Answer:
F1 = K Q1 Q2 / R1^2
F2 = K Q1 / 2 * Q2 / (2 R1)^2
F2 / F1 = 1/2 / 4 = 1/8
The new force is 5N (1/2 due to charge and 1/4 due to distance)
Answer:
The square of the orbital period of a planet is directly proportional to the cube of the semimajor axis of its orbit.
Explanation:
hope this helps.
Answer:
7772.72N
Explanation:
When u draw your FBD, you realize you have 3 forces (ignore the force the car produces), gravity, normal force and static friction. You also realize that gravity and normal force are in our out of the page (drawn with a frame of reference above the car). So that leaves you with static friction in the centripetal direction.
Now which direction is the static friction, assume that it is pointing inward so
Fc=Fs=mv²/r=1900*15²/55=427500/55=7772.72N
Since the car is not skidding we do not have kinetic friction so there can only be static friction. One reason we do not use μFn is because that is the formula for maximum static friction, and the problem does not state there is maximum static friction.
Strength of induced current increased when strength of magnetic field increases. It will also increase when the number of turns are increased or if the speed of conductor increases
Answer:
a) the magnitude of r is 184.62
b) the direction is 37.74° south of the negative x-axis
Explanation:
Given the data in the question;
as illustrated in the image blow;
To find the the magnitude of r, we will use the Pythagoras theorem
r² = y² + x²
r = √( y² + x²)
we substitute
r = √((-113)² + (-146)²)
r = √(12769 + 21316 )
r = √(34085 )
r = 184.62
Therefore, the magnitude of r is 184.62
To find its direction, we need to find ∅
from SOH CAH TOA
tan = opposite / adjacent
tan∅ = -113 / -146
tan∅ = 0.77397
∅ = tan⁻¹( 0.77397 )
∅ = 37.74°
Therefore, the direction is 37.74° south of the negative x-axis
