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Anna71 [15]
3 years ago
8

The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10

5 N/m2. If this volume of liquid is now
placed at a depth where the pressure is P = 2.30 x 10' N/m², what will be the change in volume of the liquid (in mº)? The bulk modulus of the liquid is 8.0 x 100 N/m².
(Include the appropriate sign with your answer.)
m
Additional Materials
Physics
1 answer:
Alexus [3.1K]3 years ago
3 0

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

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A load of 800 N is lifted using a block and tackle having 5 pulleys. If the applied effort is 200 N, calculate
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Explanation:

Load=800N

Effort=200N

1. Mechanical Advantage = LOAD/EFFORT

= 800N/200N

= 4

2 Velocity Ratio = no. Of pulleys =5

3. Efficiency = Mechanical advantage / velocity ratio × 100%

= (4/5)×100%

=80%

4. output work= load×load distance

= 800N × 5m

= 4 × 1000J

5. Efficiency = (output work/input work) ×100%

Or, 80% = (4000J/input work) ×100%

Or, 80%/100% = 4000J/inputwork

Or, 4/5 = 4000J/inputwork

Or, input work =4000J × 5/4

Input work = 5×1000J

I hope it helped! ;-)

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3 years ago
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What is the role of MnO2 in a dry cell?​
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Explanation:

MNO2 is depolarizer in a dry cell. depolarizers are chemicals used to reduce or prevent polarization of cells

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In what way is the structure of the Milky Way galaxy similar to the structure of the solar system?
MissTica
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Consider Compton Scattering with visible light.A photon with wavelength 500nm scatters backward(theta=180degree) from a free ele
JulijaS [17]

Answer: 4.86(10)^{-12}m

Explanation:

The Compton Shift \Delta \lambda in wavelength when photons are scattered is given by the following equation:

\Delta \lambda=\lambda' - \lambda_{o}=\lambda_{c}(1-cos\theta) (1)  

Where:  

\lambda'=500 nm=500(10)^{-9} m is the wavelength of the scattered photon

\lambda_{o}  is the wavelength of the incident photon

\lambda_{c}=2.43(10)^{-12} m is a constant whose value is given by \frac{h}{m_{e}.c}, being h=4.136(10)^{-15}eV.s the Planck constant, m_{e} the mass of the electron and c=3(10)^{8}m/s the speed of light in vacuum.  

\theta=180\° the angle between incident phhoton and the scatered photon.  

\Delta \lambda=2.43(10)^{-12} m (1-cos(180\°)) (2)

\Delta \lambda=4.86(10)^{-12}m (3)  This is the shift in wavelength

5 0
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lora16 [44]

Answer:

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