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faltersainse [42]

2 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]2 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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7. A box is being pushed from the left at 15N and pushed from the right at 10N,

professor190 [17]

It is 5N to the left

8 0

2 years ago

An object begins at position x = 0 and moves one-dimensionally along the x-axis witļi a velocity v

Liula [17]

Answer:

The answer is "between 20 s and 30 s".

Explanation:

Calculating the value of positive displacement:

$\ (x_{+ve}) = \frac{1}{2} \times 15 \times 20 \\\\$

$= \frac{1}{2} \times 300 \\\\= 150 \\\\$

Calculating the value of negative displacement upon the time t:

$(x_{-ve}) = \frac{1}{2} \times 5 \times 20- 20(t-20) \\\\$

$= \frac{1}{2} \times 100- 20t+ 400 \\\\= 50- 20t+ 400 \\\\$

$\to X= X_{+ve} + X_{-ve} \\\\$

$\to 150 - 50 -20t+400 =0\\\\\to 100 -20t+400 =0 \\\\\to 500 -20t =0\\\\\to 20t =500 \\\\\to t=\frac{500}{20}\\\\\to t=\frac{50}{2}\\\\\to t= 25$

That's why its value lie in "between 20 s and 30 s".

6 0

3 years ago

What is the magnification of an astronomical telescope whose objective lens has a focal length of 71 cm and whose eyepiece has a

Rus_ich [418]

To solve this problem it is necessary to apply the concepts related to optical magnification (is the process of enlarging the apparent size, not physical size, of something.). Specifically the angular magnification of an optical telescope is given by

$M = -\frac{f_o}{f_e}$

Where,

$f_o =$ Focal length of the objective lens in a refractor

$f_e =$ Focal length of the eyepiece

Our values are given as

$f_o =$ 71cm

$f_e =$ 2.1cm

Replacing we have

$M = -\frac{f_o}{f_e}$

$M = -\frac{71}{2.1}$

$M = - 33.81$

Therefore the magnification of this astronomical telescope is -33.81

7 0

3 years ago

A scientist adds different amounts of salt to 5 bottles of water. She then measures how long it takes for the water to boil. Wha

postnew [5]

Answer:

the responding variable is the water boiling

Explanation:

a responding variable is the same thing as a dependent variable and an independent variable you change the independent variable is the amount of salt, the control group is how long water takes to boil without adding salt, and a constant is the same amount of water

8 0

3 years ago

Conversations with astronauts on the lunar surface were characterized by a kind of echo in which the earthbound person’s voice w

abruzzese [7]

Answer:

3.85*10^8m

Explanation:

We know that

Speed v = distance/time t

So distance = v x t

So = 3*10^8 x 2.56s

= distatance = 7.7*10^8m

However this is double the distance from the earth to the moon, so to get distance from earth to the moon we divide by 2

7.7*10^8/2= 3.85*10^8m

8 0

2 years ago