(13.558 gm) · (1 L / 0.089 gm) = 152.34 L (rounded)
(fraction equal to ' 1 ') ^
Answer:
52.9 N, 364.7 N
Explanation:
First of all, we need to resolve both forces along the x- and y- direction. We have:
- Force A (178 N)

- Force B (259 N)

So the x- and y- component of the total force acting on the block are:

Answer:
b.
Explanation:
i just finished a chapter about this
It's 3.6 meters per second less than my speed was
at 4:19 PM last Tuesday.
Does that tell you anything ?
Why not ?
Answer:
T = 92.8 min
Explanation:
Given:
The altitude of the International Space Station t minutes after its perigee (closest point), in kilometers, is given by:

Find:
- How long does the International Space Station take to orbit the earth? Give an exact answer.
Solution:
- Using the the expression given we can extract the angular speed of the International Space Station orbit:

- Where the coefficient of t is angular speed of orbit w = 2*p / 92.8
- We know that the relation between angular speed w and time period T of an orbit is related by:
T = 2*p / w
T = 2*p / (2*p / 92.8)
Hence, T = 92.8 min