Answer:
Q must be placed at 0.53 L
Explanation:
Given data:
q_1 = 4.0 μC , q_2 = 3.0μC
Distance between charge is L
third charge q be placed at distance x cm from q1
The force by charge q_1 due to q is
----1
The force by charge q_2 due to q is
--2
we know that net electric force is equal to zero
F_1 = F_2
x = 0.53 L
Q must be placed at 0.53 L
Answer:
Magnification= -image distance/object distance
.253=image distance/33.5
image distance= 8.48 cm
Answer:
Initial velocity, U = 28.73m/s
Explanation:
Given the following data;
Final velocity, V = 35m/s
Acceleration, a = 5m/s²
Distance, S = 40m
To find the initial velocity (U), we would use the third equation of motion.
V² = U² + 2aS
Where;
V represents the final velocity measured in meter per seconds.
U represents the initial velocity measured in meter per seconds.
a represents acceleration measured in meters per seconds square.
S represents the displacement measured in meters.
Substituting into the equation, we have;
35² = U + 2*5*40
1225 = U² + 400
U² = 1225 - 400
U² = 825
Taking the square root of both sides, we have;
Initial velocity, U = 28.73m/s
3 as a single number is considered a sf
Kinetic energy is energy of motion. Pick choice-A, at the top of the swing, where she stops moving & then goes the other way.
