F_P + F_Q = M g
F_P = M g - F_Q
Torque, or moment of force:
∑ M_P = 0
∑ M_P = M g L - F_Q · 3 L
0 = M g L - 3 F_Q L / : L
0 = M g - 3 F_Q
3 F_Q = M g
F_Q = M g /3
Finally:
F_P = M g - M g/3
F_P = 4 M g / 3
Answer:
C
Explanation:
repeated in intervals of time
Answer:
linear charge density = -9.495 ×
C/m
Explanation:
given data
revolutions per second = 1.80 ×
radius = 1.20 cm
solution
we know that when proton to revolve around charge wire then centripetal force is require to be in orbit of radius around provide by electric force
so
- q × E = m × w² × r ..................1
- 9 ×
×
q = m × w² × r ............2
and w =
w =
w = 1.80 ×
×
w = 11304000 rad/s
so here from equation 2
- 9 ×
×
1.80 ×
= 1.672 ×
× 11304000² × 0.0120
linear charge density = -9.495 ×
C/m
Answer:
30N*s
Explanation:
Given the following data;
Force = 10N
Time = 3 seconds
To find the impulse;
Impulse = force * time
Substituting into the equation, we have;
Impulse = 10 * 3
Impulse = 30Ns