

Answer:

A) reduced air pressure on the ball.

Explanation:

3 0

3 years ago

The distance between the crests of waves

stepladder [879]

It depends, You have to have the length and the width of the crest wave.

8 0

3 years ago

Read 2 more answers

A telephone line has a signal-to-noise ratio of 1000 and a bandwidth of 4 KHz. What is the maximum data rate supported by this l

ivolga24 [154]

Answer:

The maximum data rate supported by this line is 39900 bps

Explanation:

The maximum data rate supported by this line can be obtained using the formula below

c = W*log2(S/N+1)

where;

c is the maximum data rate supported by the line

W is the bandwidth = 4kHz

S/N+1 is the signal to noise ratio = 1001

c = 4*log2(1001)

c = 39868.9 ≅ 39900 bps

Therefore, the maximum data rate supported by this line is 39900 bps

5 0

4 years ago

Can i please get some help with this.

Ivenika [448]

the answer is D cuz electricity is a conductive

5 0

4 years ago

Oil having a density of 926 kg/m3 floats on water. A rectangular block of wood 3.69 cm high and with a density of 974 kg/m3 floa

zloy xaker [14]

Answer:

the position of the wood below the interface of the two liquids is 2.39 cm.

Explanation:

Given;

density of oil, $\rho _o$ = 926 kg/m³

density of the wood, $\rho _{wood}$ = 974 kg/m³

density of water, $\rho _w$ = 1000 kg/m³

height of the wood, h = 3.69 cm

Based on the density of the wood, it will position across the two liquids.

let the position of the wood below the interface of the two liquids = x

Let the wood be in equilibrium position;

$F_{wood} - F_{oil} - F_{water} = 0\\\\\rho _{wood} .gh - \rho _o .g(h-x) - \rho_w .gx = 0\\\\\rho _{wood} .h - \rho _o (h-x) - \rho_w .x = 0\\\\\rho _{wood} .h -\rho _o h + \rho _o x - \rho_w .x =0\\\\h (\rho _{wood} -\rho _o ) = x( \rho_w - \rho _o)\\\\x =h[\frac{ \rho _{wood} -\rho _o }{\rho_w - \rho _o} ]\\\\x = 3.69\ cm \times [\frac{974 - 926}{1000-926} ]\\\\x = 2.39 \ cm$

Therefore, the position of the wood below the interface of the two liquids is 2.39 cm.

6 0

3 years ago