The distance in the x direction between two control points on a vertical aerial photograph is 4.5". If the distance between these same two points is 3.6" on another photograph having a scale of 1:24,000, determine the scale of the first vertical aerial photograph. Of the focal length of the camera is 6"and the average elevation at these points is 100 ft, determine the flying height from which each photograph was taken

Answers

Answer 1

Answer:

Answer:

Use proportions to find the scale of the first photo, then use that scale and other given information to fill in the equation

S=f/(H-h)

Where:

S = scale of the photo

f = focal length of the camera (in feet)

H = flying height

h = average elevation

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What is a constellation as astronomers define it today? What does it mean when an astronomer says, “I saw a comet in Orion last night”?

Answers

Explanation:

Constellation: The complete sky has been divided in 88 different areas, in a way we have divided Earth in countries, not necessarily having same shapes and size. These 88 areas are known as constellations. These contains a lot of stars. When we join the brightest stars together we can imagine a shape out of them which is called as Asterism. Most of the people are unaware of this difference. Some of the famous constellations are Orion, Taurus, Gemini, Hydra, Ursa Major etc.

When an astronomer says that there is a comet is in the Orion, he means that a comet is in the boundaries of Orion constellation.

Final answer:

A constellation is a group of stars that forms a certain pattern in the sky. 'Seeing a comet in Orion' means that the comet was observed in the region of the sky defined by the constellation Orion.

Explanation:

A constellation is a group of stars that astronomers have grouped together into patterns which represent certain symbols as a means of recognizing or remembering them. They are only human constructs not real associations of stars. When an astronomer says, “I saw a comet in Orion last night”, they are referring to a specific area in the sky defined by the constellation Orion. The constellation acted as a reference point that helped the astronomer locate and observe the comet.

Learn more about Constellation here:

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#SPJ6

18. Wind speed on Earth is reduced by which?A. temperature
B. friction
C. weather
D. convergence

Answers

Answer:

Temperaturereduces the wind speed on earth.

Explanation:

Wind speed is an atmospheric quantity that causes the wind to flow from high to low level due to a change in temperature.
Wind speed is controlled by the pressure gradient factor and it increases the wind speed with increase in pressure gradient.
Wind speed decreases with an increase in temperature value as it absorbs the heat developed in the system and wind gets slowed down.
Wind speed is measured in velocity and anemometer is used to measure the rate of wind speed.

Answer:

your answer would be A. temperature

Explanation:

Wind speed on Earth is reduced by temperature, it depends on the temperature because let's say there is a nice sunny day, then obviously it will reduce the wind speed

A forest fire sends carbon monoxide and ash into the air. This is an example of the__1__ affecting the__2___ .1.
A. Hydrosphere
B Lithosphere
C. Atmosphere
2
A Lithosphere
B Atmosphere
C Biosphere

Answers

1. B Lithosphere
2. B Atmosphere

lithosphere and biosphere

srry ik im super late but to help other people theres the answer

A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction (a) If the applied stress if 0.45 MPa, what will be the resolved shear stress, tRss, along the [101] direction within the (11 T) plane? (b) What tensile stress is required to produce a critical resolved shear stress, TcRss of 0.242 MPa

Answers

The resolved shear stress along the [101] direction is 0.3673 MPa and the tensile stress is required to produce a critical resolved is 0.2964 MPa.

What is a lattice unit cell?

The symmetrical 3-D structural arrangement of the ions, atoms, or molecules inside a crystalline lattice solid as a point.

A crystalline grain of nickel in a metal plate is situated so that a tensile load is oriented along the [111] crystal direction.

