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anzhelika [568]
4 years ago
12

A sample of a gas has a volume of 639 cm3 when the pressure is 75.9 kPa. What is the volume of the gas when the pressure is incr

eased to 125 kPa if the temperature remains constant? *
Physics
1 answer:
const2013 [10]4 years ago
7 0

Answer:

388 cm^3

Explanation:

For this problem, we can use Boyle's law, which states that for a gas at constant temperature, the product between pressure and volume remains constant:

pV=const.

which can also be rewritten as

p_1 V_1 = p_2 V_2

In our case, we have:

p_1 = 75.9 kPa is the initial pressure

V_1 = 639 cm^3 is the initial volume

p_2 = 125 kPa is the final pressure

Solving for V2, we find the final volume:

v_2 = \frac{p_1 V_1}{p_2}=\frac{(75.9)(639)}{125}=388 cm^3

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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

5 0
2 years ago
When its 75 kW (100 hp) engine is generating full power, a small single-engine airplane with mass 700 kg gains altitude at a rat
kobusy [5.1K]

Answer:

0.2289

Explanation:

Power required to climb= Fv where F is force and v is soeed. We know that F= mg hence Power, P= mgv and substituting 700 kg for m, 9.81 for g and 2.5 m/s for v then

P= 700*9.81*2.5=17167.5 W= 17.1675 kW

To express it as a fraction of 75 kw then 17.1675/75=0.2289 or 22.89%

3 0
3 years ago
A 5kg rock and a 10kg rock are dropped from a height of 10m?
fredd [130]
Look out below ! You should step nimbly to one side, to avoid being hit by one or the other of those hazardous weight objects when they arrive (at the same time).
5 0
3 years ago
To push a 26.0 kg crate up a frictionless incline, angled at 25.0° to the horizontal, a worker exerts a force of 209 N parallel
alukav5142 [94]

Answer:

(a) W = +397.1 J

(b) W = -204.6 J

(c) W = 0

(d) W= + 192.5 J

Explanation:

Work (W) is defined as the product of force (F) by the distance (d)the body travels due to this force. :

W= F*d Formula ( 1)

The forces that perform work on an object must be parallel to its displacement.

The forces perpendicular to the displacement of an object do not perform work on it.

The work is positive (W+) if the force has the same direction of movement of the object.  

The work is negative (W-) if the force has the opposite direction of the movement of the object.

Problem development

(a) Work performed by the worker's applied force on the box .

W= 209 N * 1.9 m = +397.1 J

(b) Work performed by the gravitational force on the crate

We calculate the weight component parallel to the displacement of the box:

We define the x-axis in the direction of the inclined plane ,25.0° to the horizontal.

We define the y-axis and in the direction of the plane perpendicular to the inclined plane.

W= m*g=26*9.8= 254.8N : total box weight

Wx= W*sen25.0°= 254.8*sen25.0°= 107.68 N

W = -Wx *d =107.68 N *1.9 m= -204.6 J

(c) Work performed by normal force (N) exerted by the incline on the crate

The force N is perpendicular to the displacement, then:

W=0

(d) Total work done on the crate

W = 397.1 J -204.6 J

W = 192.5 J

4 0
3 years ago
How do you find the molarity of: 48.2 g of sodium carbonate in 0.500 L of solution
romanna [79]
The answer is why lani

7 0
3 years ago
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