The distance between city a and city b is 833.345 miles.
We know that
1°=60'
The distance of city a from the initial ray is calculated as
x_a=3960*tan45.46°=4024.101 miles
The distance of city b from the initial ray is calculated as
x_b=3960*tan 38.86°=3190.75 miles
Now the distance between city a and b is equal to
4024.101-3190.75=833.345 miles
This is the vertical distance between the cities.
Answer:
Rmains constant
Explanation:
The equation of the trajectory of a projectile motion is presented as follows;
![Y = x \cdot tan \theta -\dfrac{g \cdot x^2}{2 \cdot u^2 \cdot cos^2 \theta}](https://tex.z-dn.net/?f=Y%20%3D%20x%20%5Ccdot%20tan%20%5Ctheta%20-%5Cdfrac%7Bg%20%5Ccdot%20x%5E2%7D%7B2%20%5Ccdot%20u%5E2%20%5Ccdot%20cos%5E2%20%5Ctheta%7D)
The vertical componet of the prjectile motion is
y = (u·sinθ)·t - g·t²/2
Where;
θ = The angle with which the projectile is launched
x = The horizontal distance
u = The initial velocity of the projectile
g = The acceleration due to gravity = Constant
t = The time of motion
The acceleration acting on the projectile is the 'g' which is the constant acceleration due to gravity
Therefore, for general projectile motion with no air resistance, the vertical component of the projectile acceleration <em>remains constant</em>
Answer:
Insider
Explanation:
An insider security breach or threat is the breach that originates within an organization due to some of it's own employees who have access to the confidential information of the organization. It can include both current or former employees, interns or anyone else who have access to the critical systems of the company.
In the case mentioned in the question, the employees that are involved in the breach are current employees of the same company so this type of breach will be termed as an insider security breach.
For a 650-kg satellite is in a circular orbit about Earth at a height above Earth equal to Earth’s mean radius, the resultant values are is mathematically given as
- V=5.59*10^3m/s
- T=3.98hours
- F=1.47*10^3N
<h3>What are the satellite’s orbital speed, the period of its revolution, and the gravitational force acting on it.?</h3>
Generally, the equation for satellite orbital speed is mathematically given as
a)
![V^2=\frac{Gm}{R+h}\\\\Therefore\\\\V^2=\frac{6.667*10^-11*5.98*10^{24}}{2*6.38*10^6}](https://tex.z-dn.net/?f=V%5E2%3D%5Cfrac%7BGm%7D%7BR%2Bh%7D%5C%5C%5C%5CTherefore%5C%5C%5C%5CV%5E2%3D%5Cfrac%7B6.667%2A10%5E-11%2A5.98%2A10%5E%7B24%7D%7D%7B2%2A6.38%2A10%5E6%7D)
V=5.59*10^3m/s
b)
![V=2\pi(R+h)/T\\\\Hence\\\\T=2\pi(R+h)/V\\\\T=2\pi(R+h)/2*3.14*6.38*10^6/5.59*10^3](https://tex.z-dn.net/?f=V%3D2%5Cpi%28R%2Bh%29%2FT%5C%5C%5C%5CHence%5C%5C%5C%5CT%3D2%5Cpi%28R%2Bh%29%2FV%5C%5C%5C%5CT%3D2%5Cpi%28R%2Bh%29%2F2%2A3.14%2A6.38%2A10%5E6%2F5.59%2A10%5E3)
T=3.98hours
c)
![F=\frac{GMM}{R+h^2}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7BGMM%7D%7BR%2Bh%5E2%7D)
Therefore
![F=\frac{6.67*10^{-11}*5.98*10^{24}*650}{2*6.38*10^6}](https://tex.z-dn.net/?f=F%3D%5Cfrac%7B6.67%2A10%5E%7B-11%7D%2A5.98%2A10%5E%7B24%7D%2A650%7D%7B2%2A6.38%2A10%5E6%7D)
F=1.47*10^3N
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