I'm not that smart but I think it is c I really hope It helps
Explanation:
It is based upon the fact that " The light travels faster then sound." As the speed of light is faster then the speed of sound, light travels 300,000 km per second and sound travels 1192 km per hour. That is why we observe the lightening first and hear the the sound of thunder later.
You can do this experiment by yourself. Once you see the lightening start counting the seconds until you hear the sound of thunder.Then divide the seconds by 5, you will find out how many miles away the lightening strike was.
All these resistors are in series so we can take the sum of them by:
Rtotal = R1 + R2 + R3......
So...
Rtotal = 2 + 3 + 4 + 6
Rtotal = 15
So now the total resistance in the circuit is 15 ohms and the potential difference applied to the circuit is 45 volts
Now we can use:
V = IR
Isolate for I
V/R = I
45/15 = I
I = 3 amps (A)
The classical physics works on the Newton's laws of motion. It is applicable on heavenly bodies which are governed by the gravitational force. On the other hand, Quantum Physics is applicable for very low mass and sized bodies like electron, protons etc. The classical physics would accurately describe the motion of satellite moving with speed 7500 m/s using the following formula:
where G is the gravitational constant, M is the mass of the planet and v is the orbital speed. Then radius of the orbit can be described by this formula.
Answer:
a) L = 440 cm
Explanation:
In the open tube on one side and cowbell on the other, we have a maximum in the open part and a node in the closed part, therefore the resonance frequencies are
λ₁ = 4L fundamental
λ₃ = 4L / 3 third harmonic
λ₅ = 4L / 5 five harmonic
The violin string is a fixed cure in its two extracts, so both are nodes, their length from resonance wave are
λ₁ = 2L fundamental
λ₂ = 2L / 2 second harmonic
λ₃ = 2L / 3 third harmonic
λ₄= 2L / 4 fourth harmonic
They indicate that resonance occurs in the fourth harmonic, let's look for the frequency
v =λ f
for the fundamental
v = λ₀ f₀
V = 2L f₀
for the fourth harmonica
v = λ₄ f ’
v = L / 2 f'
2L f₀ = L / 2 f ’
f ’= 4 f₀
f ’= 4 440
f ’= 1760 Hz
for this frequency it has the resonance with the tube
f ’= 4L
L = f ’/ 4
L = 1760/4
L = 440 cm
b) let's find the frequency of the next harmonic in the tube
λ₃ = 4L / 3
λ₃ = 4 400/3
λ₃ = 586.6 cm
v = λf
f = v / λlam₃
f₃3 = 340 / 586.6
f3 = 0.579
as the minimum frequency on the violin is 440 Beam there is no way to reach this value, therefore there are no higher resonances