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kykrilka [37]
2 years ago
6

The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant

times the density of the liquid, p = ghd. how high a column of methanol (d = 0.79 g/ml) would be supported by a pressure that supports a 713 mm column of mercury (d = 13.6 g/ml)?
Physics
2 answers:
Valentin [98]2 years ago
6 0

Answer:

Height, h = 12.27 m

Explanation:

The pressure exerted by a column of liquid is equal to the product of the height of the column times the gravitational constant times the density of the liquid i.e.

P = d g h

Pressure, P = 713 mm column of mercury

Density, d = 0.79 g/ml = 790 kg/m³

We have to find how high a column of methanol would be supported by a pressure P. Let the height is h. So,

h=\dfrac{P}{dg}

g = acceleration due to gravity 9.8 m/s²

We know that, 1 mm of mercury = 133.32 Pa

713 mm of mercury = 95058.9 Pa

h=\dfrac{95058.9\ Pa}{790\ km/m^3\times 9.8\ m/s^2}

h = 12.27 m

Hence, this is the required solution.

Lesechka [4]2 years ago
3 0

We know that the pressure at the bottom of the 713 mm column of mercury is equal to an amount of 713 mmHg. The heights will be now, inversely proportional to the densities: So the computation is:(713 mm) (13.63 g/ml) / (0.79 g/ml) = 12,302 mm, or 12.3 meters.
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The correct answer to the question is that the lost mass has been converted into energy.

EXPLANATION:

From Einstein's theory, we know that energy and mass are inter convertible .

When some amount of mass is lost, same amount of energy equivalent to mass is produced.

Let us consider m is the mass lost during any reaction. Hence, the amount of energy produced will be-

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Here, c is the velocity of light i.e c = 3\times 10^8\ m/s

As per the question, uranium-235 undergoes fission. The amount of mass defect is 0.1 %.

The mass defect is defined as the difference between mass of reactants and products. During the fission, energy is produced.

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A baseball weighs 5.19 oz. what is the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher a
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Kinetic energy =0.5*mas*velocity^2
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What is a the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at 96.0 mi/h?

Answer: KE=0.5m*v^2
=0.5*(5.12 o *0.02835 kg/1 ounce)* (95 miles/h*1609.34m/1 miles* 1hr/3600s^)2
131kg*m^2/s^2= 131 joules

By what factor with the kinetic energy change if the speed of the baseball is decreased to 55.0 mi/h?

Answer: KE=0.5*m*v^2
=0.5*(5.13 o*0.02835kg/1 ounce)*(55 miles/ h*1609.34m/1 mile*1 hr/3600s)^2
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Two weights are connected by a very light cord that passes over an 80.0Nfrictionless pulley of radius 0.300m. The pulley is a so
Citrus2011 [14]

Answer:

The force does the ceiling exert on the hook is 269.59 N

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F = m*a

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m_{2} a=W_{2} -T_{2}

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m_{2} a+m_{1} a=T_{1} -W_{1} +W_{2} -T_{2} \\m_{1} =\frac{W_{1} }{g} \\m_{2} =\frac{W_{2} }{g} \\Replacing\\T_{2}-T_{1}=W_{2}   -W_{1} -(\frac{W_{1} }{g} +\frac{W_{2} }{g} )a  (eq. 1)

The torque:

\tau =I\alpha

Where

I = moment of inertia

α = angular acceleration

If the linear acceleration is

a=r\alpha \\\alpha =\frac{a}{r} \\I=\frac{1}{2} mr^{2} \\\tau =\frac{mra}{2}

Torque due the tension is equal:

\tau =r(T_{2} -T_{1} )

Substituting torque, mass, in equation 1, the expression respect the acceleration is:

a=\frac{g*(W_{2}-W_{1})}{W_{1}+W_{2} +\frac{W}{2} }

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a=\frac{9.8*(125-75)}{75+125+\frac{80}{2} } =2.04m/s^{2}

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2 years ago
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Inessa [10]

Answer:

The average rate of energy transfer to the cooker is 1.80 kW.

Explanation:

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Pressure of boiled water = 300 kPa

Mass of water = 3 kg

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Suppose, what is the average rate of energy transfer to the cooker?

We know that,

The specific enthalpy of evaporate at 300 kPa pressure

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h_{1}=561.47\ kJ/kg

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Using formula of enthalpy

h_{2}=h_{f}+xh_{fg}

Put the value into the formula

h_{2}=561.47+0.5\times2163.8

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Q=1.80\ kW

Hence, The average rate of energy transfer to the cooker is 1.80 kW.

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2 years ago
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