The magnitude<span> of a </span>velocity<span> vector is </span>called<span> speed. Supposethat a wind is blowing in from the direction at a speed of 50 km/h. (This meansthat the direction from which the wind blows is west of the northerly direction.) Apilot is steering a plane in the direction at an airspeed (speed in still air) of250 km/h
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A) The mass is an intrinsic property of an object: it means it depends only on the properties of the object, so it does not depend on the location of the object. Therefore, Gary's mass at 300 km above Earth's surface is equal to his mass at the Earth's surface.
b) The weight of an object is given by

where
m is the mass

is the gravitational acceleration at the location of the object, with G being the gravitational constant, M the mass of the planet and r the distance of the object from the center of the planet.
At the Earth's surface,

, so Gary's weight is

(1)
where m is Gary's mass.
Then, we must calculate the value of g at 300 km above Earth's surface. the Earth's radius is

So the distance of Gary from the Earth's center is

The Earth's mass is

, so the gravitational acceleration is

Therefore, Gary's weight at 300 km above Earth's surface is

(2)
If we compare (1) and (2), we find that Gary's weight has changed by

So, Gary's weight at 300 km above Earth's surface is 91% of his weight at the surface.
<span>=1330 ft^2 x (1m^2/3.25 ft^2)
=124 m^2
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Answer:
v = 45.37 m/s
Explanation:
Given,
angle of inclination = 8.0°
Vertical height, H = 105 m
Initial K.E. = 0 J
Initial P.E. = m g H
Final PE = 0 J
Final KE = 
Using Conservation of energy




v = 45.37 m/s
Hence, speed of the skier at the bottom is equal to v = 45.37 m/s
Answer:
what that guy said
Explanation:
because he provides evidence