You can just use basic
trigonometry to solve for the x & y components.
<span>vector a = 10cos(30) i +
10sin(30) j = <5sqrt(3), 5></span>
vector b is only slightly harder because the angle is relative
to vector a, and not the positive x-axis. Anyway, this just makes vector b with
an angle of 135deg to the positive x-axis.
<span>vector b = 10cos(135) i +
10sin(135) j = <-5sqrt(2), 5sqrt(2)></span>
So
now we can do the questions:
r = a + b
r = <5sqrt(3)-5sqrt(2), 5+5sqrt(2)>
(a)
5sqrt(3)-5sqrt(2)
(b)
5+5sqrt(2)
(c)
|r|
= sqrt( (5sqrt(3)-5sqrt(2))2 + (5+5sqrt(2))2 )
=
12.175
(d)
θ = tan-1 (
(5+5sqrt(2)) / (5sqrt(3)-5sqrt(2)) )
θ
= 82.5deg
<span> </span>
So that the correct representative of the people gets elected to head the people- so that a person who is thought to be capable enough to rule over people and take the correct decisions may be representative to the people.
The Moon is 3.8 108 m from Earth and has a mass of 7.34 1022 kg. 5.97 1024 kg is the mass of the Earth.
<h3>What kind of gravitational pull does the moon have on the planet?</h3>
On the surface of the Moon, the acceleration caused by gravity around 1.625 m/s2 which is 16.6% greater than on the surface of the Earth 0.166.
<h3>What does the Earth's center's gravitational pull feel like?</h3>
Gravity is zero if you are in the centre of the earth since everything around you is pulling "up" (up is the only direction).
<h3>Where is the Earth's and the moon's gravitational centre?</h3>
It is around 1700 kilometres below Earth's surface.
To know more about gravitational force visit:-
brainly.com/question/12528243
#SPJ4
Answer:
The resultant velocity is <u>169.71 km/h at angle of 45° measured clockwise with the x-axis</u> or the east-west line.
Explanation:
Considering west direction along negative x-axis and north direction along positive y-axis
Given:
The car travels at a speed of 120 km/h in the west direction.
The car then travels at the same speed in the north direction.
Now, considering the given directions, the velocities are given as:
Velocity in west direction is, 
Velocity in north direction is, 
Now, since
are perpendicular to each other, their resultant magnitude is given as:

Plug in the given values and solve for the magnitude of the resultant.This gives,

Let the angle made by the resultant be 'x' degree with the east-west line or the x-axis.
So, the direction is given as:

Therefore, the resultant velocity is 169.71 km/h at angle of 45° measured clockwise with the x-axis or the east-west line.