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3 years ago

Which of these is a type of telescope?

2 answers:

8 0

I've heard of radio telescopes before on tv, I think they emit waves and measure how they are refracted/diffracted.

Thanks,
Gavin M George

eimsori [14]3 years ago

8 0

The answer to this problem is B hope it helps

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What is the bending of a wave around a barrier?

Mamont248 [21]

That's wave 'diffraction'.

3 0

3 years ago

A schoolbus accelerates to 65 mph and enters the freeway. It travels for 2.3 hours at that speed while on the freeway. What's th

PtichkaEL [24]

Answer:

The distance travelled on the freeway is 149.5 miles.

Explanation:

The school bus travels on the freeway at constant speed. According to the statement, we need to calculate the distance travelled by the vehicle by means of the following formula:

$x = v\cdot t$ (1)

Where:

$x$ - Traveled distance, in miles.

$v$ - Speed, in miles per hour.

$t$ - Time, in hours.

If we know that $v = 65\,\frac{mi}{h}$ and $t = 2.3\,h$ , then the distance travelled by the school bus is:

$x = v\cdot t$

$x = \left(65\,\frac{mi}{h} \right)\cdot (2.3\,h)$

$x = 149.5\,mi$

The distance travelled on the freeway is 149.5 miles.

5 0

2 years ago

What motion makes the acceleration unchanged?

Rashid [163]

Answer:

The acceleration is equal to the net force divided by the mass. If the net force acting on an object doubles, its acceleration is doubled. If the mass is doubled, then acceleration will be halved. If both the net force and the mass are doubled, the acceleration will be unchanged.

7 0

2 years ago

Read 2 more answers

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

Which of the following statements about electromagnetic radiation are TRUE? a. Wavelength and frequency are inversely proportion

Kobotan [32]

Answer: A) Wavelength and frequency are inversely proportional.

Explanation:

From the wave equation;

Velocity= frequency × wavelength

If the above equation is rearranged making the frequency the subject of formula, it would give;

Frequency= velocity/ wavelength.

From the above equation we see that frequency is inversely proportional to the wavelength. This means that for every increase in wavelength there would be a decrease in frequency, and for every increase in frequency there is a reduction in wavelength.

7 0

3 years ago