Answer: The correct answer is "Instrument A is placed closer to Sam than instrument B".
Explanation:
The sound can be soft or loud. Loudness depends on the amplitude of the sound wave. Higher the amplitude, more will be loudness. Lower the amplitude, lesser will be loudness.
Pitch depends on the frequency.
In the given problem, the instruments A and B generate sound waves of the same amplitude and at the same time.
Loudness depends on the sound energy produced as the energy of the sound is directly proportional to the square of the amplitude. It also depends on the distance between the source and the receiver.
Sam records a louder sound from instrument A than from instrument B. It means that there is mismatch in loudness. It can happen due to the placement of the instrument A closer to Sam than instrument B.
Therefore, the correct option is "Instrument A is placed closer to Sam than instrument B".
<h2>Answer:</h2>
The refractive index is 1.66
<h2>Explanation:</h2>
The speed of light in a transparent medium is 0.6 times that of its speed in vacuum
.
Refractive index of medium = speed of light in vacuum / speed of light in medium
So
RI = 1/0.6 = 5/3 or 1.66
Answer:
During charging by conduction, both objects acquire the same type of charge. If a negative object is used to charge a neutral object, then both objects become charged negatively. ... In this case, electrons are transferred from the neutral object to the positively charged rod and the sphere becomes charged positively.
Answer:
λ = V / f the wavelength versus the frequency
V = f λ and V (speed) proportional λ for a fixed frequency
F = f^2 * (M / L) * λ^2 = (f * λ)^2 * (M / L)^2 force (tension) on string at a given frequency
F2 / F1 = (λ2 / λ1)^2 other items are constant
Let λ1 = 6 then λ2 must be 3/2 λ1 for a constant length
F2 / F1 = (6 / 4)^2 = 9/4
The tension must be increased to 9 / 4 of the original tension
Check: if the frequency is fixed then V will be larger for a larger wavelength (situation 2)
One can also write V = (F / (M / L))^1/2
Then for fixed M L
F2 / F1 = (V2 / V1)^2
Since V = f λ Velocity is proportional to λ for a fixed frequency
Then if V2 / V1 = 3 / 2 F2 = 9/4 F1
Answer:
The work done is -209.42 J.
Explanation:
F(x) = (- 20 - 3 x ) N
x = 0 to x = 6.9 m
Here, the force is variable in nature, so the work done by the variable force is given by
![W =\int F dx\\\\W =\int_{0}^{6.9}(-20- 3x dx )\\\\W= \left [ - 20 x - 1.5 x^2 \right ]_{0}^{6.9}\\\\W = - 20 (6.9 - 0) - 1.5(6.9\times 6.9 - 0)\\\\W =- 138 - 71.42\\\\W = - 209.42 J](https://tex.z-dn.net/?f=W%20%3D%5Cint%20F%20dx%5C%5C%5C%5CW%20%3D%5Cint_%7B0%7D%5E%7B6.9%7D%28-20-%203x%20dx%20%29%5C%5C%5C%5CW%3D%20%5Cleft%20%5B%20-%2020%20x%20-%201.5%20x%5E2%20%5Cright%20%5D_%7B0%7D%5E%7B6.9%7D%5C%5C%5C%5CW%20%3D%20-%2020%20%286.9%20-%200%29%20-%201.5%286.9%5Ctimes%206.9%20-%200%29%5C%5C%5C%5CW%20%3D-%20138%20-%2071.42%5C%5C%5C%5CW%20%3D%20-%20209.42%20%20J)
