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ivolga24 [154]
2 years ago
12

4. A cylindrical tube has a length of 14.4cm and a radius of 1.5cm and is filled with a colorless gas. If the density of the gas

is known to be 0.00123g/cm', what is the mass of the gas in the tube in mg?​
Physics
1 answer:
professor190 [17]2 years ago
5 0

Answer:

Mass, m of gas is 0.2504 grams.

Explanation:

First, we need to solve for the volume of the cylindrical tube.

Volume of cylinder is given by the formula;

V = 2\Pi r^{2}h

Where, V represents volume.

π represents pie

r represents radius.

h represents height or length.

Given the following data;

Radius, r = 1.5cm

Length, h = 14.4cm

Density, d = 0.00123g/cm³

Substituting into the equation;

V = 2 * 3.142 * (1.5)^{2}14.4

V = 2 * 3.142 * 2.25 * 14.4

V = 203. 6016

Therefore, the volume of the cylindrical tube is 203. 6016cm³

Density can be defined as mass all over the volume of an object.

Simply stated, density is mass per unit volume of an object.

Mathematically, density is given by the formula;

Density = \frac{mass}{volume}

Mass = density  *  volume

Substituting into the equation, we have;

Mass = 0.00123 * 203. 6016

Mass = 0.2504g.

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The answer is not correct.

Explanation:

Stu's answer is not correct, the equation to use is known as the law of ohm. In which the voltage is defined as the product of the current by the resistance, then we will see this equation.

V = I*R\\where:\\I = current [amp]\\R = resistance [ohm]\\V = voltage [volts]\\

In order to find resistance, this term is found multiplying the current on the right side of the equation, therefore the current will be divided on the left side of the equation.

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That is the reason that the result found by Stu is not correct.

6 0
2 years ago
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A single crystal of aluminum is oriented for a tensile test such that its slip plane normal makes angle of 28.1 with the tensil
NISA [10]

Answer:

we have to find out the critical resolved shear stress. As it it given in the question

Ф = 28.1°and the possible values for λ are 62.4°, 72.0° and 81.1°.

a) Slip will occur in the direction where cosФ cosλ are maximum. Cosine for all possible λ values are given as follows.

cos(62.4°) = 0.46

cos(72.0°) = 0.31

cos(81.1°) = 0.15

Thus, the slip direction is at the angle of 62.4° along the tensile axis.

b) now the critical resolved shear stress can be find out by the following equation.

τ_{crss} = σ_{Y} ( cosФ cosλ)_{max}

now by putting values,

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exis [7]

Answer:

2046.37 kPa

Explanation:

Given:

Number of moles, n = 125

Temperature, T = 20° C = 20 + 273 = 293 K

Radius of the cylinder, r = 17 cm = 0.17 m

Height of the cylinder, h = 1.64 m

thus,

volume of the cylinder, V = πr²h

= π × 0.17² × 1.64

= 0.148 m³

Now,

From the ideal gas law

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PV = nRT

here,

P is the pressure

R is the ideal gas constant = 8.314  J / mol. K

thus,

P × 0.148 = 125 × 8.314 × 293

or

P × 0.148 = 304500.25

or

P = 2046372.64 Pa = 2046.37 kPa

6 0
2 years ago
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