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Doss [256]
3 years ago
14

A container is filled to a depth of 22.0 cm with water. On top of the water floats a 28.0-cm-thick layer of oil with specific gr

avity 0.500. What is the absolute pressure at the bottom of the container? 1 Pa
Physics
1 answer:
Ilya [14]3 years ago
5 0

Answer:104.853 kPa

Explanation:

Given

Depth of water h_2=22 cm

on the top of water a layer of h_1=28 cm oil layer is present

specific gravity of oil \rho _o=0.5\rho _w

Assuming Atmospheric Pressure to be 101.325 KPa

We know Pressure due to Pressure Difference is given by

\Delta P=\rho \times g\times h

\Delta P=\rho _0\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\rho _w\times g\times h_1+\rho _w\times g\times h_2

\Delta P=0.5\times 1000 \times 9.8\times 0.28+1000\times 9.8\times 0.22

\Delta P=1.372\times 10^3+2.156\times 10^3

\Delta P=3.528 kPa

Absolute\ Pressure=Atmospheric\ Pressure+\Delta P=101.325+3.528

Absolute\ Pressure =101.325+3.528=104.853 kPa

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A 10-kg projectile is fired straight up with an initial velocity of 500 m/s. (a) What is the projectile’s potential energy at th
Naily [24]
(a) the initial kinetic energy of the projectile is equal to:
K_i= \frac{1}{2}mv^2= \frac{1}{2}(10 kg)(500 m/s)^2=1.25 \cdot 10^6 J
The projectile is fired straight up, so at the top of its trajectory, its velocity is zero; this means that it has no kinetic energy left, so for the law of conservation of energy, all its energy has converted into potential energy, which is equal to
U_f=K_i= 1.25 \cdot 10^6 J

b) If the projectile is fired with an angle of 45^{\circ}, its velocity has 2 components, one in the x-direction and one in the y-direction:
v_x = v_0 \cos 45^{\circ} =(500 m/s) \cos 45^{\circ} =353.6 m/s
v_y = v_0 \sin 45^{\circ} = (500 m/s)(\sin 45^{\circ} )=353.6 m/s

This means that at the top of its trajectory, only the vertical velocity will be zero (because the horizontal velocity is constant, since the motion on the x-axis is a uniform motion). Therefore, at the top of the trajectory, the projectile will have some kinetic energy left:
K_f =  \frac{1}{2}m v_x^2 = \frac{1}{2} (10kg)(353.6 m/s)^2=6.25 \cdot 10^5 J
For the conservation of energy, the initial energy mechanical energy must be equal to the mechanical energy at the highest point:
K_i = K_f + U_f
the initial kinetic energy is the same as point a), so we can re-arrange this equation to find the new potential energy at the top of the trajectory:
U_f = K_i - K_f = 1.25 \cdot 10^6 J - 6.25 \cdot 10^5 J = 6.25 \cdot 10^5 J
7 0
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