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Sliva [168]
3 years ago
6

The bulk modulus of water is B = 2.2 x 109 N/m2. What change in pressure ΔP (in atmospheres) is required to keep water from expa

nding when it is heated from 10.9 °C to 40.0 °C?
Physics
1 answer:
Aneli [31]3 years ago
6 0

Answer:

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

Explanation:

The bulk modulus of water (B), in newtons per square meters, can be estimated by means of the following model:

B = \rho_{o}\cdot \frac{\Delta P}{\rho_{f} - \rho_{o}} (1)

Where:

\rho_{o} - Water density at 10.9 °C, in kilograms per cubic meter.

\rho_{f} - Water density at 40 °C, in kilograms per cubic meter.

\Delta P - Pressure change, in pascals.

If we know that \rho_{o} = 999.623\,\frac{kg}{m^{3}}, \rho_{f} = 992.219\,\frac{kg}{m^{3}} and B = 2.2\times 10^{9}\,\frac{N}{m^{2}}, then the bulk modulus of water is:

\Delta P = B\cdot \left(\frac{\rho_{f}}{\rho_{o}}-1 \right)

\Delta P = \left(2.2\times 10^{9}\,\frac{N}{m^{3}} \right)\cdot \left(\frac{992.219\,\frac{kg}{m^{3}} }{999.623\,\frac{kg}{m^{3}} }-1 \right)

\Delta P = -16294943.19\,Pa \,(-160.819\,atm)

A change of 160.819 atmospheres is required to keep water from expanding when it is heated from 10.9 °C to 40.0 °C.

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lys-0071 [83]

Answer:

9.62 minutes 0r 0.16 of an hour

Explanation:

Speed = distance/time

300mph = 48.1 m/t

xt

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Hope this helps!

8 0
2 years ago
A friend tells you that a lunar eclipse will take place the following week, and invites you to join him to observe the eclipse t
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Answer:

y = 80.2 mille

Explanation:

The minimum size of an object that can be seen is determined by the diffraction phenomenon, if we use the Rayleigh criterion that establishes that two objects can be distinguished without the maximum diffraction of a body coincides with the minimum of the other body, therefore so much for the pupil of the eye that it is a circular opening

          θ = 1.22 λ/ d

in a normal eye the diameter of the pupils of d = 2 mm = 0.002 m, suppose the wavelength of maximum sensitivity of the eye λ = 550 nm = 550 10⁻⁹ m

         θ = 1.22 550 10⁻⁹ / 0.002

         θ = 3.355 10⁻⁴ rad

Let's use trigonometry to find the distance supported by this angle, the distance from the moon to the Earth is L = 238900 mille = 2.38900 10⁵ mi

       tan θ = y / L

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3 years ago
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4 0
3 years ago
A wheel has a radius of r = 2.0 m and it rolls down a smooth incline. The height of the incline is h = 8.0 m . What is the angul
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The angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

The angular velocity (ω) of an object is the rate at which the object's angle position is changing in relation to time.

For a wheel attached to an incline angle, the angular velocity can be computed by considering the conservation of energy theorem.

As such the total kinetic energy (K.E) and rotational kinetic energy (R.K.E) at a point is equal to the total potential energy (P.E) at the other point.

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\mathbf{gh = \dfrac{1}{2}(r \times \omega)^2 + \dfrac{1}{2}\times r^2 \times \omega^2}

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\mathbf{\omega=4.429 \ rad/sec}

Therefore, we can conclude that the angular velocity of the wheel at the bottom of the incline is 4.429 rad/sec

Learn more about angular velocity here:

brainly.com/question/1452612

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