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kumpel [21]
3 years ago
9

Imagine another solar system, with a star of the same mass as the Sun. Suppose a planet with a mass twice that of Earth (2 Earth

) orbits at a distance of 1 AU from the star. What is the orbital period of this planet?
Physics
1 answer:
sashaice [31]3 years ago
4 0

Answer:

Time period, T = 403.78 years

Explanation:

It is given that,

Orbital distance, a=1\ AU=1.496\times 10^{11}\ m

Mass of the Earth, m_e=5.972\times 10^{24}\ kg

Mass of the planet, m_p=2m_e=11.944\times 10^{24}\ kg

Let T is the orbital period of this planet. The Kepler's third law of motion gives the relation between the orbital period and the orbital distance.

T^2=\dfrac{4\pi^2}{Gm_p}a^3

T^2=\dfrac{4\pi^2}{6.67\times 10^{-11}\times 11.944\times 10^{24}}\times (1.496\times 10^{11})^3

T=1.28\times 10^{10}\ s

or

T = 403.78 years

So, the orbital period of this planet is 404 years. Hence, this is the required solution.

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A large fake cookie sliding on a horizontal surface is attached to one end of a horizontal spring with spring constant k = 440 N
irinina [24]

Answer:

a) 0.275 m b) 13.6 J

Explanation:

In absence of friction, the energy is exchanged between the spring (potential energy) and the cookie (kinetic energy), so at any point, the sum of both energies must be the same:

E = ½ kx2 + ½ mv2

If we take as initial state, the instant when the cookie is passing through the spring’s equilibrium position, all the energy is kinetic, and we know that is equal to 20.0 J.

After sliding to the right, while is being acted on by a friction force, it came momentarily at rest. At this point, the initial kinetic energy, has become potential elastic energy, in part, and in thermal energy also, represented by the work done by the friction force.

So, for this state, we can say the following:

Ki = Uf + Eth = ½* k*d2 + Ff*d

20.0J = ½ *440 N/m* d2 + 11.0 *d, where d is the compressed length of the spring, which is equal to the distance travelled by the cookie before coming momentarily at rest.

We have a quadratic equation, that, after simplifying terms, can be solved as follows, applying the quadratic formula:

d = -0.05/2 +/- √0.090625 = -0.025 +/- 0.3 = 0.275 m (we take the positive root)

b) If we take as our new initial status the moment at which the spring is compressed, and the cookie is at rest, all the energy is potential:

E = Ui = 1/2 k d²

In this case, d is the same value that we got in a), i.e., 0.275 m (as the distance travelled by the cookie after going through the equilibrium point is the same length that the spring have been compressed).

E= 1/2 440 N/m . (0.275)m² = 16.6 J

When the cookie passes again through the equilibrium position, the energy will be in part kinetic, and in part, it will have become thermal energy again.

So, we can write the following equation:

Kf = Ui - Ff.d = 16.6 J - 11.0 (0.275) m = 16.6 J - 3.03 J = 13.6 J

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7 0
2 years ago
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A 70mm long blockhas cross-section of 50mm by 10mm the block is subjected to forces 60KN (tension) on the 50mm by 10mm face and
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Answer:

970 kN

Explanation:

The length of the block = 70 mm

The cross section of the block = 50 mm by 10 mm

The tension force applies to the 50 mm by 10 mm face, F₁ = 60 kN

The compression force applied to the 70 mm by 10 mm face, F₂ = 110 kN

By volumetric stress, we have that for there to be no change in volume, the total pressure applied by the given applied forces should be equal to the pressure removed by the added applied force

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lisabon 2012 [21]

Answer:

1. The length is 8.35m

2. The period on the moon is 14.05 secs

Explanation:

1. Data obtained from the question. This includes the following:

Period (T) = 5.8 secs

Acceleration due to gravity (g) = 9.8 m/s2

Length (L) =...?

The length can be obtained by using the formula given below:

T = 2π√(L/g)

5.8 = 2π√(L/9.8)

Take the square of both side

(5.8)^2 = 4π^2 x L/ 9.8

Cross multiply

4π^2 x L = (5.8)^2 x 9.8

Divide both side by 4π^2

L = (5.8)^2 x 9.8 / 4π^2

L= 8.35 m

2. Data obtained from the question. This includes the following:

Acceleration due to gravity (g) = 1.67 m/s2

Length (L) = 8.35m (the length remains the same)

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The period can be obtained as follow:

T = 2π√(L/g)

T = 2π√(8.35/1.67)

T = 14.05 secs

Therefore, the period on the moon is 14.05 secs

4 0
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