Answer:
The final charges of each sphere are: q_A = 3/8 Q
, q_B = 3/8 Q
, q_C = 3/4 Q
Explanation:
This problem asks for the final charge of each sphere, for this we must use that the charge is distributed evenly over a metal surface.
Let's start Sphere A makes contact with sphere B, whereby each one ends with half of the initial charge, at this point
q_A = Q / 2
q_B = Q / 2
Now sphere A touches sphere C, ending with half the charge
q_A = ½ (Q / 2) = ¼ Q
q_B = ¼ Q
Now the sphere A that has Q / 4 of the initial charge is put in contact with the sphere B that has Q / 2 of the initial charge, the total charge is the sum of the charge
q = Q / 4 + Q / 2 = ¾ Q
This is the charge distributed between the two spheres, sphere A is 3/8 Q and sphere B is 3/8 Q
q_A = 3/8 Q
q_B = 3/8 Q
The final charges of each sphere are:
q_A = 3/8 Q
q_B = 3/8 Q
q_C = 3/4 Q
Answer:
when you get to fall for free
Explanation:
Answer:
Explanation:
This is an application of Newton's second Law.
Formula
F = m * a
F = 300 N
m = 100 kg
a = ?
F = m * a
300N = 100 kg * a Divide by 100
300N/100kg = a
a = 3 m/sec^2
Answer:
Explanation:
Given
Initial reading on scale =40 N
So, we can conclude that weight of the sack is 40 N
After this a 10 N force is applied upward on the sack such that the net force becomes (40-10) N downward (because downward force is more)
This net downward force is the resultant of earth graviational pull and the applied upward force.
So, this downward force acts on the machine which inturn applies an upaward force of same magnitude called Normal reaction.
This situation can be diagramatically represented by figure given below
Answer:
I_syst = 278.41477 kg.m²
Explanation:
Mass of platform; m1 = 117 kg
Radius; r = 1.61 m
Moment of inertia here is;
I1 = m1•r²/2
I1 = 117 × 1.61²/2
I1 = 151.63785 kg.m²
Mass of person; m2 = 62.5 kg
Distance of person from centre; r = 1.05 m
Moment of inertia here is;
I2 = m2•r²
I2 = 62.5 × 1.05²
I2 = 68.90625 kg.m²
Mass of dog; m3 = 28.3 kg
Distance of Dog from centre; r = 1.43 m
I3 = 28.3 × 1.43²
I3 = 57.87067 kg.m²
Thus,moment of inertia of the system;
I_syst = I1 + I2 + I3
I_syst = 151.63785 + 68.90625 + 57.87067
I_syst = 278.41477 kg.m²
