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nataly862011 [7]
3 years ago
11

The barometric pressure at sea level is 30 in of mercury when that on a mountain top is 29 in. If the specific weight of air is

assumed constant at 0.0075 lb/ft3 , calculate the elevation of the mountain top.
Physics
1 answer:
stealth61 [152]3 years ago
7 0

To solve this problem we will apply the concepts related to pressure, depending on the product between the density of the fluid, the gravity and the depth / height at which it is located.

For mercury, density, gravity and height are defined as

\rho_m = 846lb/ft^3

g = 32.17405ft/s^2

h_1 = 1in = \frac{1}{12} ft

For the air the defined properties would be

\rho_a = 0.0075lb/ft^3

g = 32.17405ft/s^2

h_2 = ?

We have for equilibrium that

\text{Pressure change in Air}=\text{Pressure change in Mercury}

\rho_m g h_1 = \rho_a g h_2

Replacing,

(846)(32.17405)(\frac{1}{12}) = (0.0075)(32.17405)(h_2)

Rearranging to find h_2

h_2 = \frac{(846)(32.17405)(\frac{1}{12}) }{(0.0075)(32.17405)}

h = 9400ft

Therefore the elevation of the mountain top is 9400ft

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Anarel [89]

Answer:

D. infinitely extended in all directions

Explanation:

A semi infinite solid is infinitely extended in every direction. It has a single surface and can extend when heat is applied.

The body of a semi infinite solid is idealised, that is, when there is heat present, it expands in all directions to infinity. It can be used for a thick wall because its shape can be changed when subjected to different levels of heat near its surface.

It is also expands as heat is applied because its thickness is negligible.

This idealized body is used for earth, thick wall, steel piece of any shaped quenched rapidly etc indetermining variation of temperature near its surface & other surface being too far to have any impact on the region in short period of time since heat doesn’t have sufficient time to penetrate deep into body thus thickness can be neglected

8 0
3 years ago
A particle moves along the x axis from the origin. The magnitude of the position vector at time t is
Maurinko [17]

1) The average velocity is -2.1\cdot 10^5 m/s

2) The instantaneous velocity is 64t-260t^3

Explanation:

1)

The average velocity of an object is given by

v=\frac{d}{t}

where

d is the displacement

t is the time elapsed

In this problem, the position of the particle is given by the function

x(t) = 32t^2 - 65t^4

where t is the time.

The position of the particle at time t = 6 sec is

x(6) = 32(6)^2 - 65(6)^4=-83,088 m

While the position at time t = 12 sec is

x(12)=32(12)^2-65(12)^4=-1,343,232 m

So, the displacement is

d=x(12)-x(6)=-1,343,232-(-83,088)=-1,260,144 m

And therefore the average velocity is

v=\frac{-1,260,144 m}{12 s- 6 s}=-2.1\cdot 10^5 m/s

2)

The instantaneous velocity of a particle is given by the derivative of the position vector.

The position vector is

x(t) = 32t^2 - 65t^4

By differentiating with respect to t, we find the velocity vector:

v(t) = x'(t) = 2\cdot 32 t - 4\cdot 65 t^3 = 64t - 260 t^3

Therefore, the instantaaneous velocity at any time t can be found by substituting the value of t in this expression.

Learn more about velocity:

brainly.com/question/5248528

#LearnwithBrainly

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3 years ago
A pump jack scaffold must be fitted with two ______ gripping mechanisms to prevent slippage.
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A pump jack scaffold must be fitted with two positive gripping mechanisms to prevent slippage. Pump jacks are a uniquely designed scaffold consisting of a platform supported by movable brackets on vertical poles. The brackets are designed to be raised and lowered in a manner similar to an automobile jack. It is important to make sure that pump jack brackets have two positive gripping mechanisms to prevent any failure or slippage. 
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In Europe, gasoline efficiency is measured in km/L. If your car's gas mileage is 36.0mi/gal , how many liters of gasoline would
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3 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

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3 years ago
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