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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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faltersainse [42]

Answer:

a) 7200 ft/s²

b) 140 ft

c) 3.7 s

Explanation:

(a) Average acceleration is the change in velocity over change in time.

a_avg = Δv / Δt

We need to find what velocity the puck reached after it was hit by the hockey player.

We know it reached 40 ft/s after traveling 90 feet over rough ice at an acceleration of -20 ft/s².  Therefore:

v² = v₀² + 2a(x − x₀)

(40 ft/s)² = v₀² + 2(-20 ft/s²)(100 ft − 10 ft)

v₀² = 5200 ft²/s²

v₀ = 20√13 ft/s

So the average acceleration impacted to the puck as it is struck is:

a_avg = (20√13 ft/s − 0 ft/s) / (0.01 s)

a_avg = 2000√13 ft/s²

a_avg ≈ 7200 ft/s²

(b) The distance the puck travels before stopping is:

v² = v₀² + 2a(x − x₀)

(0 ft/s)² = (5200 ft²/s²) + 2(-20 ft/s²)(x − 10 ft)

x = 140 ft

(c) The time the puck takes to travel 10 ft without friction is:

t = (10 ft) / (20√13 ft/s)

t = (√13)/26 s

The time the puck travels over the rough ice is:

v = at + v₀

(0 ft/s) = (-20 ft/s²) t + (20√13 ft/s)

t = √13 s

So the total time is:

t = (√13)/26 s + √13 s

t = (27√13)/26 s

t ≈ 3.7 s

8 0
3 years ago
Take a close look at the energy transfers and transformations shown in the above diagram. Which type of energy is transformed in
labwork [276]

Kinetic Energy

How to calculate

Kinetic energy is given by K.E.

\\ \sf\longmapsto K.E=\dfrac{1}{2}mv^2

Where

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4 0
3 years ago
g is incident on 3 successive sheets of polarizing material. The transmission axis of the first sheet is vertical. The transmiss
murzikaleks [220]

Answer:

The intensity of light passing from the third polarizer is 3Io/16.

Explanation:

The law of Malus is given by

I=I_o cos^2\theta

Let the incident intensity of light is Io.

The intensity of light passing from the first polarizer is

I' = \frac{I_o}{2}

The intensity of light passing from the second polarizer is

I''=\frac{I_o}{2}\times cos^230 =\frac{3I_o}{8}

The intensity of light passing from the third polarizer is

I''' = \frac{3I_o}{8}\times cos^2 60\\\\\\I''' = \frac{3I_o}{16}

6 0
3 years ago
A ball is launched at an angle of 10 degrees from a 20 meter tall building with a speed of 4 m/s. How long is the ball in the ai
shutvik [7]

Answer:

d) None of the above

Explanation:

v_{o} = inituial velocity of launch = 4 m/s

θ = angle of launch = 10 deg

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y = vertical displacement = - 20 m  

t = time of travel

using the equation

y=v_{oy} t+(0.5)a_{y} t^{2}

- 20 = (0.695) t + (0.5) (- 9.8) t²

t = 2.1 sec

consider the motion along the horizontal direction

x = horizontal displacement

v_{ox} = initial velocity along horizontal direction = 4 Cos10 = 3.94 m/s

a_{x} = acceleration along the horizontal direction = 0 m/s²

t = time of travel =  2.1 s

Using the kinematics equation

x =v_{ox} t+(0.5)a_{x} t^{2}

x = (3.94) (2.1) + (0.5) (0) (2.1)²

x = 8.3 m

Consider the motion along the vertical direction

v_{oy} = initial velocity along vertical direction = 4 Sin10 = 0.695

m/s

a_{y} = acceleration along the vertical direction = - 9.8 m/s²

y_{o} =initial vertical position at the time of launch = 20 m  

y = vertical position at the maximum height = 20 m

v_{fy} = final velocity along vertical direction at highest point = 0 m/s

using the equation

{v_{fy}}^{2}= {v_{oy}}^{2} + 2 a_{y}(y - y_{o})

0^{2}= 0.695^{2} + 2 (- 9.8)(y - 20)

y = 20.02 m

h = height above the starting height

h = y - y_{o}

h = 20.02 - 20

h = 0.02 m

7 0
3 years ago
A 20-kg block slides down a fixed rough curved track The block has a speed of 5 0 m/s after its height above a horizontal surfac
Crank

Answer:

U = 102.8 J (100 J to two significant digits)

Explanation:

potential energy converted = 20(9.8)(1.8) = 352.8 J

kinetic energy at base of track = ½(20)5.0² = 250 J

energy (work) of friction 352.8 - 250 = 102.8 J

8 0
3 years ago
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