answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
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

You might be interested in
By examining a wave's pattern after it interacts with a barrier or gap, measurements can be made to better understand certain wa
rjkz [21]

Sample Response: How does diffraction occur?

8 0
2 years ago
Read 2 more answers
Military specifications often call for electronic devices to be able to withstand accelerations of 10 g. to make sure that their
trapecia [35]
The solution for this problem is:
(10 x 9.8) = 98.1 m/sec^2 acceleration. Time, to travel 9.4cm or (.094m.), at acceleration of 98m/sec^2

= sqrt(2d/a), = sqrt (98.1 m/sec^2/0.094m) = 32.3050619 sec per cycle

Frequency = (w/2pi), = 32.3050619/2pi
= 32.3050619/6.28318531
= 5.14 Hz would be the answer
6 0
2 years ago
In the system shown above, the pulley is a uniform disk with a mass of .75 kg and a radius of 6.5 cm. The coefficient of frictio
lord [1]

Answer:

i am answering the same question 3rd time

please find the answer in the images attached.

5 0
1 year ago
Consider the waveform expression. y (x, t) = ym sin (0.333x + 5.36 + 585t) The transverse displacement (y) of a wave is given as
Sonja [21]

Explanation:

The waveform expression is given by :

y(x,t)=y_m\ sin(0.333x+5.36+585t)...........(1)

Where

y is the position

t is the time in seconds

The general waveform equation is given by :

y(x,t)=y_m\ sin(kx+\phi+\omega t)..........(2)

Where

k=\dfrac{2\pi}{\lambda}

\omega=2\pi f

On comparing equation (1) and (2) we get :

0.333=\dfrac{2\pi}{\lambda}

\lambda=18.86\ m

585=2\pi f

f = 93.10 Hz

Time period, T=\dfrac{1}{f}

T=\dfrac{1}{0.010}

T = 0.010 s

Phase constant, \phi=5.36\ radian

Hence, this is the required solution.

8 0
2 years ago
A certain alarm clock ticks four times each second, with each tick representing half a period. The balance wheel consists of a t
Semenov [28]

Answer:

a. I=2.77x10^{-8} kg*m^2

b. K=4.37 x10^{-6} N*m

Explanation:

The inertia can be find using

a.

I = m*r^2

m = 0.95 g * \frac{1 kg}{1000g}=9.5x10^{-4} kg

r=0.54 cm * \frac{1m}{100cm} =5.4x10^{-3}m

I = 9.5x10^{-4}kg*(5.4x10^{-3}m)^2

I=2.77x10^{-8} kg*m^2

now to find the torsion constant can use knowing the period of the balance

b.

T=0.5 s

T=2\pi *\sqrt{\frac{I}{K}}

Solve to K'

K = \frac{4\pi^2* I}{T^2}=\frac{4\pi^2*2.7702 kg*m^2}{(0.5s)^2}

K=4.37 x10^{-6} N*m

3 0
1 year ago
Other questions:
  • What is a limitation of the electron cloud model theory that a law about electrons would not have?
    11·2 answers
  • What is the value of the composite constant (gmer2e), to be multiplied by the mass of the object mo in the equation above? expre
    9·2 answers
  • A 1.7-kg block of wood rests on a rough surface. A 0.011-kg bullet strikes the block with a speed of 670 m/s and embeds itself.
    5·1 answer
  • Jack and Jill are maneuvering a 3100 kg boat near a dock. Initially the boat's position is < 2, 0, 3 > m and its speed is
    9·1 answer
  • John is running down the street and hears dogs barking in the distance. How do the sound waves change as John approaches the bar
    12·1 answer
  • A 3.00-kg ball swings rapidly in a complete vertical circle of radius 2.00 m by a light string that is fixed at one end. The bal
    5·1 answer
  • A 0.25 kg ideal harmonic oscillator has a total mechanical energy of 9.8 J. If the oscillation amplitude is 20.0 cm, what is the
    12·1 answer
  • A finite rod of length L has total charge q, distributed uniformly along its length. The rod lies on the x -axis and is centered
    5·1 answer
  • 7. Imagine you are pushing a 15 kg cart full of 25 kg of bottled water up a 10o ramp. If the coefficient of friction is 0.02, wh
    8·1 answer
  • The Bohr model pictures a hydrogen atom in its ground state as a proton and an electron separated by the distance a0 = 0.529 × 1
    15·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!