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Nadusha1986 [10]

4 years ago

9

Spymaster Tim, flying a constant vh = 193.6 km/hr horizontally in a low-flying helicopter, wants to drop secret documents into h

is contact's open car which is traveling va = 141.7 km/hr on a level highway h = 71.3 m below. At what angle (with the horizontal) should the car be in his sights when the packet is released?

1 answer:

Vinvika [58]4 years ago

5 0

Answer:

θ = 51.21°

Explanation:

Vh= 193.6 km/h

Va= 141.7 km/hr

Relative velocity ,Vr= 193.6 - 141.7 = 51.9 km/hr

Time taken to cover 71.3 m

$h=\dfrac{1}{2}gt^2$

$71.3=\dfrac{1}{2}\times 9.81\times t^2$

t= 3.81 s

t= 0.00105 hr

So the horizontal distance x

x= Vr .t

x= 51.9 x 0.00105 km

x=0.0549 m

x= 54.96 m

Lets take angle is θ

tan θ = h/x

tan θ = 71.3/54.96

θ = 51.21°

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4 years ago

The Hall effect is to be used to find the sign of charge carriers in a semiconductor sample. The probe is placed between the pol

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Answer:

the sign of these carriers is POSITIVE

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3 years ago

Consult Multiple-Concept Example 15 to review the concepts on which this problem depends. Water flowing out of a horizontal pipe

kiruha [24]

Answer:

The pressure of the water in the pipe is 129554 Pa.

Explanation:

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We start usign the continuity equation, and always considering point 1 a point inside the pipe and point 2 a point in the nozzle:

$A_1v_1=A_2v_2$

We want $v_2$ , and take into account that the areas are circular:

$v_2=\frac{A_1v_1}{A_2}=\frac{\pi r_1^2 v_1}{\pi r_2^2}=\frac{r_1^2 v_1}{r_2^2}$

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For determining the absolute pressure of the water in the pipe we use the Bernoulli equation:

$P_1+\frac{\rho v_1^2}{2}+\rho gh_1=P_2+\frac{\rho v_2^2}{2}+\rho gh_2$

Since the tube is horizontal $h_1=h_2$ and those terms cancel out, so the pressure of the water in the pipe will be:

$P_1=P_2+\frac{\rho v_2^2}{2}-\frac{\rho v_1^2}{2}=P_2+\frac{\rho (v_2^2-v_1^2)}{2}$

And substituting for the values we have, considering the pressure in the nozzle is the atmosphere pressure since it is exposed to it we obtain:

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3 years ago

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weqwewe [10]

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3 years ago

Suppose that a public address system emits sound uniformly in all directions and that there are no reflections. The intensity at

dybincka [34]

Answer:

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Explanation:

Given:

Initial intensity, $I_{1}=3.0*10^{-4} Wm^{-2}$ at a distance, $d_{1} = 26 m$

Required:

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Using the inverse square law,

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$I_{2}=4.02*10^{-5} Wm^{-2}$

Thus, the intensity at a spot that is 71 m away is $4.02*10^{-5} Wm^{-2}$

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4 years ago