1) First of all, we need to find the distance between the two charges. Their distance on the xy plane is

substituting the coordinates of the two charges, we get

2) Then, we can calculate the electrostatic force between the two charges

and

, which is given by

where

is the Coulomb's constant.
Substituting numbers, we get

and the negative sign means the force between the two charges is attractive, because the two charges have opposite sign.
Answer:
The moment of a given force about a given axis of rotation can be decreased by decreasing the perpendicular distance of force from the axis of rotation.
Answer:
Explanation:
Kinetic energy= K.E = (1/2) × mv²
K.E = 0.5× 0.8kg× 100m²/s² = 40 N
Given
Weight of the block A, Wa = 20 lb, weight of block B Wb = 50 lb. Applied
force to block A, P = 6lb, coefficient of static friction µs = 0.4, coefficient
of kinetic friction µk = 0.3. If a force P
is applied to the body, no relative motion will take place until the applied
force is equal to the force of friction Ff, which is acting opposite to the
direction of motion. Magnitude of static force of friction between block A and
block B, Fs = µsN, where N is
reaction force acting on block A. Now, resolve the forces Fx = max. P = (mA +
mB)a,
6 = (20 / 32.2 + 50 / 32.2)a
2.173a = 6
A = 2.76 ft/s^2
To check slipping occurs between block A and block B, consider block A:
P – Ff = mAaA
6 – Ff = 1.71
Ff = 4.29 lb
And also,
N = wA. We know static friction,
Fs = µsN
Fs = 0.4 x 20
Fs = 8lb
Frictional force is less than static friction. Ff < Fs
<span>Therefors, acceleration of block A, aA = 2.76 ft/s^2, acceleration of
block B aB = 2.76 ft/s^2</span>
Answer:
T = 8.55 N
Explanation:
When string makes an angle 40 degree with the vertical then it will have two forces on it
1) gravitational force (mg)
2) Tension force in string (T)
now we know that net force towards the center of the path is known as centripetal force and it is given as




