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3 years ago

Atmospheric pressure at the top of mount everest.

Physics

1 answer:

Mumz [18]3 years ago

6 0

Answer:

Mt. Everest (elevation of 8848 m), the air pressure is 253 mmHg. The summit of Mt. Everest, the highest elevation on earth, where the pressure averages around 300 mb." As altitude increases, the air becomes thinner, the density of air decreases, and the pressure of the air decreases as wel

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A 15.5 kg block is pulled by two forces. The first is 11.8N at a 53.7 angle and the second is 22.9 N at a -15.8 angle. What is t

anygoal [31]

Answer:

α = 6.43° (Angle respect to the horizontal axis-x)

Explanation:

To solve this problem we must use Newton's second law which tells us that the sum of forces must be equal to the product of mass by acceleration.

Since we have two forces and we know their magnitudes and their directions, we can sum these into their X & y components, in this way we can find the resulting force and apply it in Newton's second law.

F1x = 11.8*cos(53.7) = 6.98 [N]

F1y = 11.8*sin(53.7) = 9.5 [N]

Now with the second force:

F2x = 22.9*cos (15.8) = 22.03 [N]

F2y = - 22.9*sin (15.8) = - 6.23 [N]

Now we sum the forces in the x and y axes:

Fx = F1x + F2x = 6.98 + 22.03 = 29.01 [N]

Fy = F1y + F2y = 9.5 - 6.23 = 3.27 [N]

Now using the Pythagorean theorem we can find the resulting force.

F = √(Fx² + Fy²)

F = √ (29.01)² + (3.27)²

F = 29.19 [N]

Using Newton's second law, we have:

F = m*a

where:

F = force = 29.19 [N]

m = mass = 15.5 [kg]

a = acceleration [m/s²]

a = 29.19/15.5

a = 1.88 [m/s²]

The direction of the acceleration is the same direction of the force, therefore we need to find the angle.

tan(α) = Fy/Fx

tan(α) = 3.27/29.01

α = tan⁻¹(0.1127)

α = 6.43° (Angle respect to the horizontal axis-x)

3 0

3 years ago

Which of the following describes a displacement vs. time graph that looks like this?

lubasha [3.4K]

constant non zero acceleration

6 0

3 years ago

A 2kg object is moving horizontally with a speed of 4m/s 2. How much net force is required t keep the object moving at this spee

dezoksy [38]

To continue moving at constant speed in a straight line requires NO net force. Zero. Nada. If there IS any net force on the object, then its speed or direction will change.

8 0

3 years ago

From a window 20 feet above the ground, the angle of elevation to the top of a building across the street is 78°, and the angle

Papessa [141]

The answer would be 371.16 ft.

Refer to the diagram for the scenario. The drawing is not to scale.

Notice the diagram that you will find to adjacent right triangles. All you need to do is first solve for the needed side of one triangle, by using one side.

It sounds confusing I know, but let's do this step by step.

Angle of depression is 15°.

We use this first because we do not have enough data to use the angle of elevation.

To get the adjacent side, we use the trigonometric function TOA which means:

$tan\theta= \dfrac{opposite}{adjacent}$

We use the height of the first building as our reference, so our given will be:

Opposite = 20ft

θ = 15°

Now we put that into our formula:

$tan\theta= \dfrac{opposite}{adjacent}$

$Tan15 = \dfrac{20ft}{adjacent}$

$adjacent = \dfrac{20ft}{Tan15}$

Adjacent = 74.64 ft

Angle of elevation is 78°:

Now that we have that we can solve for the height from the point of 20ft. We use the same formula, because we are looking for the opposite, given the adjacent.

$tan\theta= \dfrac{opposite}{adjacent}$

$Tan78 = \dfrac{opposite}{74.64ft}$

$(74.64ft)(Tan78) = opposite$

351.16ft = Opposite

Now that's the height from the top of the first building. To get the total height of the building we just add the 20ft.

351.16ft + 20 ft = 371.16ft

<em>The building across the street is 371.16 ft tall.</em>

5 0

3 years ago

What are the Characteristics of theta waves

stiv31 [10]

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4 0

3 years ago