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3 years ago

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What are the strengths and limitations of the doppler and transit methods? What kind of planets are easiest to detect with each

method?
help me please :( i will but brainliest

1 answer:

Arturiano [62]3 years ago

3 0

$\huge\mathfrak\red{✔Answer:-}$

Strength: able to detect planets in a wide range of orbits, as long as orbits aren't face on

Limitations: yield only planet's mass and orbital properties

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The law of reflection states that if the angle of incidence is 19 degrees, the angle of reflection is ___ degrees.

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using the law of refraction, the incidence is equal to the reflection, but not refraction

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2 years ago

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A block–spring system vibrating on a frictionless, horizontal surface with an amplitude of 7.0 cm has an energy of 14 J. If the

Bingel [31]

Answer:

$E_T= 28J$

Explanation:

The energy of Mass-Spring System the sum of the potential energy of the block plus the kinetic energy of the block:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} mv^2$

Where:

$\Delta x=Amplitude\hspace{3}or\hspace{3}d eformation\hspace{3} of\hspace{3} the\hspace{3} spring\\m=Mass\hspace{3}of\hspace{3}the\hspace{3}block\\k=Constant\hspace{3}of\hspace{3}the\hspace{3}spring\\v=Velocity\hspace{3}of\hspace{3}the\hspace{3}block$

There are two cases, the first case is when the spring is compressed to its maximum value, in this case the value of the kinetic energy is zero, since there is no speed, so:

$E_T=\frac{1}{2} k \Delta x^2\\\\14=\frac{1}{2} k7^2\\\\Solving\hspace{3} for\hspace{3} k\\\\k=\frac{28}{49} =\frac{4}{7}$

The second case is when the block passes through its equilibrium position, in this case the elastic potential energy is zero since $\Delta x=0$ , so:

$E_T=\frac{1}{2} mv^2\\\\14=\frac{1}{2} mv^2\\\\Solving\hspace{3} for\hspace{3} v\\\\v^2=\frac{28}{m}$

Now, let's find the energy of the system when the block is replaced by one whose mass is twice the mass of the original block using the previous data:

$E_T=U+K=\frac{1}{2} k \Delta x^2+\frac{1}{2} m_2v^2$

Where in this case:

$m_2=New\hspace{3}mass=Twice\hspace{3} the\hspace{3} mass \hspace{3}of\hspace{3} the\hspace{3} original=2m$

Therefore:

$E_T=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{m_2})=\frac{1}{2} (\frac{4}{7} ) (7^2)+\frac{1}{2} (2m)(\frac{28}{2m})=14+14=28J$

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3 years ago

Two pebbles are dropped into a pool of water as shown. If a new wave with a

Kazeer [188]

Answer:

Constructive interference

Explanation:

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2 years ago

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A violinist is tuning her instrument to concert A (440 Hz). She plays the note while listening to an electronically generated to

Soloha48 [4]

To solve the problem it is necessary to take into account the concepts related to beat frequency, i.e., The number of those wobbles per second.

The equation that describes the beat frequency is

$f_{beat} = |f_2-f_1|$

For our given case we have that the frequency of the instrument is 440Hz and the Beat frequency is 5Hz therefore,

A) The frequency of the violin would be given by

$f_{beat} = |f_2-f_1|$

$5Hz = |f_2-440Hz|$

$f_2 = 440 \pm 5$

$f_2 = 445Hz or 435Hz$

B) <em>The violinist must loosen the string.</em> As the tightening increases the frequency, thereby increasing the number of beats from 5 to 6, i. e, on thightening the string, the frequency further increases as high frequency will be produced by short trings.

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3 years ago

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the dist

Sidana [21]

Complete Question

The spaceship Intergalactica lands on the surface of the uninhabited Pink Planet, which orbits a rather average star in the distant Garbanzo Galaxy. A scouting party sets out to explore. The party's leader–a physicist, naturally–immediately makes a determination of the acceleration due to gravity on the Pink Planet's surface by means of a simple pendulum of length 1.08m. She sets the pendulum swinging, and her collaborators carefully count 101 complete cycles of oscillation during 2.00×102 s. What is the result? acceleration due to gravity:acceleration due to gravity: m/s2

Answer:

The acceleration due to gravity is $g = 167.2 \ m/s^2$

Explanation:

From the question we are told that

The length of the simple pendulum is $L = 1.081.08 \ m$

The number of cycles is $N = 101$

The time take is $t = 2.00 *10^{2 \ }s$

Generally the period of this oscillation is mathematically evaluated as

$T = \frac{N}{t }$

substituting values

$T = \frac{101}{2.0*10^2 }$

$T = 0.505 \ s$

The period of this oscillation is mathematically represented as

$T = 2 \pi \sqrt{\frac{l}{g} }$

making g the subject of the formula we have

$g = \frac{L}{[\frac{T}{2 \pi } ]^2 }$

$g = \frac{4 \pi ^2 L }{T^2 }$

Substituting values

$g = \frac{4 * 3.142 ^2 * 1.08 }{505.505^2 }$

$g = \frac{4 * 3.142 ^2 * 1.08 }{0.505^2 }$

$g = 167.2 \ m/s^2$

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3 years ago