Answer:
Explanation:
a. The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N, 
#substitute for actual values for the lowest frequency.
#n=1, lowest frequency
Hence, the lowest frequency for standing waves is 7.9057Hz
b.The wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N,
#The second lowest frequency happens at 
:
Hence, the second lowest frequency is 15.8114Hz
c.Given that the wire's length is 10m long and has a mass 100g and a tension of 250N.
Frequency is given by the equation:
#where t=250N*10=2500N,
The third lowest frequency happens at 
Hence, the third lowest frequency is 23.7171Hz
Answer:
Q = 4 Q₀
Explanation:
This is an exercise on capacitors, where the capacitance is
C = 
if we apply the given conditions
C = \epsilon_{o} \ \frac{2A}{0.5d}
C = 4 \epsilon_{o} \ \frac{A}{d}
let's call the capacitance Co with the initial values
C₀ = \epsilon_{o} \ \frac{A}{d}
C = 4 C₀
The charge on each plate of a capacitor is
Q = C ΔV
If the potential difference is maintained, the new charge is
Q = 4 C₀ ΔV
let's call
Q₀ = C₀ ΔV
we substitute
Q = 4 Q₀
Answer:
connect two 9 ohms resistance in series now it becomes 18 ohm
If 50 identical light bulbs are connected in series across
a single power source, then the voltage across each bulb
is ( 1/50 ) of the voltage delivered by the power source.
