Answer;
C. The brightness of each bulb would remain the same even though the total resistance of the circuit would decrease.
Explanation;
-If light bulbs are connected in parallel to a voltage source, the brightness of the individual bulbs remains more-or-less constant as more and more bulbs are added to the circuit.
-The current increases as more bulbs are added to the circuit and the overall resistance decreases. In addition, if one bulb is removed from the circuit the other bulbs do not go out. Each bulb is independently linked to the voltage source
Answer:
P(bat) = V²r/(R+r)²
Explanation:
Let the resistance of the coil be R
Internal resistance of the battery be r
Emf of the battery = V
Power dissipated in the internal resistance of the battery is normally given as P = I²r
where I is the current flowing in the circuit.
From Ohm's law,
V = I R(eq)
R(eq) = (R + r)
I = V/(R+r)
P = I²r
P = [V/(R+r)]²r
P = V²r/(R+r)²
Hope this Helps!!!
Answer:
Normal stress = 66/62.84 = 1.05kips/in²
shearing stress = T/2 = 0.952/2 = 0.476 kips/in²
Explanation:
A steel pipe of 12-in. outer diameter d₂ =12in d₁= 12 -4in = 8in
4 -in.-thick
angle of 25°
Axial force P = 66 kip axial force
determine the normal and shearing stresses
Normal stress б = force/area = P/A
= 66/ (П* (d₂²-d₁²)/4
=66/ (3.142* (12²-8²)/4
= 66/62.84 = 1.05kips/in²
Tangential stress T = force* cos ∅/area = P/A
= 66* cos 25/ (П* (d₂²-d₁²)/4
=59.82/ (3.142* (12²-8²)/4
= 59.82/62.84 = 0.952kips/in²
shearing stress = tangential stress /2
= T/2 = 0.952/2 = 0.476 kips/in²
Answer:
Cell Membrane
Explanation:
The cell membrane contains a phospholipid bilayer.
