Answer:
A) mass = 3121.58 kg
B) tension = 25940.37 N
C) tension = 25940.37 N (tension on both sides will be the same)
Explanation:
Weight of elevator = 22500 N
Distance = 6.75 m
Time = 3 sec
Since it started from rest, initial speed is zero.
Using Newton's equation of motion we have,
S = ut + 0.5at^2
S = distance covered = 6.75 m
t = time = 3 s
a = acceleration upwards
u = initial velocity = 0
Substituting values, we have,
6.75 = 0(3) + (0.5 x a x 3^2)
6.75 = 4.5a
a = 6.75/4.5 = 1.5 m/s^2 (acceleration of the elevator upwards)
Mass of the elevator = Weight/g
Where g = acceleration due to gravity 9.81 m/s
Mass = 22500/9.81 = 2293.58 kg
From the image below we solve from
T - 22500 = ma
T - 22500 = 2293.58 x 1.5
T - 22500 = 3440.37
T = 3440.37 + 22500 = 25940.37 N (this is the tension on the rope)
On the other side,
mg - T = ma
9.81m - 25940.37 = 1.5m
(9.81 - 1.5)m = 25940.37
8.31m = 25940.37
m = 3121.58 kg (mass of counter weight)
See image below
the equation of the tangent line must be passed on a point A (a,b) and
perpendicular to the radius of the circle. <span>
I will take an example for a clear explanation:
let x² + y² = 4 is the equation of the circle,
its center is C(0,0). And we assume that the tangent line passes to the point
A(2.3).
</span>since the tangent passes to the A(2,3), the line must be perpendicular to the radius of the circle.
<span>Let's find the equation of the line parallel to the radius.</span>
<span>The line passes to the A(2,3) and C (0,0). y= ax+b is the standard form of the equation. AC(-2, -3) is a vector parallel to CM(x, y).</span>
det(AC, CM)= -2y +3x =0, is the equation of the line // to the radius.
let's find the equation of the line perpendicular to this previous line.
let M a point which lies on the line. so MA.AC=0 (scalar product),
it is (2-x, 3-y) . (-2, -3)= -4+4x + -9+3y=4x +3y -13=0 is the equation of tangent
Answer:
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Answer: 
Explanation:
According to the <u>Third Kepler’s Law</u> of Planetary motion:
(1)
Where;:
is the period of the satellite
is the Gravitational Constant and its value is 
is the mass of the Earth
is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).
On the other hand, the orbital velocity 
is given by:
(2)
Now, from (1) we can find
, in order to substitute this value in (2):
![a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7BT%5E%7B2%7DGM%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(3)
![a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}](https://tex.z-dn.net/?f=a%3D%5Csqrt%5B3%5D%7B%5Cfrac%7B%281.26%2810%29%5E%7B4%7Ds%29%5E%7B2%7D%286.674%2810%29%5E%7B-11%7D%5Cfrac%7Bm%5E%7B3%7D%7D%7Bkgs%5E%7B2%7D%7D%29%285.98%2810%29%5E%7B24%7Dkg%29%7D%7B4%5Cpi%7D%5E%7B2%7D%7D)
(4)
(5)
Substituting (5) in (2):
(6)
(7) This is the speed at which the satellite travels
