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trapecia [35]
4 years ago
6

PLEASE HELP!! 50 PTS!!

Physics
2 answers:
eduard4 years ago
4 0

Answer

The direction of the sun’s pull relative to the planet’s motion changes.

Explanation:

REY [17]4 years ago
3 0
The speed of the planet decreases as it moves away from the sun because The suns gravitational pull in the sun decreases
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According to Newton’s second law of motion, force equals mass times acceleration (F = ma). Given that a car has a mass of 3,000
jek_recluse [69]

Answer:

B

Explanation:

3000 x 2 = 6000

6 0
4 years ago
If A and B are two objects with masses 6 kg and 34 kg respectively then
Mademuasel [1]

Answer:

the object B has more mass (= 34 kg) than the object A (= 6 kg)................................ ᕦ( ᐛ )ᕡ

4 0
3 years ago
Please help, very confused!
PtichkaEL [24]
D. the last choice because the info above tells u so
4 0
3 years ago
Read 2 more answers
Two planets P1 and P2 orbit around a star S in circular orbits with speeds v1 = 40.2 km/s, and v2 = 56.0 km/s respectively. If t
Readme [11.4K]

Answer: 3.66(10)^{33}kg

Explanation:

We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity \omega of the planet P1 with a period T=750years=2.36(10)^{10}s:

\omega=\frac{2\pi}{T}=\frac{V_{1}}{R} (1)

Where:

V_{1}=40.2km/s=40200m/s is the velocity of planet P1

R is the radius of the orbit of planet P1

Finding R:

R=\frac{V_{1}}{2\pi}T (2)

R=\frac{40200m/s}{2\pi}2.36(10)^{10}s (3)

R=1.5132(10)^{14}m (4)

On the other hand, we know the gravitational force F between the star S with mass M and the planet P1 with mass m is:

F=G\frac{Mm}{R^{2}} (5)

Where G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

In addition, the centripetal force F_{c} exerted on the planet is:

F_{c}=\frac{m{V_{1}}^{2}}{R^{2}} (6)

Assuming this system is in equilibrium:

F=F_{c} (7)

Substituting (5) and (6) in (7):

G\frac{Mm}{R^{2}}=\frac{m{V_{1}}^{2}}{R^{2}} (8)

Finding M:

M=\frac{V^{2}R}{G} (9)

M=\frac{(40200m/s)^{2}(1.5132(10)^{14}m)}{6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}} (10)

Finally:

M=3.66(10)^{33}kg (11) This is the mass of the star S

4 0
4 years ago
Using the sample in question 3 how many counts per second would be observed when the detector is 19 meters away from the sample
lara [203]
Okay. I'm willing to do that. But first you'll have to tell me something about the sample in question 3.
6 0
3 years ago
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