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VLD [36.1K]
3 years ago
15

Suppose a scoentist was able to construct a barometer with a liquid being denser than mercury , then how high would the liquid r

aise at standard pressure
Physics
1 answer:
ki77a [65]3 years ago
7 0

Answer:

the liquid has less height than the mercury

      h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

Explanation:

The pressure as a function of the height is given by

        P = ρ g h

where ρ is the density of the liquid, g the acceleration of gravity and h the height reached by the column of the liquid

In that case they say that the pressure is the standard one that is P = 1.01 10⁵ Pa = 760 mmHg

The first way to give the pressure is in SI units and the second way is the height that the mercury column reaches

In the case of building a barometer with a liquid that has a density greater than that of mercury

            ρ_liquid > ρ_Hg

             

the pressure

              P =ρ_lquid g h_liquid

if we have the same pressure

            ρ_{Hg} g h_{Hg} = ρ_{liquid}  g h_{liquid}

            h_{ liquid} = \frac{\rho_{Hg} }{\rho_{liqid}}  \  h_{Hg}

therefore the liquid has less height than the mercury

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Answer: The value of x is -6.

Explanation:

To calculate the value of 'x', we need to solve each function happening inthe equation.

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6x-8x=12\\\\-2x=12\\\\x=\frac{12}{-2}\\\\x=-6

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A car is traveling at 15m/s on a horizontal road. the brakes are applied and the car skids to a stop in 4.0s . the coefficient o
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Option A) .38 is the correct answer

Explanation:

Given that;

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acceleration  a = ?

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the negative sign tells us that its a  deacceleration so the sign can be ignored.

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<img src="https://tex.z-dn.net/?f=%5Chuge%20%5Csf%E0%BC%86%20%5C%3A%20%20question%20%20%5C%3A%20%E0%BC%84" id="TexFormula1" titl
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The man is standing on the plank. Then he is pulling the rope towards himself. He pushes the plank forward with the legs so that he can pull the rope backwards.

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The man is standing on the plank. Then he is pulling the rope towards himself. He pushes the plank forward with the legs so that he can pull the rope backwards. So the friction f acts in the forward direction for the plank. Also f acts in the opposite direction on the man. The tension T in the rope is 100 N (given). Let the friction force = f Newtons.Let the common acceleration = a m/s^2

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<em>net force = T - f = m a = 50 a </em>

<em>Plank: net force = T + f = m a = 100 a</em>

<em>net force = T - f = m a = 50 a </em>

<em>Plank: net force = T + f = m a = 100 a</em>

<em>As T = 100 N, a = 4/3 m/s^2 and f = 100/3 Newtons. </em>

[correct me if I'm wrong]:)

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