Answer:
Diabetes
Explanation:
Insulin is an important hormone that helps the body convert sugar, starches, and other foods into energy.
The area enclosed by the wire is

By using Faraday-Neumann-Lenz law, we can find the emf induced in the circuit, whose magnitude is equal to the variation of magnetic flux on the wire:

And then, by using Ohm's law, we can find the induced current:
We want to find the work done and power exerted, let’s start with work first.
We know that the equation for work is: W = F * D. We need to find the force which we can find by using: F = M * A.
Mass: 300kg
Acceleration (using equation from photo): 1.25 m/s^2
(The equation says x but can be used with y values)
If you are confused about how I found the acceleration; I plugged in 2.5 for the final y value, 0 for the initial y value, 0 for the initial velocity and 4 for t squared.
To solve, for acceleration it’s a matter of simple algebra. You can subtract the initial y position and the initial velocity from the final y position because they are 0. This leaves you with 2.5 m = 1/2a * t^2, from here I multiplied 2.5 by 2 to get rid of the 1/2. Now I have 5 = a * t^2. T^2 is just 2 squared, so four. Simply divide 5 by 4, and boom, you get 1.25 m/s^2.
Force = 300 kg * 1.25 m/s^2 = 375 Newtons
So, work = 500 N * 2.5 m = 1000 Joules
Power: W/t
So, Power = 1000 J / 2 seconds = 500 Watts
Hope this helps!
Answer: b. The flow of air is neither toward the warm air mass nor toward the cold air mass.
A stationary front forms between two air masses. A stationary front results when the warm front or cold front air stops moving. This occurs due to the fact that warm front and cold front air masses being opposite to each other but neither of them are able to repel the other. This affects the climatic conditions of the region. The weather is often cloudy along a stationary front and also supported with fall of rain and snow especially if the air in the front is cold with low atmospheric pressure.
<em>Therefore, along a stationary front the flow of air is neither toward the warm air mass nor toward the cold air mass.
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They transfer energy. Not sure what your choices are but this is a common question