Answer:
Explanation:
The time lag between the arrival of transverse waves and the arrival of the longitudinal waves is defined as:
Here d is the distance at which the earthquake take place and 
is the velocity of the transverse waves and longitudinal waves respectively. Solving for d:
Answer:
The slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.
Explanation:
Given;
The slope, dy/dx = 2x(y-6) - 4
2x(y-6) - 4 = 2xy - 12x - 4, divide through by 'x'
dy/dx = 2y -12 - 4/x
The slopes of the linear elements on the lines, x =0, y = 5, y = 6, y = 7.
At x = 0, and y = 5
dy/dx = 2y -12 - 4/x
dy/dx = 2(5) - 12 = -2
At x = 0, and y = 6
dy/dx = 2y -12 - 4/x
dy/dx = 2(6) - 12 = 0
At x = 0, and y = 7
= 2y -12 - 4/x
dy/dx = 2(7) - 12 = 2
Therefore, the slopes shows that the direction of the field is from -2 to +2, with three point charges, q₁, q₂ and q₃ at -2, 0 and +2 respectively.
Answer:
Sound wave X amplitude is greater than 'A' and its frequency is lesser than
'f'
Explanation:
The pitch of a sound is dictated by the frequency of the sound wave, while the loudness is dictated by the amplitude.
A high pitch sound corresponds to a high frequency and a low pitch sound corresponds to a low frequency.
The larger the amplitude of the waves, the louder the sound and vice-versa.
From the question,
Sound wave W has amplitude ‘A' and frequency 'f' and
Sound wave X is louder and lower in pitch than sound wave W.
Since sound wave X is louder, this means its amplitude is greater than 'A'.
Also, since sound wave X is lower in pitch, this means its frequency is lesser than 'f'.
Answer:
f = v / 4L
the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
Explanation:
In wind instruments the wave speed must meet
v = λ f
λ = v / f
from v is the speed of sound that depends on the temperature
v = v₀ ![\sqrt{1+ \frac{T [C]}{273} }](https://tex.z-dn.net/?f=%5Csqrt%7B1%2B%20%5Cfrac%7BT%20%5BC%5D%7D%7B273%7D%20%7D)
where I saw the speed of sound at 0ºC v₀ = 331 m/s the temperature is in degrees centigrade, we can take the degrees Fahrenheit to centigrade with the relation
(F -32) 5/9 = C
76ºF = 24.4ºC
45ºF = 7.2ºC
With this relationship we can see that the speed of sound is significantly reduced when leaving the house to the outside
at T₁ = 24ºC v₁ = 342.9 m / s
at T₂ = 7ºC v₂ = 339.7 m / s
To satisfy this speed the wavelength of the sound must be reduced, so the resonant frequencies change
λ / 4 = L
λ= 4L
v / f = 4L
f = v / 4L
Therefore, the frequency of the instruments is reduced by the decrease in the speed of the wave with the temperature.
