The formula for computing the orbital time period of a body is given as:
T² = 4π²r³ / GM
where T is the time period, r is the distance between the two bodies, G is the gravitational constant and M is the mass of the body that is being orbited. If we compute this time using SI units, the working is:
9.58 AU is 1.43 x 10¹² meters
T = √[(4*π²*(1.43 x 10¹²)³) / (6.67 × 10⁻¹¹ * 2 x 10³⁰)]
T = 9.30 x 10⁸ seconds which is approximately 29 years
Using the astronomical units, distance is in astronomical units and the mass is in solar masses. In these conditions, the ratio:
4π²/G = 1 so
T² = a³ (since the solar mass of the sun is 1)
T = √(9.58)³
T = 27 years
Answer:
a. 0.000002 m
b. 0.00000182 m
Explanation:
36 cm = 0.36 m
15 cm = 0.15 m
a) We can start by calculating the air-water pressure of the bucket submerged 20m below the water surface:

Suppose air is ideal gas, then if the temperature stays the same, the product of its pressure and volume stays the same

Where P1 = 1.105 Pa is the atmospheric pressure, V_1 is the air volume in the bucket on the suface:

As the pressure increases, the air inside the bucket shrinks. But the crossection area stays constant, so only h, the height of air, decreases:


b) If the temperatures changes, we can still reuse the ideal gas equation above:


Answer:
6.05 cm
Explanation:
The given equation is
2 aₓ(x-x₀)=( Vₓ²-V₀ₓ²)
The initial head velocity V₀ₓ =11 m/s
The final head velocity Vₓ is 0
The accelerationis given by =1000 m/s²
the stopping distance = x-x₀=?
So we can wind the stopping distance by following formula
2 (-1000)(x-x₀)=[
]
x-x₀=6.05*
m
=6.05 cm
Car with a mass of 1210 kg moving at a velocity of 51 m/s.
2. What velocity must a 1340 kg car have in order to have the same momentum as a 2680 kg truck traveling at a velocity of 15 m/s to the west? 3.0 X 10^1 m/s to the west.
Hope i helped
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