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3 years ago

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A 1-m-tall barrel is closed on top except for a thin pipe extending 4.2 m up from the top. When the barrel is filled with water

up to the base of the pipe (1 m deep) the water pressure on the bottom of the barrel is 10 kPa. What is the pressure on the bottom when water is added to fill the pipe to its top?

1 answer:

Dmitry_Shevchenko [17]3 years ago

4 0

Answer:

P = 52 kPa

Explanation:

Hidrostatic pressure is defined as the product of the height of liquid (h) by its specific weight (ρ) and by the acceleration of gravity (g).

In the first scenario, the atmospheric pressure is:

$10,000 = \rho g h_{1}\\10,000 = \rho g*1\\\rho g = 10,000$

In the second scenario, h = 4.2 + 1 m. Therefore, the pressure at the bottom of the barrel is:

$P= \rho g h_{2}\\P= 10,000*5.2\\P=52,000\\P=52kPa$

The pressure on the bottom when water is added to fill the pipe to its top is 52 kPa.

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The overall energy involved in the formation of CsCl from Cs(s) and Cl2(g) is −443 kJ/mol. Given the following information: heat

german

Answer : The magnitude of the lattice energy for CsCl is, 667 KJ/mole

Explanation :

The steps involved in the born-Haber cycle for the formation of $CsCl$ :

(1) Conversion of solid calcium into gaseous cesium atoms.

$Cs(s)\overset{\Delta H_s}\rightarrow Cs(g)$

$\Delta H_s$ = sublimation energy of calcium

(2) Conversion of gaseous cesium atoms into gaseous cesium ions.

$Ca(g)\overset{\Delta H_I}\rightarrow Ca^{+1}(g)$

$\Delta H_I$ = ionization energy of calcium

(3) Conversion of molecular gaseous chlorine into gaseous chlorine atoms.

$Cl_2(g)\overset{\frac{1}{2}\Delta H_D}\rightarrow Cl(g)$

$\Delta H_D$ = dissociation energy of chlorine

(4) Conversion of gaseous chlorine atoms into gaseous chlorine ions.

$Cl(g)\overset{\Delta H_E}\rightarrow Cl^-(g)$

$\Delta H_E$ = electron affinity energy of chlorine

(5) Conversion of gaseous cations and gaseous anion into solid cesium chloride.

$Cs^{1+}(g)+Cl^-(g)\overset{\Delta H_L}\rightarrow CsCl(s)$

$\Delta H_L$ = lattice energy of calcium chloride

To calculate the overall energy from the born-Haber cycle, the equation used will be:

$\Delta H_f^o=\Delta H_s+\Delta H_I+\Delta H_D+\Delta H_E+\Delta H_L$

Now put all the given values in this equation, we get:

$-443KJ/mole=76KJ/mole+376KJ/mole+121KJ/mole+(-349KJ/mole)+\Delta H_L$

$\Delta H_L=-667KJ/mole$

The negative sign indicates that for exothermic reaction, the lattice energy will be negative.

Therefore, the magnitude of the lattice energy for CsCl is, 667 KJ/mole

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3 years ago

A 12 volt car battery is connected to a 3 ohm brake light. What is the current carrying energy to the lights?

zloy xaker [14]

Answer:

4 A

Explanation:

V = IR, where V=voltage, I=current, R=resistance. This is Ohm's Law. (remember that for units V = volts, Ω = ohms, A = amperes.)

V = IR

12 V = I * 3 Ω

12/3 = I

<u>I = 4 A</u>

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3 years ago

A metallic wire has a diameter of 4.12mm. When the current in the wire is 8.00A, the drift velocity is 5.40×10−5m/s.What is the

podryga [215]

Answer:

$6.9\times 10^{28}m^{-3}$

Explanation:

We are given that

Diameter of wire=d=4.12 mm

Radius of wire=r $r=\frac{d}{2}=\frac{4.12}{2}=2.06mm=2.06\times 10^{-3} m$

$1mm=10^{-3} m$

Current=I=8 A

Drift velocity= $v_d=5.4\times 10^{-5} m/s$

We have to find the density of free electrons in the metal

We know that

Density of electron= $n=\frac{I}{v_deA}$

Using the formula

Density of free electrons= $\frac{8}{5.4\times 10^{-5}\times 1.6\times 10^{-19}\times 3.14\times (2.06\times 10^{-3})^2}$

By using Area of wire= $\pi r^2$

$\pi=3.14\\e=1.6\times 10^{-19} C$

Density of free electrons= $6.9\times 10^{28}m^{-3}$

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3 years ago

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A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

Olin [163]

The crate moves at constant velocity, this means that its acceleration is zero, so the net force acting on the crate is zero (Newton's second law).

There are only two forces acting on the crate: the force F applied by the worker and the frictional force, acting in the opposite direction: $\mu m g$ , where $\mu=0.25$ is the coefficient of friction and $m=30.0 kg$ is the mass of the crate. Since the net force should be equal to zero, the two forces must have same magnitude, so we have:
$F=\mu m g=(0.25)(30.0 kg)(9.81 m/s^2)=73.8 N$
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2. Jenny picks up a guitar and strums. If the sound from the string emanates out at 440 Hz,

erik [133]

Answer:

6.8×10^6 m

Explanation:

f = velocity of light/ wavelength

440= 3×10^8 / wavelength

so

wavelength = 3×10^7/ 44

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3 years ago