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Alex787 [66]
3 years ago
7

Imagine a star very similar to our own Sun in both size and mass. This star moves very slightly back and forth in the sky once e

very 4 months, and you attribute this motion to the effect of an orbiting planet. What can you conclude about the orbiting planet?
(a)-The planet must be farther from the star than Neptune is from the Sun.
(b)-You do not have enough information to say anything at all about the planet.
(c)-The planet must have a mass about the same as the mass of Jupiter.
(d)-The planet must be closer to the star than Earth is to the Sun.
Physics
1 answer:
Inessa05 [86]3 years ago
6 0

Answer:

(c) The planet must have a mass about the same as the mass of Jupiter,

(d) The planet must be closer to the star than Earth is to the Sun.

Explanation:

Astrometry is the ideal method to detect high-mass planets that are close to their star. That is because the gravitational effect that it will have the planet over its host star will be greater. This effect can be seen as a wobble in the star as a consequence of how they orbit a common center of mass¹. The center of mass will be closer to the most massive object, So, in the case of an extrasolar planet with masses like Jupiter (Jovian), this point will be a little bit farther from the star, making the wobble more notable than in a system with a low-mass planet.          

Key terms:

Astrometry: study of the position of the stars over time in the sky.

¹Center of mass: a geometrical point in which the mass from a whole system is summed.

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Please verify these are correct. No work needs to be shown unless I've made a mistake. Thank you.
nevsk [136]

Answer:

Those are correct, well done.

8 0
3 years ago
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A straight wire that is 0.60 m long is carrying a current of 2.0 A. It is placed in a uniform magnetic field of strength 0.30 T.
Degger [83]

Answer:

The angle the wire make with respect to the magnetic field is 30°

Explanation:

It is given that,

Length of straight wire, L = 0.6 m

Current carrying by the wire, I = 2 A

Magnetic field, B = 0.3 T

Force experienced by the wire, F = 0.18 N

Let θ be the angle the wire make with respect to the magnetic field. Magnetic force is given by :

F=ILB\ sin\theta

\theta=sin^{-1}(\dfrac{F}{ILB})

\theta=sin^{-1}(\dfrac{0.18\ N}{2\ A\times 0.6\times 0.3\ T})

\theta=30^{\circ}

So, the angle the wire make with respect to the magnetic field is 30°. Hence, this is the required solution.

5 0
3 years ago
Suppose that 2.5 moles of an ideal gas are in a chamber in equilibrium at temperature 310 K and volume 0.5 m3. 1) What is the pr
Basile [38]

Answer:

The pressure in the chamber are of p= 0.127 atm.

Explanation:

n= 2.5 moles

T= 310 K

V= 0.5 m³ = 500 L

R= 0.08205746 atm. L /mol . K

p= n*R*T/V

p= 0.127 atm

3 0
4 years ago
What is an emergent curriculum
MariettaO [177]

Answer:

Emergent curriculum is an early education approach where teachers design projects unique to a child or group of kids.

5 0
3 years ago
The efficiency of a device such as a lamp can be calculated using this equation:
loris [4]

efficiency = (useful energy transferred ÷ energy supplied) × 100

It's easy to use this formula, but we have to know both the useful energy and the energy supplied.  The drawing doesn't tell us the useful energy, so we have to find a clever way to figure it out.  I see two ways to do it:

<u>Way #1:</u>

We all know about the law of conservation of energy.  So we know that the total energy coming out must be  250J, because that's how much energy is going in.  The wasted energy is 75J, so the rest of the 250J must be the useful energy . . . (250J - 75J) = 175J useful energy.

(useful energy) / (energy supplied) =  (175J) / (250J) = <em>70% efficiency</em>

================================

<u>Way #2: </u>

How much of the energy is wasted ? . . . 75J wasted

What percentage of the Input is that 75J ? . . . 75/250 = 30% wasted

30% of the input energy is wasted.  That leaves the other <em>70%</em> to be useful energy.

6 0
3 years ago
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