Ocean currents<span> act much like a conveyer belt, transporting warm water and precipitation from the equator toward the poles and cold water from the poles back to the tropics. Therefore, </span>currents<span> regulate global </span>climate<span>, helping to counteract the uneven distribution of solar radiation reaching Earth's </span>surface<span>.</span>
Answer : The mass of ice melted can be, 3.98 grams.
Explanation :
First we have to calculate the moles of ice.

where,
Q = energy absorbed = 27.2 kJ
= enthalpy of fusion of ice = 6.01 kJ/mol
n = moles = ?
Now put all the given values in the above expression, we get:


Now we have to calculate the mass of ice.

Molar mass of ice = 18.02 g/mol

Thus, the mass of ice melted can be, 3.98 grams.
Weight = (mass) x (acceleration of gravity)
Acceleration of gravity = 9.81 m/s² on Earth, 1.62 m/s² on the Moon.
The feather's weight is . . .
On Earth: (0.0001 kg) x (9.81 m/s²) = <em>0.000981 Newton </em>
On the Moon: (0.0001 kg) x (1.62 m/s²) = <em>0.000162 N</em>
The presence or absence of atmosphere makes no difference. In fact, the numbers would be the same if the feather were sealed in a jar, or spinning wildly in a tornado, or hanging by a thread, or floating in a bowl of water or chicken soup. Weight is just the force of gravity between the feather and the Earth. It's not affected by what's around the feather, or what's happening to it.
Answer:
it depends on a person's own weight
Answer:
The minimum coefficient of friction is 0.27.
Explanation:
To solve this problem, start with identifying the forces at play here. First, the bug staying on the rotating turntable will be subject to the centripetal force constantly acting toward the center of the turntable (in absence of which the bug would leave the turntable in a straight line). Second, there is the force of friction due to which the bug can stick to the table. The friction force acts as an intermediary to enable the centripetal acceleration to happen.
Centripetal force is written as

with v the linear velocity and r the radius of the turntable. We are not given v, but we can write it as

with ω denoting the angular velocity, which we are given. With that, the above becomes:

Now, the friction force must be at least as much (in magnitude) as Fc. The coefficient (static) of friction μ must be large enough. How large?

Let's plug in the numbers. The angular velocity should be in radians per second. We are given rev/min, which can be easily transformed by a factor 2pi/60:

and so 45 rev/min = 4.71 rad/s.

A static coefficient of friction of at least be 0.27 must be present for the bug to continue enjoying the ride on the turntable.