As per the question the distance of venus from sun is given as 0.723 AU
We have been asked to calculate the time period of the planet venus.
As per kepler's laws of planetary motion the square of time period of planet is directly proportional to the cube of semi major axis. mathematically
⇒ 
where is k is the proportionality constant
We may solve this problem by comparing with the time period of the earth . We know that time period of earth is 365.5 days
Hence 
The distance of sun from earth is taken as 1 AU i.e the mean distance of earth from sun
Hence 
The distance of venus from sun is 0.723 AU i.e
From keplers law we know that-
⇒
Putting the values mentioned above we get-
⇒ 
⇒
Hence the time period of venus is 224.388352752710 days
Yes, friction eventually slows down objects, unless you are in space. Hope this helps!
Answer:
The correct option is;
D. There is not enough information to answer this question
Explanation:
The universal gravitational constant = 6.67408 × 10⁻¹¹ 3³/(kg·s²)
For an in between distance of 1 m and equal masses of 60 kg, we have;
The gravitational attraction ≈ 2.403 × 10⁻⁷ N, which does not correspond with the answers, therefore, the best option is that there is not enough information to answer this question.
Answer:
kinetic Energy = 12.58 GJ
Explanation:
1 metric ton is equal to 1000 kg
then,
93000 metric ton
Mass is m = 93000 x 1000 kg
speed is v = 32 knots
1 knot is 0.514 m/s
then
,
v = 32 x 0.514 = 16.448 m/s
To solve for the Kinetic Energy (KE), we have;
KE = 0.5 x m x v²
KE = 0.5*93000*1000*(16.448)²
= 12.58 x 
J
= 12.58 GJ
Answer: D. movement of material along a coast by waves that approach at an angle to the shore.
Explanation:
Longshore drift is also referred to as the littoral druft and it means the sediment that is moved by the longshore current.
Longshore drift is the movement of material along a coast by waves which approach at an angle to the shore but then recede down the beach.
Therefore, the correct option is D.
