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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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3 years ago
Locate the complete verbal phrase and identify its type.
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7 0
3 years ago
Read 2 more answers
A woman can row her canoe at 2.5 m/s. If she faces an opposing current of 3.0 m/s, how fast will she go forward?
taurus [48]

Answer:

V = - 0.5 [m/s]

Explanation:

In order to solve this problem, we must use the principle of relative speeds. This is for an observer who is on the edge of the river he can see how the river moves to the left and the woman tries to move to the right but can not since:

V_{total}=-3+2.5\\V_{total}=-0.5 [m/s]

That is, the person sees how the woman moves to the left but with avelocity of 0.5 [m/s] to the left

8 0
2 years ago
An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F
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Answer

given,

mass of jogger  = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed  = 1.3 m/s  in  61 ° north of east.

jogger one

P_1 = m_1 v_1 \hat{i}

P_1 = 67 \times 2.3\hat{i}

P_1 = 154.1 \hat{i}

P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}

P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}

P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}

P_2 = 44.12\hat{i} +79.59\hat{j}

now

P = P₁ + P₂

P = 198.22 \hat{i} +79.59 \hat{j}

magnitude

P = \sqrt{198.22^2 + 79.59^2}

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\theta = tan^{-1}\dfrac{79.59}{198.22}

\theta = 21.87

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7 0
3 years ago
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IrinaVladis [17]

Answer:

176.58 m

Explanation:

t = Time taken = 6 seconds

u = Initial velocity = 0

v = Final velocity

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Equation of motion

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The object travels 176.58 m from the cliff in 6 seconds.

8 0
3 years ago
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