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3 years ago

9

Sonar signals and infrared light are are used to send messages to submarines deep under water. If we compare the two signals, wh

ich statement describes a difference between the two?

1 answer:

WARRIOR [948]3 years ago

6 0

What are the statement choices?

You might be interested in

An object whose specific gravity is 0.850 is placed in water. What fraction of the object is below the surface of the water?

Fynjy0 [20]

Answer:

The fraction of the object that is below the surface of the water is ¹⁷/₂₀

Explanation:

Given;

specific gravity of the object, γ = 0.850

Specific gravity is given as;

$specific \ gravity = \frac{density \ of the \ object}{density \ of \ water}\\\\0.85= \frac{density \ of the \ object}{1000 \ kg/m^3} \\\\density \ of the \ object = 850 \ kg/m^3$

Fraction of the object's weight below the surface of water is calculated as;

$= \frac{850}{1000} \ \times\ 100\%\\\\= 85 \% \\\\= \frac{17}{20}$

Therefore, the fraction of the object that is below the surface of the water is ¹⁷/₂₀

8 0

3 years ago

Although it shouldn’t have happened, on a dive i fail to watch my spg and run out of air. if my buddy is close by, my best optio

leva [86]

Answer:

B ) Ascend using my buddy alternative air source / make an emergency Ascent

Explanation:

From the description it can be seen his buddy is close by of which he can easily use the alternative air source. Also we can see that he is closer to the water surface than his buddy, of which controlled emergency swimming ascent is highly favourable in this condition.

5 0

3 years ago

Read 2 more answers

Assume the Earth is a ball of perimeter 40, 000 kilometers. There is a building 20 meters tall at point a. A robot with a camera

torisob [31]

Answer:

Approximately 21 km.

Explanation:

Refer to the not-to-scale diagram attached. The circle is the cross-section of the sphere that goes through the center C. Draw a line that connects the top of the building (point B) and the camera on the robot (point D.) Consider: at how many points might the line intersects the outer rim of this circle? There are three possible cases:

No intersection: There's nothing that blocks the camera's view of the top of the building.
Two intersections: The planet blocks the camera's view of the top of the building.
One intersection: The point at which the top of the building appears or disappears.

There's only one such line that goes through the top of the building and intersects the outer rim of the circle only once. That line is a tangent to this circle. In other words, it is perpendicular to the radius of the circle at the point A where it touches the circle.

The camera needs to be on this tangent line when the building starts to disappear. To find the length of the arc that the robot has travelled, start by finding the angle $\angle \mathrm{B\hat{C}D}$ which corresponds to this minor arc.

This angle comes can be split into two parts:

$\angle \mathrm{B\hat{C}D} = \angle \mathrm{B\hat{C}A} + \angle \mathrm{A\hat{C}D}$ .

Also,

$\angle \mathrm{B\hat{A}C} = \angle \mathrm{D\hat{A}C} = 90^{\circ}$ .

The radius of this circle is:

$\displaystyle r = \frac{c}{2\pi} = \rm \frac{4\times 10^{7}\; m}{2\pi}$ .

The lengths of segment DC, AC, BC can all be found:

$\rm DC = \rm \left(1.75 \displaystyle + \frac{4\times 10^{7}\; m}{2\pi}\right)\; m$ ;
$\rm AC = \rm \displaystyle \frac{4\times 10^{7}}{2\pi}\; m$ ;
$\rm BC = \rm \left(20\; m\displaystyle +\frac{4\times 10^{7}}{2\pi} \right)\; m$ .

In the two right triangles $\triangle\mathrm{DAC}$ and $\triangle \rm BAC$ , the value of $\angle \mathrm{B\hat{C}A}$ and $\angle \mathrm{A\hat{C}D}$ can be found using the inverse cosine function:

$\displaystyle \angle \mathrm{B\hat{C}A} = \cos^{-1}{\rm \frac{AC}{BC}}$

$\displaystyle \angle \mathrm{D\hat{C}A} = \cos^{-1}{\rm \frac{AC}{DC}}$

$\displaystyle \angle \mathrm{B\hat{C}D} = \cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}$ .

The length of the minor arc will be:

$\displaystyle r \theta = \frac{4\times 10^{7}\; \rm m}{2\pi} \cdot (\cos^{-1}{\rm \frac{AC}{BC}} + \cos^{-1}{\rm \frac{AC}{DC}}) \approx 20667 \; m \approx 21 \; km$ .

5 0

3 years ago

When Sweden is playing Denmark, it is SWE-DEN. The remaining letters, not used, is DEN-MARK

iren [92.7K]

Yes it is! It would be SWE vs. DEN

4 0

3 years ago

A solenoid of radius r 5 1.25 cm and length , 5 30.0 cm has 300 turns and carries 12.0 A. (a) Calculate the flux through the sur

Elina [12.6K]

Answer:

$\phi = 7.4 \times 10^{-6} Wb$

Explanation:

Magnetic field due to a long solenoid at the center is given by

$B = \frac{\mu_o N i}{L}$

here we know that

N = 300

L = 30 cm

i = 12 A

now magnetic field is given as

$B = \frac{(4\pi \times 10^{-7})(300)(12)}{0.30}$

$B = 0.015 T$

Now magnetic flux through the disc is given as

$\phi = B.A$

$\phi = (0.015)(\pi r^2)$

$\phi = (0.015)(\pi)(0.0125)^2$

$\phi = 7.4 \times 10^{-6} Wb$

8 0

3 years ago