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
You mean like a box sitting on a table.
One force is the force of gravity, pulling downward on the box.
Now, you know that the forces acting on the box must be balanced, because
if they're not, then the box would be accelerating. But it's just sitting there, so
there must be some other force, just exactly the right strength and direction to
exactly cancel the force of gravity on the box, so that the net force on it is zero.
The other force is the force of the table pushing upward on the box. It's called
the "normal force".
Answer:
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Explanation:
The force by the tugboats acting on the supertanker is constant and the displacement of the supertanker is along a straight line.
The angle between the 2 forces and displacement is ∅ = 15°.
First we have to calculate the work done by the individual force and then we can calculate the total work.
The work done on a particle by a constant force F during a straight line displacement s is given by following formula:
W = F*s
W = F*s*cos∅
With ∅ = the angles between F and s
The magnitude of the force acting on the supertanker is F of tugboat1 = F of tugboat 2 = F = 2.2 * 10^6 N
The total work done can be calculated as followed:
Wtotal = Ftugboat1 s * cos ∅1 + Ftugboat2 s* cos ∅2
Wtotal = 2Fs*cos∅
Wtotal = 2*2.2*10^6 N * 0.81 *10³ m s *cos15°
Wtotal = 3.44*10^9 Nm = <u>3.44 *10^9 J</u>
<u />
The total work done by the two tugboats on the supertanker is 3.44 *10^9 J
Different densities have to have a reason - different pressure and/or humidity etc. If there is a different pressure, there is a mechanical force that preserves the pressure difference: think about the cyclones that have a lower pressure in the center. The cyclones rotate in the right direction and the cyclone may be preserved by the Coriolis force.
If the two air masses differ by humidity, the mixing will almost always lead to precipitation - which includes a phase transition for water etc. It's because the vapor from the more humid air mass gets condensed under the conditions of the other. You get some rain. In general, intense precipitation, thunderstorms, and other visible isolated weather events are linked to weather fronts.
At any rate, a mixing of two air masses is a nontrivial, violent process in general. That's why the boundary is called a "front". In the military jargon, a front is the contested frontier of a conflict. So your idea that the air masses could mix quickly and peacefully - whatever you exactly mean quantitatively - either neglects the inertia of the air, a relatively low diffusion coefficient, a low thermal conductivity, and/or high latent heat of water vapor. A front is something that didn't disappear within minutes so pretty much tautologically, there must be forces that make such a quick disappearance impossible.
