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faltersainse [42]

3 years ago

6

during the course of songraphic exam, you notice lateral splaying of echoes in the far field. what can you do to improve the ima

ge

1 answer:

horrorfan [7]3 years ago

8 0

Answer:

lateral splaying of echoes in the far field can be improved by Increasing the maximum number of transmit focal zones and optimize their location.

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An explorer is caught in a whiteout (in which the snowfall is so thick that the ground cannot be distinguished from the sky) whi

Veronika [31]

Answer:

$s=5.79\ km$

$\theta=47^{\circ}$ east of south

Explanation:

Given:

distance of the person form the initial position, $d'=8.4\ km$

direction of the person from the initial position, $47^{\circ}$ north of east

distance supposed to travel form the initial position, $d=5.3\ km$

direction supposed to travel from the initial position, due North

<u>Now refer the schematic for visualization of situation:</u>

$y=d'.\sin47^{\circ}-d$

$y=8.4\times \sin47-5.3$ ...............(1)

$x=d'.\cos47^{\circ}$

$x=8.4\times \cos47^{\circ}$ .................(2)

<u>Now the direction of the desired position with respect to south:</u>

$\tan\theta=\frac{y}{x}$

$\tan\theta=\frac{8.4\times \sin47}{8.4\times \cos47}$

$\theta=47^{\circ}$ east of south

<u>Now the distance from the current position to the desired position:</u>

$s=\sqrt{x^2+y^2}$

$s=\sqrt{(8.4\times \cos47)^2+(8.4\times \sin47-5.3)^2}$

$s=5.79\ km$

4 0

3 years ago

Some populations have cooperative relationships. This is where:

Ilia_Sergeevich [38]

C. Members of the same species work together for survival

5 0

3 years ago

(i). A ball of mass 1.500 kg is attached to the end of a cord 1.50 m long. The ball moves in a horizontal circle. If the cord ca

Aleks04 [339]

(a) Let $v$ be the maximum linear speed with which the ball can move in a circle without breaking the cord. Its centripetal/radial acceleration has magnitude

$a_{\rm rad} = \dfrac{v^2}R$

where $R$ is the radius of the circle.

The tension in the cord is what makes the ball move in its plane. By Newton's second law, the maximum net force on it is

$F = (1.500\,\mathrm{kg}) a_{\rm rad}$

so that

$(1.500\,\mathrm{kg}) \dfrac{v^2}{1.50\,\rm m} = 64.0\,\mathrm N$

Solve for $v$ :

$v^2 = \dfrac{(64.0\,\mathrm N)(1.50\,\mathrm m)}{1.500\,\rm kg} \\\\ \implies \boxed{v = 8.00 \dfrac{\rm m}{\rm s}}$

(b) The net force equation in part (a) leads us to the relation

$F = \dfrac{mv^2}R \implies v = \sqrt{\dfrac{FR}m}$

so that $v$ is directly proportional to the square root of $R$ . As the radius $R$ increases, the maximum linear speed $v$ will also increase, so the cord is less likely to break if we keep up the same speed.

6 0

2 years ago

Phantasy [73]

Answer:

Explanation:

lll

5 0

3 years ago

Read 2 more answers

How do you find density of a regular solid?

Effectus [21]

Put the object or material on a scale to figure out<span> its mass. 3. Divide the mass by the volume to </span>figure out the density<span> (p = m / v). You may also need to know </span>how to calculate<span> the volume of a </span>solid s<span>o use the formula</span>

3 0

3 years ago