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SVEN [57.7K]
3 years ago
8

A 7.59-m-tall container is filled to the brim, partway with mercury and the rest of the way with water. The container is open to

the atmosphere. What must be the depth of the mercury so that the absolute pressure on the bottom of the container is twice the atmospheric pressure?
Physics
2 answers:
mixer [17]3 years ago
6 0

Answer:

1.0335 m

Explanation:

Atmospheric pressure, Po = 1.01 x 10^5 Pa

Let the depth of mercury is h

the depth of water is 7.59 - h

density of mercury = 13.6 x 10^3 kg/m^3

density of water = 1000 kg/m^3

Pressure due to mercury + pressure due to water = 2 x Po

height of mercury x density of mercury x g + height of water x density x g

                                        = 2Po

h x 13.6 x 1000 x g + (7.59 - h) x 1000 x g = 2 x 1.01 x 10^5

h x 13.6 x 9.8 + (7.59 - h) x 9.8 = 2 x 1.01 x 100

133.28 h + 74.382 - 9.8 h = 202

123.48 h = 127.618

h = 1.0335 m

Thus, the height of mercury column is 1.0335 m.

hoa [83]3 years ago
5 0

Answer:

0.22 m

Explanation:

P_{o} = Atmospheric pressure = 101325 Pa

P = Pressure at the bottom = tex]2 P_{o}[/tex] = 2 (101325) = 202650 Pa

h = height of the container = 7.59 m

x = depth of the mercury

Pressure at the bottom = Atmospheric pressure + Pressure due to mercury + Pressure due to water

P = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\2 P_{o} = P_{o} + \rho _{w} g (h - x) + \rho _{hg} g x\\P_{o} = \rho _{w} g (h - x) + \rho _{hg} g x\\101325 = (1000) (9.8) (7.59 - x) + (13600) (9.8) x\\x = 0.22 m

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expeople1 [14]

Answer:

<h3>The answer is 2.51 s</h3>

Explanation:

The time taken can be found by using the formula

t =  \frac{d}{v}  \\

d is the distance

v is the velocity

From the question we have

t =  \frac{1256}{500}  \\  = 2.512

We have the final answer as

<h3>2.51 s</h3>

Hope this helps you

4 0
2 years ago
An object in the shape of a thin ring has radius a and mass M. A uniform sphere with mass m and radius R is placed with its cent
madreJ [45]

Answer:

F = GMmx/[√(a² + x²)]³

Explanation:

The force dF on the mass element dm of the ring due to the sphere of mass, m at a distance L from the mass element is

dF = GmdM/L²

Since the ring is symmetrical, the vertical components of this force cancel out leaving the horizontal components to add.

So, the horizontal components add from two symmetrically opposite mass elements dM,

Thus, the horizontal component of the force is

dF' = dFcosФ where Ф is the angle between L and the x axis

dF' = GmdMcosФ/L²

L² = a² + x² where a = radius of ring and x = distance of axis of ring from sphere.

L = √(a² + x²)

cosФ = x/L

dF' = GmdMcosФ/L²

dF' = GmdMx/L³

dF' = GmdMx/[√(a² + x²)]³

Integrating both sides we have

∫dF' = ∫GmdMx/[√(a² + x²)]³

∫dF' = Gm∫dMx/[√(a² + x²)]³    ∫dM = M

F = GmMx/[√(a² + x²)]³  

F = GMmx/[√(a² + x²)]³

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F = GMmx/[√(a² + x²)]³

3 0
3 years ago
A 300-kg piano being held by a crane is accidentally dropped from a height of 15 meters. a. What is the speed of the piano just
FinnZ [79.3K]

Answer:

a) 17.16m/s

b) 44,145J

c) Sound the piano makes when hitting the ground, vibration of the ground, heat.

d) i) It's smaller due to the energy dissipated by the friction between air and the parachute.

ii) It stays the same, the only difference is that the dissipated energy is distributed between air resistance and the kinetic energy dissipated by the ground whent he piano hits it.

Explanation:

a)

In order to solve this problem we must start by doing a drawing of the situation, which will help us visualize the problem better. (See attached picture).

So, in this problem we can ignore air resistance so we can say that the energy is conserved, this is the total initial energy is the same as the total final energy, so we get that:

U_{0}+K_{0}=U_{f}+K_{f}

When the piano is released it has an initial speed of zero, so the initial kinetic energy is zero. When the piano hits the ground it will have a height of 0m, so the final potential energy is zero as well. This will simplify our equation:

U_{0}=K_{f}

We know that potential energy is given by the formula:

U=mgh

and kinetic energy is given by the formula:

K=\frac{1}{2}mv^{2}

which can be substituted in our equation:

mgh=\frac{1}{2}mv^{2}

we can divide both sides of the equation into the mass of the piano, so we get:

gh=\frac{1}{2}v^{2}

which can be solved for the final velocity which yields:

v=\sqrt{2gh}

we can now substitute the data provided by the problem so we get:

v=\sqrt{2(9.81m/s^{2})(15m)}

which yields:

v=17.16m/s

b)

Since energy is conserved, this means that the total dissipated energy will be the same as the potential energy, so we get that:

E=mgh

so

E=(300kg)(9.81m/s^{2})(15m)

which yields:

E=44,145J

c)

When the piano hits the ground, the kinetic energy it had will be transformed to other types of energy, mostly vibration and heat. The vibration will turn to sound due to the movement of air created by the piano itself and the ground. And heat is created by the friction between the molecules created by the vibrations and the collition itself. So some of the indicators of this release of energy could be:

-Sound

-Vibration

-Heat.

d)

i) The amount of inetic energy dissipated would decrease due to the friction between air and the parachute. Since air is resisting the movement of the piano, this will translate into a loss of energy, if we did an energy balance we would get that:

U_{0}=K_{f}+E_{p}

The total amount of energy is conserved but it will be distributed between the energy lost due to air resistance and the kinetic energy the piano has at the time it hits the ground.

ii) So the total amount of energy dissipated remains the same, the only difference is that it will be distributed between air resistance and the kinetic energy of the piano.

3 0
3 years ago
The charges of two particles are as follows: Q1=2 x 10 -8 C and Q2 = 3 x 10 -7 C. Find the magnitude of the force between these
Fantom [35]

Answer:

F = 3.86 x 10⁻⁶ N

Explanation:

First, we will find the distance between the two particles:

r = \sqrt{(x_{2}-x_{1})^2+(y_{2}-y_{1})^2+(z_{2}-z_{1})^2}\\

where,

r = distance between the particles = ?

(x₁, y₁, z₁) = (2, 5, 1)

(x₂, y₂, z₂) = (3, 2, 3)

Therefore,

r = \sqrt{(3-2)^2+(2-5)^2+(3-1)^2}\\r = 3.741\ m\\

Now, we will calculate the magnitude of the force between the charges by using Coulomb's Law:

F = \frac{kq_{1}q_{2}}{r^2}\\

where,

F = magnitude of force = ?

k = Coulomb's Constant = 9 x 10⁹ Nm²/C²

q₁ = magnitude of first charge = 2 x 10⁻⁸ C

q₂ = magnitude of second charge = 3 x 10⁻⁷ C

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Therefore,

F = \frac{(9\ x\ 10^9\ Nm^2/C^2)(2\ x\ 10^{-8}\ C)(3\ x\ 10^{-7}\ C)}{(3.741\ m)^2}\\

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3 years ago
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