Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant
Given:
P = 687 days
A = 1.52 AU
Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³
Answer: 1.3439 x 10⁵ (days²/AU³)
The sound wave will have traveled 2565 m farther in water than in air.
Answer:
Explanation:
It is known that distance covered by any object is directly proportional to the velocity of the object and the time taken to cover that distance.
Distance = Velocity × Time.
So if time is kept constant, then the distance covered by a wave can vary depending on the velocity of the wave.
As we can see in the present case, the velocity of sound wave in air is 343 m/s. So in 2.25 s, the sound wave will be able to cover the distance as shown below.
Distance = 343 × 2.25 =771.75 m
And for the sound wave travelling in fresh water, the velocity is given as 1483 m/s. So in a time interval of 2.25 s, the distance can be determined as the product of velocity and time.
Distance = 1483×2.25=3337 m.
Since, the velocity of sound wave travelling in fresh water is greater than the sound wave travelling in air, the distance traveled by sound wave in fresh water will be greater.
Difference in distance covered in water and air = 3337-772 m = 2565 m
So the sound wave will have traveled 2565 m farther in water than in air.
Answer:
Earth is nearest the Sun in July and farthest away in July.
Explanation:
Answer:
when a two bodies A and B are moving at an angle 180° with each other then the relative velocity is the sum of bodies the velocity .i.e,
when the bodies move the opposite direction
then their relative velocity is the sum of individual velocities.
Answer:
Explanation:
Firstly, when you measure the voltage across the battery, you get the emf,
E = 13.0 V
In order to proceed we have to assume that the voltmeter offers no loading effect, which is a valid assumption since it has a very high resistance.
Secondly, the wires must be uniform. So the resistance per unit length is constant (say z). Now, even though the ammeter has very little resistance it cannot be ignored as it must be of comparable value/magnitude when compared to the wires. This is can seen in the two cases when currents were measured. Following Ohm's law and the resistance of a length of wire being proportional to it's length, we should have gotten half the current when measuring with the 40 m wire with respect to the 20 m wire (
). But this is not the case.
Let the resistance of the ammeter be r
Hence, using Ohm's law we get the following 2 equations:
.......(1)
......(2)
Substituting the value of r from (2) in (1), we have,
which simplifying gives us, 
(which is our required solution)
putting the value of z in either (1) or (2) gives us, r = 0.5325 
