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spin [16.1K]
2 years ago
7

On planet X, the absolute pressure at a depth of 2.00 m below the surface of a liquid nitrogen lake is 5.00 × 105 N/m2 . At a de

pth 5.00 m below the surface, the absolute pressure is 8.00 × 105 N/m2 . The density of liquid nitrogen is 808 kg/m3 . a. What is the atmospheric pressure on planet X? b. What is the acceleration due to gravity on planet X?
Physics
1 answer:
eimsori [14]2 years ago
7 0

Answer:

300000.01008 Pa

123.76237 m/s²

Explanation:

\rho = Density of liquid nitrogen = 808 kg/m³

h = Depth

g = Acceleration due to gravity

P = Atmospheric pressure

Absolute Pressure is given by

Below 2 m from surface

P_a=P+\rho gh\\\Rightarrow 5\times 10^5=P+808g\times 2

Below 5 m from surface

P_a=P+\rho gh\\\Rightarrow 8\times 10^5=P+808g\times 5

Subtracting the above equations we get

-3\times 10^5=-808g3\\\Rightarrow g=\frac{3\times 10^5}{808\times 3}\\\Rightarrow g=123.76237\ m/s^2

The acceleration due to gravity on the planet is 123.76237 m/s²

Equating the value of g in the first equation

5\times 10^5=P+808\times 123.76237\times 2\\\Rightarrow P=5\times 10^5-808\times 123.76237\times 2\\\Rightarrow P=300000.01008\ Pa

The atmospheric pressure on the planet is 300000.01008 Pa

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Answer:

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Explanation:

Using Lambert's law

Let dI / dt = kI, where k is a proportionality constant, I is intensity of incident light and t is thickness of the medium

then dI / I = kdt

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ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

k = ln(0.25)/3

k = -1.386/3

k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
2 years ago
A battery charges a parallel-plate capacitor fully and then is removed. The plates are then slowly pulled apart. What happens to
White raven [17]

Answer:

<h2>The potential difference increases </h2>

Explanation:

from the relation E= \frac{V}{d}

where E= electric field (force per coulomb)

            V= voltage

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Hence the voltage is going to be V= E×d.

Therefore this means that increasing the distance increases the voltage.

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2 years ago
Two small diameter, 10gm dielectric balls can slide freely on a vertical channel each carry a negative charge of 1microcoulomb.
dimulka [17.4K]

Answer:

The distance of separation is d = 0.092 \ m

Explanation:

The mass of the each ball is  m= 10 g  =  0.01 \ kg

 The negative charge on each ball is q_1 =q_2=q =  1 \mu C  =  1 *10^{-6} \ C

Now we are told that the lower ball is  restrained from moving this implies that the net force acting on it is  zero

Hence the gravitational force acting on the lower ball is equivalent to the electrostatic force i.e

          F =  \frac{kq_1 * q_2}{d}

=>       m* g  =  \frac{kq_1 * q_2}{d}

here k the the coulomb's  constant with a value  k = 9*10^{9} \ kg\cdot m^3\cdot s^{-4}\cdot A^2.

So  

      0.01 * 9.8  =  \frac{ 9*10^9 *[1*10^{-6} * 1*10^{-6}]}{d}

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5 0
2 years ago
A 7.5 nC point charge and a - 2.9 nC point charge are 3.2 cm apart. What is the electric field strength at the midpoint between
Oduvanchick [21]

Answer:

Net electric field, E_{net}=91406.24\ N/C

Explanation:

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Charge 1, q_1=7.5\ nC=7.5\times 10^{-9}\ C

Charge 2, q_2=-2.9\ nC=-2.9\times 10^{-9}\ C

distance, d = 3.2 cm = 0.032 m

Electric field due to charge 1 is given by :

E_1=\dfrac{kq_1}{r^2}

E_1=\dfrac{9\times 10^9\times 7.5\times 10^{-9}}{(0.032)^2}

E_1=65917.96\ N/C

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E_2=\dfrac{kq_2}{r^2}

E_2=\dfrac{9\times 10^9\times 2.9\times 10^{-9}}{(0.032)^2}

E_2=25488.28\ N/C

The point charges have opposite charge. So, the net electric field is given by the sum of electric field due to both charges as :

E_{net}=E_1+E_2

E_{net}=65917.96+25488.28

E_{net}=91406.24\ N/C

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E = 5356800J \frac{1KCal}{4.184*10^{6}J}

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4 0
2 years ago
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