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3 years ago

An object is fired from the ground at 100. m/s at an angle of 30° with the horizontal.

Physics

1 answer:

Softa [21]3 years ago

7 0

Answer:

a. V_y = 50m/s and V_x = 86.6m/s

b. d_x = 173.2m

c. d_y = 80.38m

Explanation:

V_x = 100m/s*cos(30°)

V_x = 86.6m/s

V_y = 100m/s*sin(30°)

V_y = 50m/s

D_x = v_x*t

d_x = 86.6m/s*2s

d_x = 173.2m

v_f = 50m/s+(-9.81m/s²)*2s

v_f = 30.38m/s

d_y = .5*(30.38m/s+50m/s)*2s

d_y = 80.38m

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The two speakers at S1 and S2 are adjusted so that the observer at O hears an intensity of 6 W/m² when either S1 or S2 is sounde

Zanzabum

Answer:

The minimum frequency is 702.22 Hz

Explanation:

The two speakers are adjusted as attached in the figure. From the given data we know that

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Here

I is the intensity at O when both speakers are on which is given as 6 $W/m^2$

I1 is the intensity of one speaker on which is 6 $W/m^2$

δ is the Path difference which is given as

$\delta=S_2O-S_1O\\\delta=5-4\\\delta=1 m$

λ is wavelength which is given as

$\lambda=\frac{v}{f}$

Here

v is the speed of sound which is 320 m/s.

f is the frequency of the sound which is to be calculated.

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where k=0,1,2

for minimum frequency $f_1$ , k=1

${f}=320 \times (\frac{7 }{36}+2 \times 1 )\\\\{f}=320 \times (\frac{79 }{36} )\\\\ f=702.22 Hz$

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