<span>A large body of air hats has similar properties of a temperature and moisture is an air mass.</span>
To answer the two questions, we need to know two important equations involving centripetal movement:
v = ωr (ω represents angular velocity <u>in radians</u>)
a =
Let's apply the first equation to question a:
v = ωr
v = ((1800*2π) / 60) * 0.26
Wait. 2π? 0.26? 60? Let's break down why these numbers are written differently. In order to use the equation v = ωr, it is important that the units of ω is in radians. Since one revolution is equivalent to 2π radians, we can easily do the conversion from revolutions to radians by multiplying it by 2π. As for 0.26, note that the question asks for the units to be m/s. Since we need meters, we simply convert 26 cm, our radius, into meters. The revolutions is also given in revs/min, and we need to convert it into revs/sec so that we can get our final units correct. As a result, we divide the rate by 60 to convert minutes into seconds.
Back to the equation:
v = ((1800*2π)/60) * 0.26
v = (1800*2(3.14)/60) * 0.26
v = (11304/60) * 0.26
v = 188.4 * 0.26
v = 48.984
v = 49 (m/s)
Now that we know the linear velocity, we can find the centripetal acceleration:
a =
a =
a = 9234.6 (m/)
Wow! That's fast!
<u>We now have our answers for a and b:</u>
a. 49 (m/s)
b. 9.2 * (m/)
If you have any questions on how I got to these answers, just ask!
- breezyツ
Answer:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
Explanation:
The bulk modulus is represented by the following differential equation:
Where:
- Bulk module, measured in pascals.
- Sample volume, measured in cubic meters.
- Local pressure, measured in pascals.
Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:
This resultant expression is solved by definite integration and algebraic handling:
The final volume is predicted by:
If , and , then:
Change in volume due to increasure on pressure is:
A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.
1 mile = 1.609 km
(135,000 km) x (1 mile / 1.609 km) = 83,885.1 miles