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Kisachek [45]
3 years ago
10

What is the increase in pressure required to decrease volume of mercury by 0.001%

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

Answer:

Explanation:

Using Boyles law

Boyle's law states that, the volume of a given gas is inversely proportional to it's pressure, provided that temperature is constant

V ∝ 1 / P

V = k / P

VP = k

Then,

V_1 • P_1 = V_2 • P_2

So, if we want an increase in pressure that will decrease volume of mercury by 0.001%

Then, let initial volume be V_1 = V

New volume is V_2 = 0.001% of V

V_2 = 0.00001•V

Let initial pressure be P_1 = P

So,

Using the equation above

V_1•P_1 = V_2•P_2

V × P = 0.00001•V × P_2

Make P_2 subject of formula by dividing be 0.00001•V

P_2 = V × P / 0.00001 × V

Then,

P_2 = 100000 P

So, the new pressure has to be 10^5 times of the old pressure

Now, using bulk modulus

Bulk modulus of mercury=2.8x10¹⁰N/m²

bulk modulus = P/(-∆V/V)

-∆V = 0.001% of V

-∆V = 0.00001•V

-∆V = 10^-5•V

-∆V/V = 10^-5

Them,

Bulk modulus = P / (-∆V/V)

2.8 × 10^10 = P / 10^-5

P = 2.8 × 10^10 × 10^-5

P = 2.8 × 10^5 N/m²

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Aleonysh [2.5K]

Answer:  53.31\° East of North

Explanation:

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Now, we need to find the direction the bee should fly directly to the flower (due North):

sin \theta=\frac{Windspeed-from-East-to-West}{Speed-bee-relative-to-air}

sin \theta=\frac{6.68 m/s}{8.33 m/s}

Clearing \theta:

\theta=sin^{-1} (\frac{6.68 m/s}{8.33 m/s})

\theta=53.31\°

6 0
3 years ago
A new prototype cup has been designed to keep liquids, such as hot coffee or cold annk, near their original temperature for long
san4es73 [151]

Answer:

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If a dog walks north for 10 meters and then east for 10 meters, what is the direction of its displacement?
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The direction of its displacement wil be

c.northeast

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A tank of gasoline (n = 1.40) is open to the air (n = 1.00). A thin film of liquid floats on the gasoline and has a refractive i
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A normal walking speed is around 2.0 m/s . how much time t does it take the box to reach this speed if it has the acceleration 5
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Given:

u(initial velocity)=0

a=5.54m/s^2

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Where v is the final velocity.

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2=0+5.54t

t=2/5.54

t=0.36 sec


6 0
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