Kepler's third law hypothesizes that for all the small bodies in orbit around the
same central body, the ratio of (orbital period squared) / (orbital radius cubed)
is the same number.
<u>Moon #1:</u> (1.262 days)² / (2.346 x 10^4 km)³
<u>Moon #2:</u> (orbital period)² / (9.378 x 10^3 km)³
If Kepler knew what he was talking about ... and Newton showed that he did ...
then these two fractions are equal, and may be written as a proportion.
Cross multiply the proportion:
(orbital period)² x (2.346 x 10^4)³ = (1.262 days)² x (9.378 x 10^3)³
Divide each side by (2.346 x 10^4)³:
(Orbital period)² = (1.262 days)² x (9.378 x 10^3 km)³ / (2.346 x 10^4 km)³
= 0.1017 day²
Orbital period = <u>0.319 Earth day</u> = about 7.6 hours.
Answer:I think the answer is 5.43 I don’t rlly know
Explanation:
Answer:
distance= velocity ×time
distance= 62×10
distance=620m
hope it helps you mate please mark me as brainliast
Answer:
(a) The absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b) The force exerted by the water on the window is 36101.5 N
Explanation:
(a)
The absolute pressure is given by the formula

Where
is the absolute pressure
is the atmospheric pressure
is the density
is the acceleration due to gravity (Take
)
h is the height
From the question
h = 30.0 m
= 1.00 × 10³ kg/m³ = 1000 kg/m³
= 101.3 kPa = 101300 Pa
Using the formula
P = 101300 + (1000×9.8×30.0)
P = 101300 + 294000
P =395300 Pa
P = 395.3 kPa
Hence, the absolute pressure at the bottom of the freshwater lake is 395.3 kPa
(b)
For the force exerted
From
P = F/A
Where P is the pressure
F is the force
and A is the area
Then, F = P × A
Here, The area will be area of the window of the underwater vehicle.
Diameter of the circular window = 34.1 cm = 0.341 m
From Area = πD²/4
Then, A = π×(0.341)²/4 = 0.0913269 m²
Now,
From F = P × A
F = 395300 × 0.0913269
F = 36101.5 N
Hence, the force exerted by the water on the window is 36101.5 N