Answer:
I believe Mercury has the most extreme temperatures in the solar system, ranging from -280?F at night to 800 degrees F during the day for parts of the surface.
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The rms current in the transmission lines is I = 487.18 A.
The root-imply-rectangular (rms) voltage of a sinusoidal supply of electromotive force is used to represent the source. it is the rectangular root of the time average of the voltage squared.
Alternating-present day circuits. the root-imply-square (rms) voltage of a sinusoidal source of electromotive force is used to symbolize the supply. it's far the square root of the time average of the voltage squared.
Electric power is by using present day or the waft of electric fee and voltage or the capacity of rate to deliver electricity. A given cost of power can be produced by using any combination of contemporary and voltage values
power = 38 M watt
rms voltage = 78 K v
power = IV
I = power/V
I = (38 * 1000000)/78*1000
I = 487.18 A.
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Answer:
Air at higher altitude is under less pressure than air at lower altitude because there is less weight of air above it, so it expands (and cools), while air at lower altitude is under more pressure so it contracts (and heats up).
Explanation:
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Answer:
Pressure will be 
Explanation:
We have given mass of the women m = 65 kg
Radius of the heels r = 0.578 cm = 0.00578 m
We have to find the pressure
We know that pressure is given by

So force F = mg = 65×9.8 = 637 N
Area 
So pressure 
The answer is photocoagulation.
The use of a laser beam to seal leaky blood vessels and to prevent the growth of new ones in diabetic retinopathy is called laser <u>photocoagulation.</u>
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What is photocoagulation?
A minimally invasive method used to treat numerous retinal illnesses is photocoagulation of the retina, also known as retinal laser photocoagulation. The retina may expand due to aberrant leaky blood vessels developing across it in a number of disorders. Laser photocoagulation uses thermal energy above 65 °C to burn the retinal tissue by creating thermal burns. This can prevent the retina from being damaged by the bleeding blood vessels. In addition to causing fibrosis, laser photocoagulation can also seal retinal tears. Laser photocoagulation is typically unable to recover already lost vision in cases of retinal disease, but it can slow the progression of the condition, lower the chance of further vision loss, and preserve residual vision. The likelihood of problems following the operation is quite minimal.
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