It is estimated that 26 large pizzas are about right to serve 66 students of a physics club meeting. How many pizzas would be re
1 answer:
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A. 0.77 A
Using the relationship:
where P is the power, V is the voltage, and R the resistance, we can find the resistance of each bulb.
For the first light bulb, P = 60 W and V = 120 V, so the resistance is
For the second light bulb, P = 200 W and V = 120 V, so the resistance is
The two light bulbs are connected in series, so their equivalent resistance is
The two light bulbs are connected to a voltage of
V = 240 V
So we can find the current through the two bulbs by using Ohm's law:
B. 142.3 W
The power dissipated in the first bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
C. 42.7 W
The power dissipated in the second bulb is given by:
where
I = 0.77 A is the current
is the resistance of the bulb
Substituting numbers, we get
D. The 60-W bulb burns out very quickly
The power dissipated by the resistance of each light bulb is equal to:
where
E is the amount of energy dissipated
t is the time interval
From part B and C we see that the 60 W bulb dissipates more power (142.3 W) than the 200-W bulb (42.7 W). This means that the first bulb dissipates energy faster than the second bulb, so it also burns out faster.
Answer:
The shear stress at a distance 0.3-in away from the pipe wall is 0.06012lb/ft²
The shear stress at a distance 0.5-in away from the pipe wall is 0
Explanation:
Given;
pressure drop per unit length of pipe = 0.6 psi/ft
length of the pipe = 12 feet
diameter of the pipe = 1 -in
Pressure drop per unit length in a circular pipe is given as;
make shear stress (τ) the subject of the formula
Where;
τ is the shear stress on the pipe wall.
ΔP is the pressure drop
L is the length of the pipe
r is the distance from the pipe wall
Part (a) shear stress at a distance of 0.3-in away from the pipe wall
Radius of the pipe = 0.5 -in
r = 0.5 - 0.3 = 0.2-in = 0.0167 ft
ΔP = 0.6 psi/ft
ΔP, in lb/ft² = 0.6 x 144 = 86.4 lb/ft²
Part (b) shear stress at a distance of 0.5-in away from the pipe wall
r = 0.5 - 0.5 = 0