Answer:
x component 60.85 m
y component 101.031 m
Explanation:
We have given distance r = 118 km
Angle which makes from ground = 58.9°
(a) X component of distance is given by 
(b) Y component of distance is given by 
These are the x and y component of position vector
Answer:48 V
Explanation:
Given
Three charged particle with charge



Electric Potential is given by

Distance of
from 



similarly 




Potential at
is

![V_{net}=k[\frac{q_1}{d_1}+\frac{q_2}{d_2}+\frac{q_3}{d_3}]](https://tex.z-dn.net/?f=V_%7Bnet%7D%3Dk%5B%5Cfrac%7Bq_1%7D%7Bd_1%7D%2B%5Cfrac%7Bq_2%7D%7Bd_2%7D%2B%5Cfrac%7Bq_3%7D%7Bd_3%7D%5D)
![V_{net}=9\times 10^9[\frac{50}{10}-\frac{80}{12}+\frac{70}{10}]\times 10^{-9}](https://tex.z-dn.net/?f=V_%7Bnet%7D%3D9%5Ctimes%2010%5E9%5B%5Cfrac%7B50%7D%7B10%7D-%5Cfrac%7B80%7D%7B12%7D%2B%5Cfrac%7B70%7D%7B10%7D%5D%5Ctimes%2010%5E%7B-9%7D)


First, you find what 20% of 10 gallons of gas would be. This will show how many gallons the car actually uses.
10 gallons x 20% =
10 x 0.20 =
2 gallons used
Then you subtract that number from the total 10 gallons to get how many gallons of gas would be wasted.
10 gallons - 2 gallons =
8 gallons of gas wasted
Answer:
a) see attached, a = g sin θ
b)
c) v = √(2gL (1-cos θ))
Explanation:
In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by
Wₓ = m a
W sin θ = m a
a = g sin θ
b) The diagram is the same, the only thing that changes is the angle that is less
θ' = 9/2 θ
c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.
The easiest way to find linear speed is to use conservation of energy
Highest point
Em₀ = mg h = mg L (1-cos tea)
Lowest point
Emf = K = ½ m v²
Em₀ = Emf
g L (1-cos θ) = v² / 2
v = √(2gL (1-cos θ))