Answer:
The speed of space station floor is 49.49 m/s.
Explanation:
Given that,
Mass of astronaut = 56 kg
Radius = 250 m
We need to calculate the speed of space station floor
Using centripetal force and newton's second law
Where, v = speed of space station floor
r = radius
g = acceleration due to gravity
Put the value into the formula
Hence, The speed of space station floor is 49.49 m/s.
Given that the function of the wave is f(x) = cos(π•t/2), we have;
a. The graph of the function is attached
b. 4 units of time
c. Even
d. 4.935 J/kg
e. 1.234 W/kg
<h3>How can the factors of the wave be found?</h3>a. Please find attached the graph of the signal created with GeoGebra
b. The period of the signal, T = 2•π/(π/2) = <u>4</u>
c. The signal is <u>even</u>, given that it is symmetrical about the y-axis
d. The energy of the signal is given by the formula;
Which gives;
E = 0.5 × 1.571² × 1² × 4 = <u>4.935 J/kg</u>
e. The power of the wave is given by the formula;
E = 0.5 × 1.571² × 1² × 4 × 0.25 = <u>1.234 W/</u><u>kg</u>
Learn more about waves here:
Answer:
0.625 c
Explanation:
Relative speed of a body may be defined as the speed of one body with respect to some other or the speed of one body in comparison to the speed of second body.
In the context,
The relative speed of body 2 with respect to body 1 can be expressed as :
Speed of rocket 1 with respect to rocket 2 :
Therefore, the speed of rocket 1 according to an observer on rocket 2 is 0.625 c
