To find 
we need to use vector addition and use the x and y components. First we subtract vector 2 from vector 5 which results in a vector with a length of 3 pointing directly east, then we use the distance formula to find the length of the net force 
which gives 
. We now have a magnitude but we also need a direction, since vector 4 and vector 5 are perpendicular. Using 
where tan^-1(y/x) we get an angle of 53 degrees. The resultant force vector is 5 distance with an angle of 53 degrees north east.
........................z......
Answer:
v = 3.08 km/s
Explanation:
Given that,
The angular velocity of the satellite = 
A satellite is in orbit 36000km above the surface of the earth.
The radius of the earth is 6400 km
Let v is the velocity of the satellite. It can be calculated as :
So, the velocity of the satellite is 3.08 km/s.
Answer:
270 m/s²
Explanation:
Given:
α = 150 rad/s²
ω = 12.0 rad/s
r = 1.30 m
Find:
a
The acceleration will have two components: a radial component and a tangential component.
The tangential component is:
at = αr
at = (150 rad/s²)(1.30 m)
at = 195 m/s²
The radial component is:
ar = v² / r
ar = ω² r
ar = (12.0 rad/s)² (1.30 m)
ar = 187.2 m/s²
So the magnitude of the total acceleration is:
a² = at² + ar²
a² = (195 m/s²)² + (187.2 m/s²)²
a = 270 m/s²
Explanation:
6000 years = 6000 x 365 x 24 x 60 x 60
= 1.892 x 10¹¹ second
gain is 1 second
1 second is equivalent to 9.193 × 10⁹ oscillations .
In 1.892 x 10¹¹ second , change in oscillation is 9.193 × 10⁹ oscillation
in one second change in oscillation = (9.193 / 1.892 ) x 10⁹⁻¹¹
= 4.859 x 10⁻² oscillations .
