Refractive Index is a ratio of two similar physical quantity which is dimension less
refractive index = sin I / sin r
therefore it doesn't have a unit.
Answer:
F= 5.71 N
Explanation:
width of door= 0.91 m
door closer torque on door= 5.2 Nm
In order to hold the door in open position we need to exert an equal and opposite torque, to the door closer torque, on the door.
so wee need to exert 5.2 Nm torque on the door.
If we want to apply minimum force to exert the required torque we need to apply force perpendicularly on the door knob (end of door) so that to to greater moment arm.
T= r x F
T= r F sin∅
F= T/ (r * sin∅)
F= 5.2/ (0.91 * 1)
F= 5.71 N
Answer:
The maximum energy stored in the combination is 0.0466Joules
Explanation:
The question is incomplete. Here is the complete question.
Three capacitors C1-11.7 μF, C2 21.0 μF, and C3 = 28.8 μF are connected in series. To avoid breakdown of the capacitors, the maximum potential difference to which any of them can be individually charged is 125 V. Determine the maximum energy stored in the series combination.
Energy stored in a capacitor is expressed as E = 1/2CtV² where
Ct is the total effective capacitance
V is the supply voltage
Since the capacitors are connected in series.
1/Ct = 1/C1+1/C2+1/C3
Given C1 = 11.7 μF, C2 = 21.0 μF, and C3 = 28.8 μF
1/Ct = 1/11.7 + 1/21.0 + 1/28.8
1/Ct = 0.0855+0.0476+0.0347
1/Ct = 0.1678
Ct = 1/0.1678
Ct = 5.96μF
Ct = 5.96×10^-6F
Since V = 125V
E = 1/2(5.96×10^-6)(125)²
E = 0.0466Joules
3. Law: Every action has a reaction equal in magnitude and opposite in direction.
Answer:
F = 37.8 × 10^(6) N
Explanation:
The charges are 0.06 C and 0.07 C.
Thus;
Charge 1; q1 = 0.06 C
Charge 2; q2 = 0.07 C
Distance between them; r = 3 m
Formula for the force in between them is;
F = kq1•q2/r²
Where k is a constant = 9 × 10^(9) N.m²/C²
Thus;
F = (9 × 10^(9) × 0.06 × 0.07)/3²
F = 37.8 × 10^(6) N
