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1 year ago

For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is:

1 answer:

4 0

For maximum radiation protection the suggested distance between array fan-beam scanner source and the operator is 2m.

The Fan beam 5 position reference system (PRS) uses accurate time-of-flight laser technology to determine vessel position relative to custom reflectors.

A fan beam allows only the measurement of the azimuth angle. A fan beam, one with a narrow beam width in azimuth and a broad beam width in elevation, can be obtained by illuminating an asymmetrical section of the paraboloid.

The operators’ desk should be positioned at least 1m away from a pencil beam, and at least 2m from a fan-beam system. Some older models, that are not now common, require a distance of 3.5 m.

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To persuade my audience that the use of mobile communication devices by drivers—even when they are hands-free—is contributing to

pickupchik [31]

Answer:

specific purpose statement

Explanation:

It is a specific purpose statement made for a persuasive speech on the question of fact.

A specific purpose statement helps to build on the general purpose (that is to inform) and to make it more specific to the audience. So if the first speech is an informative speech, our general purpose is to inform our audience about a very specific realm of knowledge.

A specific purpose statement is given to audience to persuade on specific information.

8 0

3 years ago

A gray kangaroo can bound across level ground with each jump carrying it 9.6 m from the takeoff point. Typically the kangaroo le

xz_007 [3.2K]

Answer:

(A) 11 m/s

(B) 1.3 m

Explanation:

Horizontal range, R = 9.6 m

Angle of projection, theta = 28 degree

(A)

Use the formula of horizontal range

R = u^2 Sin 2 theta / g

u^2 = R g / Sin 2 theta

u^2 = 9.6 × 9.8 / Sin ( 2 × 28)

u = 10.65 m/s

u = 11 m/s

(B)

Use the formula for maximum height

H = u^2 Sin ^2 theta / 2g

H =

10.65 × 10.65 × Sin^2 (28) / ( 2 × 9.8)

H = 1.275 m

H = 1 .3 m

4 0

2 years ago

Need some help with these two physics problems!

Juliette [100K]

The force that keeps the puck moving is 0.25 N while the velocity of the puck is 3.7 m/s.

<h3>What is the centripetal force?</h3>

We know that the centripetal force is the force that acts on a body that is moving along a circular path. In this case, we are told that the puck is moving along a circular path hence it is acted upon by the centripetal force that acts on it.

The centripetal force in this case would be supplied by the weight of the object that is moving in the circular path. Thus we can write in our equation that;

Centripetal force = Weight of object = mg

m = mass of the object

g = acceleration due to gravity

Then;

W = 0.026 Kg * 9.8 m/s^2

W = 0.25 N

To obtain the velocity of the object;

FT = mv^2/r

v = √ FT r/m

v = √0.25 * 1.4/0.026

v = 3.7 m/s

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5 0

1 year ago

Earth's neighboring galaxy, the Andromeda Galaxy, is a distance of 2.54×10^7 light-years from Earth. If the lifetime of a human

kupik [55]

Answer:

0.9999986*c

Explanation:

The ship would travel 2.54*10^7 light years, which means that at a speed close to the speed of light the trip would take 2.54*10^7 years from the point of view of an observer on Earth. However from the point of view of a passenger of that ship it will take only 70 years if the speed is close enough to the speed of light.

$\Delta t = \Delta t' * \sqrt{1 - (\frac{v}{c})^2}$

Where

Δt is the travel time as seen by a passenger

Δt' is the travel time as seen by someone on Earth

v is the speed of the ship

c is the speed of light in vacuum

We can replace the fraction v/c with x

$\Delta t = \Delta t' * \sqrt{1 - x^2}$

$\sqrt{1 - x^2} = \frac{\Delta t}{\Delta t'}$

$1 - x^2 = (\frac{\Delta t}{\Delta t'})^2$

$x^2 = 1 - (\frac{\Delta t}{\Delta t'})^2$

$x = \sqrt{1 - (\frac{\Delta t}{\Delta t'})^2}$

$x = \sqrt{1 - (\frac{70}{2.54*10^7})^2} = 0.9999986$

It would need to travel at 0.9999986*c

5 0

3 years ago

Two cyclists start at the same point, but travel along two straight streets in different directions. if they continue, how many

Mumz [18]

Since they are travelling on two different straight directions, therefore their paths will NEVER cross again. This because straight lines will keep on going forever on the same direction therefore there is 0 possibility for two straight lines to paths cross again.

Answer:

<span>zero or never</span>

8 0

3 years ago