For the given question the answer to the first part is that frequency of sound will be 404.76 Hz. and answer to the second part will be that the initial frequency for which that place will have the highest level of sound intensity
Given, the speed of sound 340 m/s
a) L1 + L2 = 3.5 m
and Ld = | L1 - L2 | = ( 1.75 + 0.21 ) - (1.75 - 0.21 ) = 1.96 - 1.54 = 0.42 m
Ld = λ/2
λ = 2Ld = 2×0.42 = 0.84 m
and finally,
f = v/λ
f = 340/0.84
f = 404.76 hertz
Frequency came out to be 404.76 hertz in this case
b) For the first frequency
0.42 = λ
f = v/λ
f = 340 / 0.42
f = 809.52 Hertz
Frequency came out to be 809.52 Hertz in this case.
To conclude with we can say that the Frequency of the sound in case on came out to be 404.76 hertz which is approximately 405 Hz after applying all the concepts and calculations, in second case first frequency for which that location will be a maximum of sound intensity came out to be 809.52 Hertz after applying all the concepts and calculations.
Learn more about sound here:
brainly.com/question/24142935
#SPJ10
Answer:
The centripetal acceleration of the rock is 10.7 m/s^2, towards the centre of the circle
Explanation:
The answer is b. because they move freely
F = 2820.1 N
Explanation:
Let the (+)x-axis be up along the slope. The component of the weight of the crate along the slope is -mgsin15° (pointing down the slope). The force that keeps the crate from sliding is F. Therefore, we can write Newton's 2nd law along the x-axis as
Fnet = ma = 0 (a = 0 no sliding)
= F - mgsin15°
= 0
or
F = mgsin15°
= (120 kg)(9.8 m/s^2)sin15°
= 2820.1 N
Answer:
Explanation:
Given
object distance u = 3 m (- ve)
Focal length f = .05 m ( + ve )
Lens formula ,
1 / v - 1 / u = 1 / f
1 / v + 1/3 = 1 / .05
1 / v = 1 / .05 - 1 / 3
1 / v = 20 -0 .333
= 19.667
v = 1 /19.667
v = .05 m = 5 cm.
magnification = v / u = .05 / 3
size of image = ( .05 / 3) x 1.75 = 2.9 cm.