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1 year ago

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The radial velocity method preferentially detects: Choose one: A. all of the above described planets equally well. B. small plan

ets far from the central star. C. small planets close to the central star. D. large planets close to the central star. E. large planets far from the central star.

1 answer:

Zina [86]1 year ago

5 0

The radial velocity method preferentially detects large planets close to the central star

what is the Radial velocity:

The radial velocity technique is able to detect planets around low-mass stars, such as M-type (red dwarf) stars.

This is due to the fact that low mass stars are more affected by the gravitational tug of planets.

When a planet orbits around a star, the star wobbles a little.

From this, we can determine the mass of the planet and its distance from the star.

hence we can say that,

option D is correct.

The radial velocity method preferentially detects large planets close to the central star

Learn more about radial velocity here:

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When the distance between two masses is doubled, the gravitational attraction between

mrs_skeptik [129]

The new gravitational attraction will be 1/4 as much

Explanation:

The magnitude of the gravitational force between two objects is given by

$F=G\frac{m_1 m_2}{r^2}$

where

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$ is the gravitational constant

m1, m2 are the masses of the two objects

r is the separation between them

In this problem, the original force between the two objects is F, when they are separated by a distance r.

Later, the distance between the two objects is doubled, so the new distance is

$r'=2r$

Therefore, the new force will be

$F'=G\frac{m_1 m_2}{(2r)^2}=\frac{1}{4}(\frac{Gm_1 m_2}{r^2})=\frac{F}{4}$

Therefore, the new force will be one-fourth as much.

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3 years ago

this is a 3 part questionOn vacation, your 1400-kg car pulls a 560-kg trailer away from a stoplight with an acceleration of 1.85

mamaluj [8]

ANSWER:

(a) 1036 N

(b) -1036 N

(c) 2590 N

STEP-BY-STEP EXPLANATION:

Given:

Mc = 1400 kg

Mt = 560 kg

a = 1.85 m/s^2

(a)

Force by car on trailer:

$\begin{gathered} F_c=m\cdot a \\ F_c=560\cdot1.85 \\ F_c=1036\text{ N} \end{gathered}$

(b)

$\begin{gathered} F_t=-F_c \\ F_t=-1036\text{ N} \end{gathered}$

(c)

$\begin{gathered} F_n=1400\cdot1.85 \\ F_n=2590\text{ N} \end{gathered}$

3 0

10 months ago

A spring with spring constant 11.5 N/m hangs from the ceiling. A 490 g ball is attached to the spring and allowed to come to res

Natalija [7]

Answer:

The time constant is $\tau = 17.5 \ s$

Explanation:

From the question we are told that

The spring constant is $k = 11.5 \ N/m$

The mass of the ball is $m_b = 490 \ g = 0.49 \ kg$

The amplitude of the oscillation t the beginning is $x = 6.70 cm = 0.067 \ m$

The amplitude after time t is $x_t = 2.20 cm = 0.022 \ m$

The number of oscillation is $N = 30$

Generally the time taken to attain the second amplitude is mathematically represented as

$t = N * T$ Here T is the period of oscillation

$t = N * 2\pi \sqrt{\frac{m}{k} }$

=> $t = 30 * 2 * 3.142 * \sqrt{\frac{ 0.490}{11.5} }$

=> $t = 38.88 \ s$

Generally the amplitude at time t is mathematically represented as

$x(t) = x e^{-\frac{at}{2m} }$

Here a is the damping constant so

at $t = 38.88 \ s$ , $x_t = 2.20 cm = 0.022 \ m$

So

$0.022 = 0.067 e^{-\frac{a * 38.88}{2 * 0.490} }$

=> $0.3284 = e^{-\frac{a * 38.88}{2 * 0.490} }$

taking natural log of both sides

=> $ln(0.3284 ) = -\frac{a * 38.88}{2 * 0.490} }$

=> $a = 0.028$

Generally the time constant is mathematically represented as

$\tau = \frac{m}{a}$

=> $\tau = \frac{0.490}{ 0.028}$

=> $\tau = 17.5 \ s$

4 0

3 years ago

Joaquin mows the lawn at his grandmother’s home during the summer months. Joaquin measured the

nordsb [41]

Velocity is define as how fast an object is moving, and in what direction, it is a vector quantity, meaning velocity has both magnitude and direction. Anything goes to the left is negative, and anything goes to the right is positive.

a. Direction from east to west, given distance 11.5 meters, and time of 7.10 s

V = displacement/time V = -11.5/7.10 S V = -1.62 m/s (going left)

b. Joaquin reaches his original position. Displacement is now zero.

Velocity of the lawnmower is equal to "zero" but if we calculate for the average speed of the lawn, you just have to add the distance covered and the time it take to go back at the original position or point of origin

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3 years ago

Assume that a 15-kg ball moving at 8 m/s strikes a wall perpendicularly and rebounds elastically at the same speed. What is the

Aleks [24]

I think u should follow the formulae F=MA. So I think the answer is 120N

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2 years ago