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3 years ago

Electromagnetic waves travel at a speed of 3.0 x 108 m/s. A radio station transmits waves

Physics

1 answer:

Sophie [7]3 years ago

6 0

Answer:

1.2×10⁶ Hz

Explanation:

Applying,

v = λf............. Equation 1

Where v = speed of wave, λ = wavelength, f = frequency.

make f the subject of the equation

f = v/λ............... Equation 2

From the question,

Given: v = 3.0×10⁸ m/s, λ = 2.5×10² m

Substitute these values into equation 2

f = 3.0×10⁸/2.5×10²

f = 1.2×10⁶ Hz

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Listed following are the names and mirror diameters for six of the world’s greatest reflecting telescopes used to gather visible

ziro4ka [17]

Answer:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope

Explanation:

How much light a telescope can collect depends on its diameter, since in a bigger area more photons will be collected.

Remember that in a circle the area is defined as:

$A = \pi r^{2}$ (1)

Where A is the area and r is its radius.

However, the radius can be determined by means of its diameter.

$d = 2r$

$r = \frac{d}{2}$ (1)

Where d is its diameter.

An example of this is when a person is collecting raindrops with a bucket and with a cup. Since the bucket has a bigger area than the cup, it will collect more raindrops by unit of time. In this scenario the raindrops represent the photons.

To determine the light collecting area of each telescope, equation 2 will be replaced in equation 1.

$A = \pi (\frac{d}{2})^{2}$ (3)

Case for Large binocular telescope:

$A_{mirror1} = \pi (\frac{8.4m}{2})^{2}$

$A_{mirror1} = 55.41m$

For the second mirror will be the same value

$A = A_{mirror1}+A_{mirror2}$

$A = 55.41m+55.41m$

$A= 110.82m$

Case for Keck 1 telescope:

$A = \pi (\frac{10m}{2})^{2}$

$A = 78.53m$

Case for Hobby-Ebberly telescope:

$A = \pi (\frac{9.2m}{2})^{2}$

$A = 66.47m$

Case for Subaru telescope:

$A = \pi (\frac{8.3m}{2})^{2}$

$A = 54.10m$

Case for Gemini North telescope:

$A = \pi (\frac{8m}{2})^{2}$

$A = 50.26m$

Case for Magellan 2 telescope:

$A = \pi (\frac{6.5m}{2})^{2}$

$A = 33.18m$

Hence, they may be rank in the following way:

Large binocular telescope, Keck 1 telescope, Hobby-Ebberly telescope, Subaru telescope, Gemini North telescope, Magellan 2 telescope.

<em>Photons: particles that constitute light. </em>

3 0

3 years ago

A copper cable needs to carry a current of 160 A with a power loss of only 2.0 W/m. What is the required radius of the copper ca

lesantik [10]

Answer:

The radius of the cable is 0.0083 m or 8.3 mm.

Explanation:

The resistance of copper cable of 1 meter length will be given by

$R_{cable} = \frac{\rho \times l}{a}$ .... (i)

where the resistivity of copper is $\rho = 1.7 \times 10^{-8} \Omega.m$ , and l is the length of the wire which is considered to be 1m, and a is the cross sectional area of the wire in $m^{2}$ .

From the formula of power we know that, $P = I^2 R$ .... (ii)

Therefore 2 W/m = $(160)^2 \times R$ .... (iii)

where the resistance,R, actually means the resistance of the cable per meter.

Therefore R ( resistance of cable per meter)

= $\frac{2}{160^2} = 7.812 \times 10^{-5}$ ohms / meter. .... (iv)

Therefore from (i)

$7.812 \times 10^{-5}$ = $\frac{1.7 \times 10^{-8} \times 1}{a} = \frac{1.7 \times 10^{-8} \times 1}{\pi r^{2} }$ ..... (v)

where cross sectional area of the cable, a = $\pi r^2$ ,

where r is the radius of the cable, and length of cable,l = 1m.

Therefore r = $\sqrt{\frac{ 1.7 \times 10^{-8}}{\pi \times 7.812 \times 10^{-5} } } = 0.0083m = 8.3 mm$

5 0

4 years ago

Note: Plagiarism will be reported

iren [92.7K]

Y no me han dicho que me ha dado un teléfono y me ha dicho la policía que han dado la cara para la que ha sido un policía que han dicho nada que no se han dado cuenta de

6 0

3 years ago

Read 2 more answers

Baseball reporters say that long fly balls that would have carried for home runs in July "die" in the cool air of Octo- ber and

anastassius [24]

Answer:

Consider the followig calculation

Explanation:

a) use deal equation:

PV = nRT

ρ = m/V,= ==> V = m/ρ

therefore,

ρ = Pm/RT

convert 95 oF in degree

95 oF = 308.15 K

1 atm = 1.013 * 105 pascal

ρ = 1.013*105 * 29 * /8.314 * 308.15

= 1.146 kg/m3

b) again use ideal gas equation:

ρ = Pm/RT

T = 50 oF = 283.15 K

1 atm = 1.013 * 105 pascal

molar mass will be same

ρ = 1.013 * 105 * 29 / 8.314 * 283.15

ρ = 1.248 kg / m3

So,

c) . more than density of the hot, dry air computed in part (a)

7 0

3 years ago

What physical characteristics of sound can be used to determine the energy carried by sound waves? Explain.

padilas [110]

Answer:

Explanation:

It transmits energy into the medium through its vibration. More energetic vibration corresponds to larger amplitude. The molecules move back and forth more vigorously. The loudness of a sound is also determined by the sensitivity of the ear.

3 0

3 years ago