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3 years ago

8

Explain the use of remote sensing in surveying.

1 answer:

Nikolay [14]3 years ago

7 0

Answer:

Remote sensing is the science of obtaining information about objects or areas from a distance typically from aircrafts and satellites

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What is Applied Science?

andreev551 [17]

Answer:

Applied science is the application of existing scientific knowledge to practical applications, like technology or inventions. Within natural science, disciplines that are basic science develop basic information to predict and perhaps explain and understand phenomena in the natural world.

6 0

3 years ago

A model of a submarine, 1:15 scale, is to be tested at 180 ft/s in a wind tunnel with standard sea-level air, while theprototype

ella [17]

Answer:

Explanation:

Given

scale i.e. $L_r=1:15$

Using Reynolds number similarity

$(Re)_m=(Re)_p$

$(\frac{Vl}{\nu })_m=(\frac{Vl}{\nu })_p$

Properties of air

$\nu _{air}=1.57\times 10^{-4} ft/s$

Properties of sea water

$\nu _{sea}=1.26\times 10^{-5} ft/s$

$\left ( \frac{V_ml_M}{\nu _m}\right )=\left ( \frac{V_pl_p}{\nu _p}\right )$

$V_p=V_m\left ( \frac{l_m}{l_p}\right )\left ( \frac{\nu _p}{\nu _m}\right )$

$V_p=180\times \frac{1}{15}\times \frac{1.26\times 10^{-5}}{1.57\times 10^{-4}}$

$V_p=0.963\ ft/s$

8 0

4 years ago

Eagletrons are all-electric automobiles produced by Mogul Motors, Inc. One of the concerns of Mogul Motors is that the Eagletron

Ronch [10]

Answer:

Mogul executives takes samples of n=8 eagletrons at a time.

They found that the average is 3.25 mph.

For n = 8,

D4 =1.864(table value)

D3=0.136(table value)

The upper and lower control limits are found as

UCL =1.864 ×3.25= 6.058

LCL = 0.136 ×3.25= 0.442

3 0

4 years ago

Select the parameters that are included in a baseline performance check. you may select more than one. select one or more: a. co

pishuonlain [190]

Answer:

A F E C D B

Explanation:

6 0

3 years ago

Air flows through a device such that the stagnation pressure is 0.4 MPa, the stagnation temperature is 400°C, and the velocity i

RoseWind [281]

To solve this problem it is necessary to apply the concepts related to temperature stagnation and adiabatic pressure in a system.

The stagnation temperature can be defined as

$T_0 = T+\frac{V^2}{2c_p}$

Where

T = Static temperature

V = Velocity of Fluid

$c_p =$ Specific Heat

Re-arrange to find the static temperature we have that

$T = T_0 - \frac{V^2}{2c_p}$

$T = 673.15-(\frac{528}{2*1.005})(\frac{1}{1000})$

$T = 672.88K$

Now the pressure of helium by using the Adiabatic pressure temperature is

$P = P_0 (\frac{T}{T_0})^{k/(k-1)}$

Where,

$P_0$ = Stagnation pressure of the fluid

k = Specific heat ratio

Replacing we have that

$P = 0.4 (\frac{672.88}{673.15})^{1.4/(1.4-1)}$

$P = 0.399Mpa$

Therefore the static temperature of air at given conditions is 72.88K and the static pressure is 0.399Mpa

<em>Note: I took the exactly temperature of 400 ° C the equivalent of 673.15K. The approach given in the 600K statement could be inaccurate.</em>

3 0

3 years ago