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adoni [48]
3 years ago
13

Rank the pressures from highest to lowest:

Physics
1 answer:
n200080 [17]3 years ago
3 0

Answer:

P₃ > P₁ > P₂

Explanation:

To rank pressure of the given situation

a) we know

  Pressure at height h below

      P = ρ g h

density of salt water, ρ = 1029 kg/m³

      P₁ = 1029 x 10 x 0.2

      P₁ = 2058 Pa

b) density of fresh water, ρ = 1000 kg/m³

      P₂ = 1000 x 10 x 0.2

      P₂ = 2000 Pa

c) density of mercury, ρ = 13593 kg/m³

      P₃ = 13593 x 10 x 0.05

      P₃ = 6796.5 Pa

Rank of Pressures from highest to lowest

        P₃ > P₁ > P₂

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What are possible formulas for impulse? Check all that apply. J = Fdeltat J = StartFraction force over change in time EndFractio
Alex

<u>The possible formulas for impulse are as follows:</u>

J = FΔt

J = mΔv

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Answer: Option  A, E and F

<u>Explanation:</u>

The quantity which explains the consequences of a overall force acting on an object (moving force) is known as impulse. It is symbolised as J. When the average overall force acting on an object than such products are formed and in given duration than the start fraction force over change in time end fraction J = FΔt.

The impulse-momentum theorem explains that the variation in momentum of an object is same as the impulse applied to it: J = Δp J = mΔv if mass is constant J = m dv + v dm if mass changes. Logically, the impulse-momentum theorem is equivalent to Newton second laws of motion which is also called as force law.

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A motorcycle accelerates from at a rate of 4m/s 2 while traveling 60m what it’s the motorcycles velocity at the end of this moti
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1.)A tank travels at a rate of 10.0 km/hr for 12.00 minutes, then at 15.0 km/hr for 8.00
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12.00 min = 0.2 hr

8.00 min = 0.15 hr

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Change in position:

(10.0 km/hr) (0.2 hr) + (15.0 km/hr) (0.15 hr) - (20.0 km/hr) (0.2 hr)

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An important diagnostic tool for heart disease is the pressure difference between blood pressure in the heart and in the aorta l
butalik [34]

Answer:

a)   f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} } , b)   Δf = 2 f₀ \frac{v}{c}

Explanation:

a) This is a Doppler effect exercise, which we must solve in two parts in the first the emitter is fixed and in the second when the sound is reflected the emitter is mobile.

Let's look for the frequency (f ’) that the mobile aorta receives, the blood is leaving the aorta or is moving towards the source

                    f ’= fo\frac{c+v}{c}

This sound wave is reflected by the blood that becomes the emitter, mobile and the receiver is fixed.

                   f ’’ = f’ \frac{c}{ c-v}

where c represents the sound velocity in stationary blood

therefore the received frequency is

                 f ’’ = f₀   \frac{c}{c-v}

let's simplify the expression

                f ’’ = f₀ \frac{c+v}{c-v}

                f ’’ = f₀ \frac{1 + \frac{v}{c} }{1- \frac{v}{c} }

         

b) At the low speed limit v <c, we can expand the quantity

                 (1 -x)ⁿ = 1 - x + n (n-1) x² + ...

                 ( 1- \frac{v}{c} ) ^{-1} = 1 + \frac{v}{c}

 

                f ’’ = fo ( 1+ \frac{v}{c}) ( 1 + \frac{v}{c} )

                f ’’ = fo ( 1 + 2 \frac{v}{c} + \frac{v^2}{ c^2} )

leave the linear term

               f ’’ = f₀ + f₀ 2\frac{v}{c}

the sound difference

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               Δf = 2 f₀ \frac{v}{c}

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