The period of rotation of Mars is 1 day and 37 minutes. Determine its frequency of rotation in Hertz.
1 answer:
The frequency of rotation of Mars is 0.0000113 Hertz.
<u>Given the following data:</u>
- Period = 1 day and 37 minutes.
To find the frequency of rotation in Hertz:
First of all, we would convert the the value of period in days and minutes to seconds because the period of oscillation of a physical object is measured in seconds.
<u>Conversion:</u>
1 day = 24 hours
24 hours to minutes = ×
=
minutes
1 minute = 60 seconds
1477 minute = X seconds
Cross-multiplying, we have:
×
X = 88620 seconds
Now, we can find the frequency of rotation of Mars by using the formula:
<em>Frequency </em><em>of rotation</em> = <em>0.0000113 Hertz</em>
Therefore, the frequency of rotation of Mars is 0.0000113 Hertz.
Read more: brainly.com/question/14708169
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Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)
where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:
0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:
Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:
5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:
10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:
10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]