vector A has magnitude 12 m and direction +y
so we can say
![\vec A = 12 \hat j](https://tex.z-dn.net/?f=%5Cvec%20A%20%3D%2012%20%5Chat%20j)
vector B has magnitude 33 m and direction - x
![\vec B = -33 \hat i](https://tex.z-dn.net/?f=%5Cvec%20B%20%3D%20-33%20%5Chat%20i)
Now the resultant of vector A and B is given as
![\vec A + \vec B = 12 \hat j - 33 \hat i](https://tex.z-dn.net/?f=%5Cvec%20A%20%2B%20%5Cvec%20B%20%3D%2012%20%5Chat%20j%20-%2033%20%5Chat%20i)
now for direction of the two vectors resultant will be given as
![\theta = tan^{-1}\frac{12}{-33}](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20tan%5E%7B-1%7D%5Cfrac%7B12%7D%7B-33%7D)
![\theta = 160 degree](https://tex.z-dn.net/?f=%5Ctheta%20%3D%20160%20degree)
so it is inclined at 160 degree counterclockwise from + x axis
magnitude of A and B will be
![R = \sqrt{A^2 + B^2}](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7BA%5E2%20%2B%20B%5E2%7D)
![R = \sqrt{12^2 + 33^2} = 35.11 m](https://tex.z-dn.net/?f=R%20%3D%20%5Csqrt%7B12%5E2%20%2B%2033%5E2%7D%20%3D%2035.11%20m)
so magnitude will be 35.11 m
Answer:
This happens in two ways. If an object is at rest and an unbalanced force pushes or pulls the object, it will move. Unbalanced forces can also change the speed or direction of an object that is already in motion.
Explanation:
Answer:
Explanation:
A common theory suggests that desert pavements are formed through gradual removal of sand and other fine particles by the wind and intermittent rains leaving behind the large fragments. The larger rock particles are shaken into place by actions of different agents such as rain, wind, gravity, and animals
Answer:
v = 7934.2 m/s
Explanation:
Here the total energy of the Asteroid and the Earth system will remains conserved
So we will have
![-\frac{GMm}{r} + \frac{1}{2}mv_0^2 = -\frac{GMm}{R} + \frac{1}{2}mv^2](https://tex.z-dn.net/?f=-%5Cfrac%7BGMm%7D%7Br%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_0%5E2%20%3D%20-%5Cfrac%7BGMm%7D%7BR%7D%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we know that
![v_0 = 660 m/s](https://tex.z-dn.net/?f=v_0%20%3D%20660%20m%2Fs)
![M = 5.98 \times 10^{24} kg](https://tex.z-dn.net/?f=M%20%3D%205.98%20%5Ctimes%2010%5E%7B24%7D%20kg)
![m = 5 \times 10^9 kg](https://tex.z-dn.net/?f=m%20%3D%205%20%5Ctimes%2010%5E9%20kg)
![r = 4 \times 10^9 m](https://tex.z-dn.net/?f=r%20%3D%204%20%5Ctimes%2010%5E9%20m)
![R = 6.37 \times 10^6 m](https://tex.z-dn.net/?f=R%20%3D%206.37%20%5Ctimes%2010%5E6%20m)
now from above formula
![GMm(\frac{1}{R} - \frac{1}{r}) + \frac{1}{2}mv_0^2 = \frac{1}{2}mv^2](https://tex.z-dn.net/?f=GMm%28%5Cfrac%7B1%7D%7BR%7D%20-%20%5Cfrac%7B1%7D%7Br%7D%29%20%2B%20%5Cfrac%7B1%7D%7B2%7Dmv_0%5E2%20%3D%20%5Cfrac%7B1%7D%7B2%7Dmv%5E2)
now we have
![2GM(\frac{1}{R} - \frac{1}{r}) + v_0^2 = v^2](https://tex.z-dn.net/?f=2GM%28%5Cfrac%7B1%7D%7BR%7D%20-%20%5Cfrac%7B1%7D%7Br%7D%29%20%2B%20v_0%5E2%20%3D%20v%5E2)
now plug in all data
![2(6.67 \times 10^{-11})(5.98 \times 10^{24})(\frac{1}{6.37 \times 10^6} - \frac{1}{4 \times 10^9}) + (660)^2 = v^2](https://tex.z-dn.net/?f=2%286.67%20%5Ctimes%2010%5E%7B-11%7D%29%285.98%20%5Ctimes%2010%5E%7B24%7D%29%28%5Cfrac%7B1%7D%7B6.37%20%5Ctimes%2010%5E6%7D%20-%20%5Cfrac%7B1%7D%7B4%20%5Ctimes%2010%5E9%7D%29%20%2B%20%28660%29%5E2%20%3D%20v%5E2)
![v = 7934.2 m/s](https://tex.z-dn.net/?f=v%20%3D%207934.2%20m%2Fs)
Answer: 17.83 AU
Explanation:
According to Kepler’s Third Law of Planetary motion <em>“The square of the orbital period of a planet is proportional to the cube of the semi-major axis (size) of its orbit”. </em>
(1)
Talking in general, this law states a relation between the <u>orbital period</u>
of a body (moon, planet, satellite, comet) orbiting a greater body in space with the <u>size</u>
of its orbit.
However, if
is measured in <u>years</u>, and
is measured in <u>astronomical units</u> (equivalent to the distance between the Sun and the Earth:
), equation (1) becomes:
(2)
This means that now both sides of the equation are equal.
Knowing
and isolating
from (2):
(3)
(4)
Finally:
(5)