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3 years ago

Why can radio telescopes be used 24 hours a day?

Physics

2 answers:

lbvjy [14]3 years ago

8 0

They are not affected by visible light

Arte-miy333 [17]3 years ago

5 0

The two reasons are one, that sunlight or darkness or clouds do not affect radio spectrum and two that emission from sun in radio wave-length spectrum is relatively weak.

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Write whether true or false...

elixir [45]

Answer:

a --> true, b --> false, c --> true, d -->false

Explanation:

a) since it stays floating the gravity force and the upqards push is the same

b) if it's balanced the rocket won't move from the ground, the force of the rocket, has to exceed the force of gravity

c) since it's going in a diretion the force of gravity is exceeding the force pushing it up

d) since that are speeding up at a rate, meaning growing, the force is unbalanced.

8 0

3 years ago

An exoplanet with one half of Earth's mass and 50% of Earth's radius is discovered.

Georgia [21]

Answer:

The space cadet that weighs 800 N on Earth will weigh 1,600 N on the exoplanet

Explanation:

The given parameters are;

The mass of the exoplanet = 1/2×The mass of the Earth, M = 1/2 × M

The radius of the exoplanet = 50% of the radius of the Earth = 1/2 × The Earth's radius, R = 50/100 × R = 1/2 × R

The weight of the cadet on Earth = 800 N

$The \ weight, W =G\dfrac{M \times m}{R^{2}} = 800 \ N$

Therefore, for the weight of the cadet on the exoplanet, W₁, we have;

$W_1 =G\dfrac{\dfrac{M}{2} \times m}{ \left ( \dfrac{R}{2} \right ) ^{2}} = G\dfrac{\dfrac{M}{2} \times m \times 4}{ R ^{2}} = 2 \times G \times \dfrac{M \times m}{R^{2}} = 2 \times 800 \, N = 1,600 \, N$

The weight of a space cadet on the exoplanet, that weighs 800 N on Earth = 1,600 N.

7 0

3 years ago

If the speed of sound in air is 343 m/s and the wavelength of this note is 2.62 m, what is the frequency of this C3 note? (Round

snow_lady [41]

Answer:

The frequency of of the note = 131 Hz.

Explanation:

Frequency: Frequency can be defined as the number of complete oscillation completed by a wave in one seconds. The S.I unit of frequency is Hertz ( Hz)

v = λf ............................ Equation 1

Making f the subject of the equation,

f = v/λ .......................... Equation 2

Where v = Speed, λ = wavelength, f = frequency

Given: v = 343 m/s, λ = 2.62 m.

Substituting these values into equation 2

f = 343/2.62

f = 131 Hz

Thus the frequency of of the note = 131 Hz.

8 0

3 years ago

In an engine, a piston oscillates with simple harmonic motion so that its position varies according to the following expression,

eduard

(a) 4.06 cm

In a simple harmonic motion, the displacement is written as

$x(t) = A cos (\omega t + \phi)$ (1)

where

A is the amplitude

$\omega$ is the angular frequency

$\phi$ is the phase

t is the time

The displacement of the piston in the problem is given by

$x(t) = (5.00 cm) cos (5t+\frac{\pi}{5})$ (2)

By putting t=0 in the formula, we find the position of the piston at t=0:

$x(0) = (5.00 cm) cos (0+\frac{\pi}{5})=4.06 cm$

(b) -14.69 cm/s

In a simple harmonic motion, the velocity is equal to the derivative of the displacement. Therefore:

$v(t) = x'(t) = -\omega A sin (\omega t + \phi)$ (3)

Differentiating eq.(2), we find

$v(t) = x'(t) = -(5 rad/s)(5.00 cm) sin (5t+\frac{\pi}{5})=-(25.0 cm/s) sin (5t+\frac{\pi}{5})$

And substituting t=0, we find the velocity at time t=0:

$v(0)=-(25.00 cm/s) sin (0+\frac{\pi}{5})=-14.69 cm/s$

(c) -101.13 cm/s^2

In a simple harmonic motion, the acceleration is equal to the derivative of the velocity. Therefore:

$a(t) = v'(t) = -\omega^2 A cos (\omega t + \phi)$

Differentiating eq.(3), we find

$a(t) = v'(t) = -(5 rad/s)(25.00 cm/s) cos (5t+\frac{\pi}{5})=-(125.0 cm/s^2) cos (5t+\frac{\pi}{5})$

And substituting t=0, we find the acceleration at time t=0:

$a(0)=-(125.00 cm/s) cos (0+\frac{\pi}{5})=-101.13 cm/s^2$

(d) 5.00 cm, 1.26 s

By comparing eq.(1) and (2), we notice immediately that the amplitude is

A = 5.00 cm

For the period, we have to start from the relationship between angular frequency and period T:

$\omega=\frac{2\pi}{T}$

Using $\omega = 5.0 rad/s$ and solving for T, we find

$T=\frac{2\pi}{5 rad/s}=1.26 s$

4 0

3 years ago

Sound waves that enter the ear canal are directed to the ____, causing it to vibrate.

andre [41]

According to the research, the correct option the eardrum. Sound waves that enter the ear canal are directed to the eardrum, causing it to vibrate.

<h3>What is the eardrum?</h3>

It is the membrane found in the middle ear of vertebrate animals, separating this sector from the external auditory canal.

When sound waves enter through the external auditory canal, the eardrum vibrates, transmitting its movement to the middle ear through a series of bones and in this way, the pressure change is transformed into a mechanical movement.

Therefore, we can conclude that according to the research, the correct option is the eardrum. Sound waves that enter the ear canal are directed to the eardrum, causing it to vibrate.

Learn more about the eardrum here: brainly.com/question/12770491

#SPJ1

4 0

1 year ago