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3 years ago

How do very long radio waves reach stations beyond the curvature of the Earth?

2 answers:

8 0

They actually bounce off a charged layer of atmosphere to reach stations beyond the curvature of the Earth. The correct option among all the options that are given in the question is the last option or option "d". The radio waves bounce off the ions present in the ionosphere of earths atmosphere and reach stations beyond the curvature of Earth.

Drupady [299]3 years ago

5 0

The correct answer for the question that is being presented above is this one: "d. They bounce off a charged layer of the atmosphere." Very long radio waves reach stations beyond the curvature of the Earth. They bounce off a charged layer of the atmosphere<span>
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A 50 Q resistor in a circuit has a current flowing through it of 2.0 A. What is

Alex17521 [72]

Hello!

We can use the following equation for calculating power dissipated by a resistor:
$P = i^2R$

P = Power (? W)
i = Current through resistor (2.0 A)
R = Resistance of resistor (50Ω)

Plug in the known values and solve.

$P = (2.0^2)(50) = \boxed{\text{ B. }200 W}$

7 0

2 years ago

A tin can has a volume of 1100 cm³ and a mass of 80 g. Approximately how many grams of lead shot can it carry without sinking in

Kruka [31]

Answer:

1020g

Explanation:

Volume of can= $1100cm^3=1100\times 10^{-6}m^3$

$1cm^3=10^{-6}m^3$

Mass of can=80g= $\frac{80}{1000}=0.08kg$

1Kg=1000g

Density of lead= $11.4g/cm^3=11.4\times 10^{3}=11400kg/m^3$

By using $1g/cm^3=10^3kg/m^3$

We have to find the mass of lead which shot can it carry without sinking in water.

Before sinking the can and lead inside it they are floating in the water.

Buoyancy force = $F_b=Weight of can+weight of lead$

$\rho_wV_cg=m_cg+m_lg$

Where $\rho_w=10^3kg/m^3=$ Density of water

$m_c=$ Mass of can

$m_l=$ Mass of lead

$V_c=$ Volume of can

Substitute the values then we get

$1000\times 1100\times 10^{-6}=0.08+m_l$

$1.1-0.08=m_l$

$m_l=1.02 kg=1.02\times 1000=1020g$

$1 kg=1000g$

Hence, 1020 grams of lead shot can it carry without sinking water.

4 0

4 years ago

The barometer of a mountain hiker reads 930 mbars at the beginning of a hiking trip and 780 mbars at the end. Neglecting the eff

lara31 [8.8K]

We could use the change of pressure to calculate for the height climbed by the mountain hiker. The change of pressure is given by

p = rho * g * h, where p is the change of pressure, rho is the air density, g is the acceleration due to gravity, and h is the height.

Using the conversion 1 mbar = 100 Pa,

(930 - 780)(100) = (1.20)(9.80)h
15000 = 1.20*9.80*h

h = 1.28 km

6 0

4 years ago

Light bulb 1 operates with a filament temperature of 2700 K whereas light bulb 2 has a filament temperature of 2100 K. Both fila

Lemur [1.5K]

Answer:

0.3659

Explanation:

The power (p) is given as:

P = AeσT⁴

where,

A =Area

e = transmittivity

σ = Stefan-boltzmann constant

T = Temperature

since both the bulbs radiate same power

P₁ = P₂

Where, 1 denotes the bulb 1

2 denotes the bulb 2

thus,

A₁e₁σT₁⁴ = A₂e₂σT₂⁴

Now e₁=e₂

⇒A₁T₁⁴ = A₂T₂⁴

$\frac{A_1}{A_2} =\frac{T_{2}^{4}}{T_{1}^{4}}$

substituting the values in the above question we get

$\frac{A_1}{A_2} =\frac{2100_{2}^{4}}{2700_{1}^{4}}$

$\frac{A_1}{A_2} }$ =0.3659

6 0

3 years ago

You have decided to study the effect of loud noise on plant growth. You put one plant in a quiet room and the other plant in you

Aleksandr [31]

B. control

also known as constant

7 0

3 years ago