Answer:
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Explanation:
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Your answer is "<span>surface of a sphere"
Hope this helps.</span>
First we have to find out the gravity on that planet. We use Newton second equation of motion. It is given as,
s = ut +(gt^2)/2
Distance s = 25m
Time t = 5 s
Velocity u = 0
By putting these values,
25 = 1/2.g.(5)²
g = 2
So the gravity on that planet is 2. Lets find out the weight of the astronaut.
Mass of the astronaut on earth m = 80 kg
Weight of astronaut on earth W = mg = (80)(9.8) = 784 N
Weight of astronaut on earth like planet = (80)(2) = 160 N
x = 160N
-- In combination with 610 Hz, the beat frequency is 4 Hz.
So the unknown frequency is either (610+4) = 614 Hz
or else (610-4) = 606 Hz.
In combination with 605 Hz, the beat frequency will be
either (614-605) = 9 Hz or else (606-605) = 1 Hz.
-- In actuality, when combined with the 605 Hz, the beat
frequency is too high to count accurately. That must be
the 9 Hz rather than the 1 Hz.
So the unknown is (605+9) = 614 Hz.
