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Tatiana [17]
2 years ago
11

The atmosphere pressure can support mercury in a tube, which the upper end is closed, up to 0.76 meter. If the mercury is replac

ed by pure water in this case, what is the maximum height the atmosphere pressure can support the water in such kind a tube? (The density of mercury is about 13.6 times larger than that of water).
Physics
1 answer:
Leni [432]2 years ago
6 0

Answer:

Maximum height the atmosphere pressure can support the

water=10.336 m

Explanation:

We know that ,

Pressure = h\cdot\rho\cdot g

Case 1 - Mercury in the tube

Density\ of\ mercury =\rho_1\\and\ height\ attained\ for\ mercury\ column = h_1

Case 2 - Water in the tube

Density\ of\ water =\rho_2\\and\ height\ attained\ for\ water\ column = h_2

Since atmospheric pressure is same

.P=h_1\cdot\rho_1\cdot g = h_2\cdot\rho_2\cdot g

or,  h_2=\frac{h_1\rho_1}{\rho_2}

Given\ h_1= 0.76\  m,\rho_1=13.6\cdot\rho_2

∴ h_2=0.76\cdot13.6=10.336\ m

Hence height of the water column =10.336 m

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Mariulka [41]

Answer: d = 4750n/3.1+95n

Explanation:

Using the principle of moment to solve the question.

Sum of clockwise moments = sum of anti clockwise moments

Since there are n identical coins with mass 3.1g placed at point 0cm, 1 coin will have mass of 3.1/n grams

Taking moment about the pivot,

Mass 3.1/n grams will move anti-clockwisely while the mass 95g will move in the clockwise direction.

Since its a meter rule (100cm) the distance from the center mass(95g) to the pivot will be 50-d (check attachment for diagram).

To get 'd'

We have 3.1/n × d = 95 × (50-d)

3.1d/n = 4750-95d

3.1d = 4750n-95dn

3.1d+95dn=4750n

d(3.1+95n) = 4750n

d = 4750n/3.1+95n

6 0
2 years ago
A mercury atom in the ground state absorbs 20.00 electronvolts of energy and is ionized by losing an electron. How much kinetic
sukhopar [10]
We are given a mercury atom in the ground state which absorbs 20 eV of energy. It is then ionized by losing an electron. We need to calculate the kinetic energy that the electron has after ionization.

The initial energy is 20 eV = 20 J/C
The electron charge is = 1.60217662 × 10-19<span> coulombs

To determine the kinetic energy, we can use this equation:

KE = 20 Joules / Coulombs * </span>1.60217662 × 10-19<span> coulombs 
KE = 1.25x10^20 Joules 

Therefore, the amount of kinetic energy that the electron has after ionization is </span>1.25x10^20 Joules or 1.25x10^17 kJ. <span />
8 0
2 years ago
The value of specific heat for copper is 390 J/kg⋅C∘, for aluminun is 900 J/kg⋅C∘, and for water is 4186 J/kg⋅C∘.
abruzzese [7]

Answer:

The equilibrium temperature is

21.97°c

Explanation:

This problem bothers on the heat capacity of materials

Given data

specific heat capacities

copper is Cc =390 J/kg⋅C∘,

aluminun Ca = 900 J/kg⋅C∘,

water Cw = 4186 J/kg⋅C∘.

Mass of substances

Copper Mc = 235g

Aluminum Ma = 135g

Water Mw = 825g

Temperatures

Copper θc = 255°c

Water and aluminum calorimeter θ1= 16°c

Equilibrium temperature θf =?

Applying the principle of conservation of heat energy, heat loss by copper equal heat gained by aluminum calorimeter and water

McCc(θc-θf) =(MaCa+MwCw)(θf-θ1)

Substituting our data into the expression we have

235*390(255-θf)=

(135*900+825*4186)(θf-16)

91650(255-θf)=(3574950)(θf-16)

23.37*10^6-91650*θf=3.57*10^6θf- +57.2*10^6

Collecting like terms and rearranging

23.37*10^6+57.2*10^6=3.57*10^6θf+91650θf

8.2*10^6=3.66*10^6θf

θf=80.5*10^6/3.6*10^6

θf =21.97°c

5 0
2 years ago
A mass weighing 20 pounds stretches a spring 6 inches. The mass is initially released from rest from a point 6 inches below the
hoa [83]

Answer:

Since the spring mass system will execute simple harmonic motion the position as a function of time can be written asx(t)=Asin(\omega t+\phi)

'A' is the amplitude = 6 inches (given)

\omega =\sqrt{\frac{k}{m}} is the natural frequency of the system

At equilibrium we have

mg=kx\\\\k=\frac{mg}{x}

Applying values we get

k=40 lb/ft

thus natural frequency equals

\omega =\sqrt{\frac{40}{\frac{20}{32}}}\\\\\omega =8s^{-1}

Thus the equation of motion becomes

x(t)=6sin(8t+\phi)

At time t=0 since mass is at it's maximum position thus we have

A=Asin(\omega t+\phi)\\\\\therefore sin(\omega\times 0+\phi)=1\\\\\phi=\frac{\pi}{2}\\\\\therefore x(t)=Asin(\omega t+\frac{\pi}{2})

Thus the position of mass at the given times is as follows

1) at \frac{\pi}{12} x(t)=5.99inches

2) at \frac{\pi}{8} x(t)=5.9909inches

3) at \frac{\pi}{6} x(t)=5.98397inches

4) at \frac{\pi}{4} x(t)=5.9639inches

5) at \frac{9\pi}{32} x(t)=5.954inches

4 0
1 year ago
You are in a hot-air balloon that, relative to the ground, has a veloc- ity of 6.0 m/s in a direction due east. You see a hawk m
Ann [662]

Answer:

6.32 m/s 18.43° northeast

Explanation:

We express the velocity of hawk as:

v_{Hawk}=v_{balloon}+v_{HawkRelativetoBalloon}=6 x+2 y

We consider positive x towards east and positive y due north. So the magnitude is simply the square root of the square components:

|v_{hawk}|=\sqrt[]{6^2+2^2}=\sqrt{40}≈6.32 m/s

And the angle with respect to the east should be with:

arctan(\frac{2}{6} )=18.43 \°

8 0
2 years ago
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