The moon clock is A) (9.8/1.6)h compared to 1 hour on Earth
Explanation:
The period of a simple pendulum is given by the equation
where
L is the length of the pendulum
g is the acceleration of gravity
In this problem, we want to compare the period of the pendulum on Earth with its period on the Moon. The period of the pendulum on Earth is
where
is the acceleration of gravity on Earth
The period of the pendulum on the Moon is
where
is the acceleration of gravity on the Moon
Calculating the ratio of the period on the Moon to the period on the Earth, we find
Therefore, for every hour interval on Earth, the Moon clock will display a time of
A) (9.8/1.6)h
#LearnwithBrainly
When lost fluid is not replaced adequately, dehydration can result.
Answer:
A) T.
Explanation:
Kepler's third law states that the orbital period (T) of a satellite is related with the radius (R) and the mass of the object (M) it orbits:
So the orbital period is independent of the mass of the satellite, that means no matter the mass every satellite at a radius R around the earth have an orbital period A.
Answer:
53.13 °
Explanation:
In order to do this, we just need to apply the following:
tanα = Dy/Dx
Where:
Vy: speed of the ball in the y axis.
Vx: speed of the ball in the x axis.
At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.
According to this, then:
tanα = (40/30)
tanα = 1.3333
α = tan⁻¹(1.3333)
<h2>
α = 53.13°</h2>
This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.
Hope this helps
