<h2>
Answer:
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The velocity of a satellite describing a circular orbit is <u>constant</u> and defined by the following expression:
(1)
Where:
is the gravity constant
the mass of the massive body around which the satellite is orbiting
the radius of the orbit (measured from the center of the planet to the satellite).
Note this orbital speed, as well as orbital period, does not depend on the mass of the satellite. I<u>t depends on the mass of the massive body.</u>
In addition, this orbital speed is constant because at all times <u>both the kinetic energy and the potential remain constant</u> in a circular (closed) orbit.
Answer:
<em>The final velocity is 20 m/s.</em>
Explanation:
<u>Constant Acceleration Motion</u>
It's a type of motion in which the velocity of an object changes by an equal amount in every equal period of time.
Being a the constant acceleration, vo the initial speed, and t the time, the final speed can be calculated as follows:
The provided data is: vo=10 m/s, 
, t=2 s. The final velocity is:
The final velocity is 20 m/s.
Because an object in rest stays in rest until an unequal force pushes it so gravity is pushing on the egg making it drop
Answer:
Explanation:
Using the pythagoras theorem, the displacement is expressed as;
d² = x²+y²
y = 36m (north)
x = 20m east
Substitute;
d² = 36²+20²
d² = 1296+400
d² = 1696
d = √1696
d = 41.18m
For the direction;
theta = tan^-1(y/x)
theta = tan^-1(36/20)
theta = tan^-1(1.8)
theta = 60.95°
Hence the magnitude is 41.18m and the direction is 60.95°
Answer:
The answer is 12.67 TMU
Explanation:
Recall that,
worker’s eyes travel distance must be = 20 in.
The perpendicular distance from her eyes to the line of travel is =24 in
What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element = ?
Now,
We solve for the given problem.
Eye travel is = 15.2 * T/D
=15.2 * 20 in/24 in
so,
= 12.67 TMU
Therefore, the MTM -1 of normal time that should be allowed for the eye travel element is = 12.67 TMU
