<h3><u>Answer;</u></h3>
= 20.436 seconds
<h3><u>Explanation;</u></h3>
Speed = Distance × time
Therefore;
Time = Distance/speed
Distance = 7.50 m, speed = 0.367 m/s
Time = 7.50/0.367
<u>= 20.436 seconds </u>
Answer:
F=ma
here F is force, m is mass and a is accelaration,
According to the question,
F=3*F= 3F
m= 1/3 of m= m/3
a= ?
so the equation becomes,
3F= m/3*a
3F*3= ma
9F=ma
F= ma/9
Therefore accelaration reduces by 1/9.
I am not very sure.
Ans; see attached file for calculation and answer
Explanation:
Answer:
r = 4.24x10⁴ km.
Explanation:
To find the radius of such an orbit we need to use Kepler's third law:
<em>where T₁: is the orbital period of the geosynchronous Earth satellite = 1 d, T₂: is the orbital period of the moon = 0.07481 y, r₁: is the radius of such an orbit and r₂: is the orbital radius of the moon = 3.84x10⁵ km. </em>
From equation (1), r₁ is:
![r_{1} = r_{2} \sqrt[3] {(\frac{T_{1}}{T_{2}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%20r_%7B2%7D%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7BT_%7B1%7D%7D%7BT_%7B2%7D%7D%29%5E%7B2%7D%7D%20)
![r_{1} = 3.84\cdot 10^{5} km \sqrt[3] {(\frac{1 d}{0.07481 y \cdot \frac{365 d}{1 y}})^{2}}](https://tex.z-dn.net/?f=%20r_%7B1%7D%20%3D%203.84%5Ccdot%2010%5E%7B5%7D%20km%20%5Csqrt%5B3%5D%20%7B%28%5Cfrac%7B1%20d%7D%7B0.07481%20y%20%5Ccdot%20%5Cfrac%7B365%20d%7D%7B1%20y%7D%7D%29%5E%7B2%7D%7D%20)
Therefore, the radius of such an orbit is 4.24x10⁴ km.
I hope it helps you!
Assuming that the students
worldwide are being considered, because of the extremely large population, this
can be considered as a binomial distribution. A normal distribution is used most
usually as a fair approximation of the binomial. The mean is the expectation,
therefore:<span>
E[x] = np = (16)(0.22) = 3.52
<span>μ = 3.52 </span></span>
