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PilotLPTM [1.2K]
10 months ago
10

True or FalseIf an object is submerged in water, on earth, at a depth of 3km, the Pressure on the object is2.94 x 10^7 Pa

Physics
1 answer:
zhenek [66]10 months ago
6 0

The pressure of a submerged object in a fluid is given by:

P=\rho gh+P_{atm}

where ρ is the density of the fluis, g is the acceleration of gravity, h is the depth of the object and Patm is the pressure of the atmosphere. In this case we know that:

• The density of water is 1000 kg/m^3

,

• The acceleration of gravity is 9.8 m/s^2

,

• The depth of the object is 3 km, that is, 3000 m.

,

• The atmospheric pressure is 101325 pascals.

Plugging these values in the equation given above we have:

\begin{gathered} P=(1000)(9.8)(3000)+101325 \\ P=2.95\times10^7 \end{gathered}

Therefore, the pressure at this depth is 2.95x10^7 Pa.

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A helium nucleus (charge = 2e, mass = 6.63 10-27 kg) traveling at 6.20 105 m/s enters an electric field, traveling from point ci
MA_775_DIABLO [31]

Answer:

v_B=3.78\times 10^5\ m/s

Explanation:

It is given that,

Charge on helium nucleus is 2e and its mass is 6.63\times 10^{-27}\ kg

Speed of nucleus at A is v_A=6.2\times 10^5\ m/s

Potential at point A, V_A=1.5\times 10^3\ V

Potential at point B, V_B=4\times 10^3\ V

We need to find the speed at point B on the circle. It is based on the concept of conservation of energy such that :

increase in kinetic energy = increase in potential×charge

\dfrac{1}{2}m(v_A^2-v_B^2)=(V_B-V_A)q\\\\\dfrac{1}{2}m(v_A^2-v_B^2)={(4\times 10^3-1.5\times 10^3)}\times 2\times 1.6\times 10^{-19}=8\times 10^{-16}\\\\v_A^2-v_B^2=\dfrac{2\times 8\times 10^{-16}}{6.63\times 10^{-27}}\\\\v_A^2-v_B^2=2.41\times 10^{11}\\\\v_B^2=(6.2\times 10^5)^2-2.41\times 10^{11}\\\\v_B=3.78\times 10^5\ m/s

So, the speed at point B is 3.78\times 10^5\ m/s.

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