Answer:Final volume after pressure is applied=4,292cm3
Explanation:
Using the bulk modulus formulae
We have that The bulk modulus of waTer is given as
K =-V dP/dV
Where K, the bulk modulus of water = 2.15 x 10^9N/m^2
2.15 x 10^9N/m^2= - 4,300 x 4 × 106N/m2 / dV
dV = - 4,300 x 4 × 10^6N/m^2/ 2.15 x 10^9N/m^2
dV (change in volume)= -8.000cm^3
Final volume after pressure is applied,
V= V+ dV
V= 4300cm3 + (-8.000cm3)
=4300cm3 - 8.000cm3
Final Volume, V =4,292cm3
Planck find the correct curve for the specturm of light emitted by a hot object by vibrational energies of the atomic resonators were quantized.
<h3>Briefing :</h3>
- The energy density of a black body between λ and λ + dλ is the energy E=hc/λ of a mode times the density of states for photons, times the probability that the mode is occupied.
- This is Planck's renowned equation for a black body's energy density.
- According to this, electromagnetic radiation from heated bodies emits in discrete energy units or quanta, the size of which depends on a fundamental physical constant (Planck's constant). The basis of infrared imaging is the correlation between spectral emissivity, temperature, and radiant energy, which is made possible by Planck's equation.
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This is for the reason that individuals are not continually taking a gander at precisely the same, and on the grounds that individuals' psyches of ten work distinctively and process data in marginally extraordinary ways getting diverse understandings of similar information.
Answer:
54.6°
Explanation:
From law of reflection i=r.
So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.
Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.
If you consider triangle AOB, one angle is ∠AOB=90°
and ∠OAB is 54.6°
From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°
So, the second incident angle will be 54.6°
Hence, the second reflected angle will be 54.6 degrees.
The force of the tripped catch exerted on the 2.5 Kg ball moving at 8.5 m/s to the Left is 160 N
<h3>Data obtained from the question </h3>
- Initial velocity (u) = 8.5 m/s
- Final velocity (v) = 7.5 m/s
- Time (t) = 5 ms = 0.25 s
- Mass (m) = 2.5 Kg
- Force (F) = ?
<h3>How to determine the force</h3>
The force exerted on the ball can be obtained as follow:
F = m(v + u) / t
F = [2.5(7.5 + 8.5)]/ 0.25
F = 40 / 0.25
F = 160 N
Thus, the force exerted on the ball is 160 N
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