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kirill [66]
3 years ago
7

g The bottom end of a long vertical tube filled with liquid is opened in a basin exposed to air having pressure 100.8 kilo-Pasca

ls. The top end of the tube remains closed. The maximum height of liquid in the tube that this air pressure can support is 10.5 meters. Determine the density of the liquid in units of bevelled fraction numerator k g over denominator m cubed end fraction.
Physics
1 answer:
Masja [62]3 years ago
3 0

Answer:

979.6 kg/m³

Explanation:

We know pressure P = hρg where h = height of liquid = 10.5 m, ρ = density of liquid and g = acceleration due to gravity = 9.8 m/s²

So, density ρ = P/hg

Since P = 100.8 kPa = 100.8 × 10³ Pa

substituting the values of the variables into the equation for ρ, we have

ρ = P/hg

= 100.8 × 10³ Pa ÷ (10.5 m × 9.8 m/s²)

= 100.8 × 10³ Pa ÷ 102.9 m²/s²

= 0.9796 × 10³ kg/m³

= 979.6 kg/m³

So, the density of the liquid is 979.6 kg/m³

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You place a point charge q = -4.00 nC a distance of 9.00 cm from an infinitely long, thin wire that has linear charge density 3.
valentinak56 [21]

Answer:

F=6\times 10^{-7}\ N

Explanation:

Given:

  • quantity of point charge, q=-4\times 10^{-9}\ C
  • radial distance from the linear charge, r=0.09\ m
  • linear charge density, \lambda=3\times 10^{-9}\ C.m^{-1}

<u>We know that the electric field by the linear charge  is given as:</u>

E=\frac{\lambda}{2\pi.\epsilon_0.r}

E=\frac{1}{2}\times 9\times 10^9\times \frac{3\times10^{-9}}{0.09}

E=150\ N.C^{-1}

<u>Now the force on the given charge can be given as:</u>

F=E.q

F=150\times 4\times 10^{-9}

F=6\times 10^{-7}\ N

3 0
3 years ago
A gas at a pressure of 2.10 atm undergoes a quasi static isobaric expansion from 3.70 to 5.40 L. How much work is done by the ga
BartSMP [9]

Answer:

Total work done in expansion will be 3.60\times 10^5J

Explanation:

We have given pressure P = 2.10 atm

We know that 1 atm =1.01\times 10^5Pa

So 2.10 atm =2.10\times 1.01\times 10^5=2.121\times 10^5Pa

Volume is increases from 3370 liter to 5.40 liter

So initial volume V_1=3.70liter

And final volume V_2=5.40liter

So change in volume dV=5.40-3.70=1.70liter

For isobaric process work done is equal to W=PdV=2.121\times 10^5\times 1.70=3.60\times 10^5J

So total work done in expansion will be 3.60\times 10^5J

5 0
3 years ago
What is the weight on the surface of Earth of an object of mass 2.00 kilograms?
Anni [7]

Answer:

D

Explanation:

19.6 newtons

A 2.00-kilogram object weighs 19.6 newtons on Earth.

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Answer:

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Please help don't for get to show work
emmasim [6.3K]
-- Bob covered a distance of (32m + 45m) = 77 meters.

-- His displacement is the straight-line distance and direction
from his starting point to his ending point.

The straight-line distance is

D = √(32² + 45²)
D = √(1,024 + 2,025)
D = √3,049 = 55.22 meters

The direction is the angle whose tangent is (32/45) south of east.

tan⁻¹(32/45) = tan⁻¹(0.7111...) = 35.42° south of east.
3 0
3 years ago
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