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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Answer:
6
Explanation:
The median is found by listing the numbers in order numerically and finding the one that is right in the middle. In this case you have 1, 3, 5, 6, 7, 12, 15 where the middle term is 6.
Answer:
Strong nuclear force is 1-2 order of magnitude larger than the electrostatic force
Explanation:
There are mainly two forces acting between protons and neutrons in the nucleus:
- The electrostatic force, which is the force exerted between charged particles (therefore, it is exerted between protons only, since neutrons are not charged). The magnitude of the force is given by
where k is the Coulomb's constant, q1 and q2 are the charges of the two particles, r is the separation between the particles.
The force is attractive for two opposite charges and repulsive for two same charges: therefore, the electrostatic force between two protons is repulsive.
- The strong nuclear force, which is the force exerted between nucleons. At short distance (such as in the nucleus), it is attractive, therefore neutrons and protons attract each other and this contributes in keeping the whole nucleus together.
At the scale involved in the nucleus, the strong nuclear force (attractive) is 1-2 order of magnitude larger than the electrostatic force (repulsive), therefore the nucleus stays together and does not break apart.
W=20 e(-kt)
A. Rearranging gives k= -(ln(w/20)/t
Substituting w= 10 and solving gives k=0.014
B. Using W=20e(-kt). After 0 hours, W=20. After 24 hours, W=14.29g. After 1 week (24x7=168h) W=1.9g
C. Rearranging gives t=-(ln(10/20)/k. Substituting w=1 and solving gives t=214 hours.
D. Differentiating gives dW/ dt = -20ke(-kt). Solving for t=100 gives dW/dt = 0.07g/h. Solving for t=1000 gives 0.0000002g/h
E. dW/dt = -20ke(-kt). But W=20e(-kt) so dW/dt = -kW