(a) If the applied stress is 0.45 MPa. Then the resolved shear stress, along the [101] direction within the (11T) plane will be

The θ be the angle between [101] and (111) will be

$\rm cos\ \theta = \frac{1*1+1*0+1*1}{√(1^2+1^2+1^2)\sqer{1^2+0^2+1^2}}\n\n\ncos\ \theta = \sqrt{ (2)/(3)}$

And ∅ be the angle between [111] and [111], then we have

$\rm cos\ \phi= \frac{1*1+1*1+1*1}{√(1^2+1^2+1^2)\sqer{1^2+1^2+1^2}}\n\n\ncos\ \phi= 1$

Now, for the resolved components along [101], we have

$\rm = \sigma cos \ \theta \ \ cos \ \phi\n\n= 0.45 *10^6*\sqrt{(2)/(3)}*1= 0.3673 \ \ MPa$

(b) The tensile stress is required to produce a critical resolved shear stress of 0.242 MPa will be

$\rm \sigma _1 = (shear\ stress)/(cos\ \theta \ cos \ \phi)\n\n\n\sigma _1 = \frac{0.242*10^6}{\sqrt{(2)/(3)} * 1}\n\n\n\sigma _1 = 0.2964 \ MPa$

More about the crystalline lattice link is given below.

brainly.com/question/10951564

Given:

Applied stress, $\sigma = 0.45 MPa$

Critical Resolved Stress, $T_(cRss)= 0.242 MPa$

Solution:

a). According to the question, orientation of tensile load is along [1 1 1],

$\sigma = 0.45 MPa$

Now, for resolved shear stress, $\t_(Rss)$ along [1 0 1] within (1 1 T)

let ' $\theta$ ' be the angle between [1 1 1] and [1 0 1], then by coordinate formula:

$cos\theta$ = $\frac{1* 1 + 1* 0 + 1* 1}{\sqrt{1^(2)+1^(2)+1^(2)\sqrt{1^(2)+0^(2)+1^(2)}}}$

$cos\theta$ = $\sqrt{(2)/(3)}$

let ' $\phi$ ' be the angle between [1 1 1] and [1 1 1], then by coordinate formula:

$cos\phi = \frac{1* 1 + 1* 1 + 1* 1}{\sqrt{1^(2)+1^(2)+1^(2)\sqrt{1^(2)+1^(2)+1^(2)}}}$

$cos\phi$ = 1

Now, for the resolved components along [1 0 1]

$\t_(Rss)$ = $\sigma cos\theta cos\phi$

$\t_(Rss)$ = $0.45* 10^(6)* \sqrt{(2)/(3)}* 1$ = 0.3673 MPa

b). For required tensile stress to produce $T_(cRss)= 0.242 MPa$ :

$\sigma _(1) = (T_(cRss))/(cos\theta cos\phi )$

$\sigma _(1) = \frac{0.242* 10^(6)}{\sqrt{(2)/(3)}* 1}$

$\sigma _(1) = 0.2964 MPa$

A typical automobile under hard braking loses speed at a rate of about 7.2 m/s2; the typical reaction time to engage the brakes is 0.55 s. A local school board sets the speed limit in a school zone such that all cars should be able to stop in 3.6 m. What maximum speed does this imply for a car in the school zone?

Answers

Answer:

4.3 m/s

Explanation:

a = rate at which the automobile loses speed = - 7.2 m/s²

v₀ = initial maximum speed of automobile

t' = reaction time for applying the brakes = 0.55 s

d = distance available for stopping the vehicle = 3.6 m

d' = distance traveled while applying the brakes = v₀ t' = (0.55) v₀

v = final speed after the vehicle comes to stop = 0 m/s

Using the equation

v² = v₀² + 2 a (d - d' )

0² = v₀² + 2 (- 7.2) (3.6 - (0.55) v₀)

v₀ = 4.3 m/s

What is the motion of the particles in this kind of wave? A hand holds the left end of a set of waves. The waves themselves make a larger set of waves in the same direction as that of the smaller waves. A label Wave motion is above the series of waves and an arrow next to the label points right. The particles will move up and down over large areas. The particles will move up and down over small areas. The particles will move side to side over small areas. The particles will move side to side over large areas.

Answers

Answer:

→A←

Explanation:

D its incorrect in edge

Answer:

Explanation:

The particles will move side to side over large areas

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