<h2>QUESTION:- </h2>
➜what is kepler's law??
Kepler gave the three laws or theorems of motion of the orbitals bodies
This law state that the celestial bodies revolves around the stars in elliptical orbit and star as a single focus.
Example :- Earth revolves around the Sun as assuming it as single focus
This also shows that earth revolves around the sun in elliptical orbit.
Area covered by the planet is equal in equal duration of time irrespective of the position of the planet.
It also states that Angular momentum is constant
As Angular momentum is constant it means areal velocity is also constant.
where:-
A is the area.
T is the time.
L is the angular momentum.
M is the mass of the body.
square of the time of the revolution is directly proportional to the cube of the distance between the planet and star in Astronomical unit.
where:-
T = time of revolution
a is the distance between the planet and star.
Answer:
a = g = 9.81[m/s^2]
Explanation:
This problem can be solve using the second law of Newton.
We know that the forces acting over the skydiver are only his weight, and it is equal to the product of the mass by the acceleration.
m*g = m*a
where:
g = gravity = 9.81[m/s^2]
a = acceleration [m/s^2]
Note: If the skydiver will be under air resistance forces his acceleration will be different.
Answer:
Negative
Explanation:
Observe that the object below moves in the positive direction with a changing velocity. An object which moves in the positive direction has a positive velocity. If the object is slowing down then its acceleration vector is directed in the opposite direction as its motion (in this case, a negative acceleration).
Answer:
86.6 lbs
Explanation:
Let the force is X.
Resultant force, R = 100 lbs
Other force is Y. Angle between resultant force and force X is 30°.
According to the diagram
X = 86.6 lbs
Other force Y
Y = 50 lbs
Explanation:
Area of ring 
Charge of on ring 
Charge on disk
![\begin{aligned}d v &=\frac{k d q}{\sqrt{x^{2}+a^{2}}} \\&=2 \pi-k \frac{a d a}{\sqrt{x^{2}+a^{2}}} \\v(1) &=2 \pi c k \int_{0}^{R} \frac{a d a}{\sqrt{x^{2}+a^{2}}} \cdot_{2 \varepsilon_{0}}^{2} R \\&=2 \pi \sigma k[\sqrt{x^{2}+a^{2}}]_{0}^{2} \\&=\frac{2 \pi \sigma}{4 \pi \varepsilon_{0}}[\sqrt{z^{2}+R^{2}}-(21)] \\&=\frac{\sigma}{2}(\sqrt{2^{2}+R^{2}}-2)\end{aligned}](https://tex.z-dn.net/?f=%5Cbegin%7Baligned%7Dd%20v%20%26%3D%5Cfrac%7Bk%20d%20q%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5C%26%3D2%20%5Cpi-k%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5C%5Cv%281%29%20%26%3D2%20%5Cpi%20c%20k%20%5Cint_%7B0%7D%5E%7BR%7D%20%5Cfrac%7Ba%20d%20a%7D%7B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%7D%20%5Ccdot_%7B2%20%5Cvarepsilon_%7B0%7D%7D%5E%7B2%7D%20R%20%5C%5C%26%3D2%20%5Cpi%20%5Csigma%20k%5B%5Csqrt%7Bx%5E%7B2%7D%2Ba%5E%7B2%7D%7D%5D_%7B0%7D%5E%7B2%7D%20%5C%5C%26%3D%5Cfrac%7B2%20%5Cpi%20%5Csigma%7D%7B4%20%5Cpi%20%5Cvarepsilon_%7B0%7D%7D%5B%5Csqrt%7Bz%5E%7B2%7D%2BR%5E%7B2%7D%7D-%2821%29%5D%20%5C%5C%26%3D%5Cfrac%7B%5Csigma%7D%7B2%7D%28%5Csqrt%7B2%5E%7B2%7D%2BR%5E%7B2%7D%7D-2%29%5Cend%7Baligned%7D)
Note: Refer the image attached
