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lapo4ka [179]
2 years ago
9

Explain why a black and white aeronautical map would be difficult to use

Physics
1 answer:
Aneli [31]2 years ago
5 0

Aeronautical maps are usually meant to be used by pilots and air aviation professionals in other to navigate or traverse though the sky. With various elements such as vegetation, hills, valleys being depicted by color coded keys or legend. Hence, the absence of color on an aeronautical map make the <em>representation of elements very difficult</em>.

Visual map interpretation is usually aided by the use of legends. The legend hold the key to the elements which are represented on the map. Usually, a combination of colors and shapes makes up the legend and makes map interpretation easy.

Therefore, the absence of various color palletes for representation on a black and white aeronautical map will make it difficult to use.

Learn more : brainly.com/question/25323763

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Which statement most accurately describes the second law of thermodynamics?
balandron [24]

Answer:

Energy cannot be changed from one form to another without a loss of usable energy

Explanation:

Second law of thermodynamics states that the total entropy or the randomness of the system remains constant over time.  It also states that the net entropy will remain the same or it will increase.

Entropy of a system is given by heat absorbed divided by temperature. It is given by :

\Delta S=\dfrac{\Delta Q}{T}

So, the correct option is (A) "Energy cannot be changed from one form to another without a loss of usable energy".

4 0
3 years ago
An airplane is moving at 350 km/hr. If a bomb is
Molodets [167]

Answers:

a) -171.402 m/s

b) 17.49 s

c) 1700.99 m

Explanation:

We can solve this problem with the following equations:

y=y_{o}+V_{oy}t-\frac{1}{2}gt^{2} (1)

x=V_{ox}t (2)

V_{f}=V_{oy}-gt (3)

Where:

y=0 m is the bomb's final jeight

y_{o}=1.5 km \frac{1000 m}{1 km}=1500 m is the bomb'e initial height

V_{oy}=0 m/s is the bomb's initial vertical velocity, since the airplane was moving horizontally

t is the time

g=9.8 m/s^{2} is the acceleration due gravity

x is the bomb's range

V_{ox}=350 \frac{km}{h} \frac{1000 m}{1 km} \frac{1 h}{3600 s}=97.22 m/s is the bomb's initial horizontal velocity

V_{f} is the bomb's fina velocity

Knowing this, let's begin with the answers:

<h3>b) Time</h3>

With the conditions given above, equation (1) is now written as:

y_{o}=\frac{1}{2}gt^{2} (4)

Isolating t:

t=\sqrt{\frac{2 y_{o}}{g}} (5)

t=\sqrt{\frac{2 (1500 m)}{9.8 m/s^{2}}} (6)

t=17.49 s (7)

<h3>a) Final velocity</h3>

Since V_{oy}=0 m/s, equation (3) is written as:

V_{f}=-gt (8)

V_{f}=-(97.22)(17.49 s) (9)

V_{f}=-171.402 m/s (10) The negative sign ony indicates the direction is downwards

<h3>c) Range</h3>

Substituting (7) in (2):

x=(97.22 m/s)(17.49 s) (11)

x=1700.99 m (12)

5 0
3 years ago
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

5 0
3 years ago
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