Answer:
Orbital speed=8102.39m/s
Time period=2935.98seconds
Explanation:
For the satellite to be in a stable orbit at a height, h, its centripetal acceleration V2R+h must equal the acceleration due to gravity at that distance from the center of the earth g(R2(R+h)2)
V2R+h=g(R2(R+h)2)
V=√g(R2R+h)
V= sqrt(9.8 × (6371000)^2/(6371000+360000)
V= sqrt(9.8× (4.059×10^13/6731000)
V=sqrt(65648789.18)
V= 8102.39m/s
Time period ,T= sqrt(4× pi×R^3)/(G× Mcentral)
T= sqrt(4×3.142×(6.47×10^6)^3/(6.673×10^-11)×(5.98×10^24)
T=sqrt(3.40×10^21)/ (3.99×10^14)
T= sqrt(0.862×10^7)
T= 2935.98seconds
Exhaling allows excess volume to escape from the lungs, and by exhaling at a suitable rate the diver can continue exhaling throughout the ascent and still have air in his or her lungs at the surface. If the diver fails to exhale during the ascent, lung over-expansion injury is likely to occur.
A small increase in aperture will result in a small increase in brightness.
Answer:
2.45 x 10^-2
Explanation:
All numbers in scientific notation or standard form are written in the form
m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.
To convert 0.0245 into scientific notation, follow these steps:
Step 1) Move the decimal 2 times to right in the number so that the resulting number, m = 2.45, is greater than or equal to 1 but less than 10
Step 2) Since we moved the decimal to the right the exponent n is negative
n = -2
Step 3) Write in the scientific notation form m × 10n
= 2.45 × 10-2
Therefore,
2.45 × 10-2 is the scientific notation form of 0.0245 number and 2.45e-2 is the scientific e-notation form for 0.0245
Here are some examples of decimal to scientific notation calculator
Since the orbit is circular and you are given the period and average distance( in other words the radius of the orbit) we can calculate the orbital speed right away. Keep in mind that ship's mass is much smaller than the mass of the moon, this is really important, as we can simply say that moon is not affected by the ship.
Angular frequency and orbital speed have the following relationship:


Let's do the calculation with rest of the numbers just to make sure.
Because you have the stable circular orbit the gravitational force and centrifugal force are in balance:

We solve this for v:

You can use this formula for any celestial body as long as the ship's mass is much smaller than the mass of the body, and this only applies to circular orbits
When we plug in the numbers we get:

As you can see the results are identical.