Answer:
The necessary separation between the two parallel plates is 0.104 mm
Explanation:
Given;
length of each side of the square plate, L = 6.5 cm = 0.065 m
charge on each plate, Q = 12.5 nC
potential difference across the plates, V = 34.8 V
Potential difference across parallel plates is given as;
Where;
d is the separation or distance between the two parallel plates;
Therefore, the necessary separation between the two parallel plates is 0.104 mm
Answer:
80.4 N
Explanation:
As the block is at rest on the slope, it means that all the forces acting on it are balanced.
We are only interested in the forces that act on the block along the direction perpendicular to the slope. Along this direction, we have two forces acting on the block:
- The normal reaction N (contact force), upward
- The component of the weight of the block, , downward, where m is the mass of the block, g is the gravitational acceleration and
is the angle of the incline
Since the block is in equilibrium along this direction, the two forces must balance each other, so they must be equal in magnitude:
And by substituting the numbers into the equation, we find the size of the contact force normal to the slope:
Answer:
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Explanation:
Given data,
The mass of the bar AB, m = 6.4 kg
The angular velocity of the bar, θ˙ = 2.7 rad/s
The angle of the bar at A, θ = 24°
Let the length of the bar be, L = l
The angular moment at point A is,
∑ Mₐ = Iα
Where, Mₐ - the moment about A
α - angular acceleration
I - moment of inertia of the rod AB
Let G be the center of gravity of the bar AB
The position vector at A with respect to the origin at G is,
The acceleration at the center of the bar
Since the point A is fixed, acceleration is 0
The acceleration with respect to the coordinate axes is,
Comparing the coefficients of i
Comparing coefficients of j
Net force on x direction
substituting the values
=1.5(14.58L+11.96)
Similarly net force on y direction
= 3.2(2.97L - 157.03) + 62.72
Where L is the length of the bar AB
Therefore the net force,
F=√[(1.5(14.58L+11.96))² + (3.2(2.97L - 157.03) + 62.72)²]
Substituting the value of L gives the force at pin A