What are the three longest wavelengths for standing sound waves in a 121-cm-long tube that is (a) open at both ends and (b) open
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:
8.8 cm
31.422 cm/s
Explanation:
m = Mass of block = 0.6 kg
k = Spring constant = 15 N/m
x = Compression of spring
v = Velocity of block
A = Amplitude
As the energy of the system is conserved we have
Amplitude of the oscillations is 8.8 cm
At x = 0.7 A
Again, as the energy of the system is conserved we have
The block's speed is 31.422 cm/s