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2 years ago

Determine the density, specific gravity, and mass of the air in a room whose dimensions are 4 m×5 m× 6 m at 100 kPa and 25 C

1 answer:

4 0

Answer : The density, specific gravity, and mass of the air in a room is, 1.16825 g/L, 0.916 and 140.19 kg respectively.

Explanation :

First we have to calculate the volume of air.

$Volume=Length\times Breadth\times Height$

$Volume=4m\times 5m\times 6m$

$Volume=120m^3=120000L$ $(1m^3=1000L)$

Now we have to calculate the mole of air.

Using ideal gas equation:

$PV=nRT$

where,

P = Pressure of air = 100 kPa = 0.987 atm (1 atm = 101.3 kPa)

V = Volume of air = 120000 L

n = number of moles of air = ?

R = Gas constant = $0.0821L.atm/mol.K$

T = Temperature of air = $25^oC=273+25=298K$

Putting values in above equation, we get:

$0.987atm\times 120000L=n\times (0.0821L.atm/mol.K)\times 298K$

$n=4841.04mol$

Now we have to calculate the mass of air.

$\text{Mass of air}=\text{Moles of air}\times \text{Molar mass of air}$

As we know that the molar mass of air is, 28.96 g/mol

$\text{Mass of air}=4841.04mol\times 28.96g/mol=140196.5184g=140.19kg$

Now we have to calculate the density of air.

$\text{Density of air}=\frac{\text{Mass of air}}{\text{Volume of air}}$

$\text{Density of air}=\frac{140.19kg}{120000L}=1.16825\times 10^{-3}kg/L=1.16825g/L$

Now we have to calculate the specific gravity of air.

$\text{Specific gravity of air}=\frac{\text{Air density at given condition}}{\text{Air density at STP}}$

As we know that air density at STP is, 1.2754 g/L

$\text{Specific gravity of air}=\frac{1.16825g/L}{1.2754g/L}=0.916$

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For a metal that has an electrical conductivity of 7.1 x 107 (Ω-m)-1, do the following: (a) Calculate the resistance (in Ω) of a

jonny [76]

Answer:

(a) 0.0178 Ω

(b) 3.4 A

(d) 9.01 x 10⁻³ V/m

Explanation:

(a)

σ = Electrical conductivity = 7.1 x 10⁷ Ω-m⁻¹

d = diameter of the wire = 2.6 mm = 2.6 x 10⁻³ m

Area of cross-section of the wire is given as

A = (0.25) π d²

A = (0.25) (3.14) (2.6 x 10⁻³)²

A = 5.3 x 10⁻⁶ m²

L = length of the wire = 6.7 m

Resistance of the wire is given as

$R=\frac{L}{A\sigma }$

$R=\frac{6.7}{(5.3\times10^{-6})(7.1\times10^{7}) }$

R = 0.0178 Ω

(b)

V = potential drop across the ends of wire = 0.060 volts

i = current flowing in the wire

Using ohm's law, current flowing is given as

$i = \frac{V}{R}$

$i = \frac{0.060}{0.0178}$

i = 3.4 A

(c)

Current density is given as

$J = \frac{i}{A}$

$J = \frac{3.4}{5.3\times10^{-6}}$

J = 6.4 x 10⁵ A/m²

(d)

Magnitude of electric field is given as

$E = \frac{J}{\sigma }$

$E = \frac{6.4 \times 10^{5}}{ 7.1 \times 10^{7}}$

E = 9.01 x 10⁻³ V/m

5 0

2 years ago

Describe the energy transformations that take place when a skier starts skiing down a hill but after a time is brought to rest b

Andrews [41]

<span>The skier will transform their gravitational energy into mostly kinetic energy (with a minor amount transformed into heat from the friction of the skis across the snow and air friction). Once the skier hits the snowdrift, their kinetic energy is transferred into the snow which moves when they strike it due to the kinetic energy that is now in the snow. Along with again a minor amount of heat energy transferred as they move through the snowdrift.</span>

6 0

2 years ago

A 150 g particle at x = 0 is moving at 8.00 m/s in the +x-direction. As it moves, it experiences a force given by Fx=(0.850N)sin

krok68 [10]

Answer:

9.98 m/s

Explanation:

The force acting on the particle is defined by the equation:

$F=(0.850) sin (\frac{x}{2.00})$ [N]

where x is the position in metres.

The acceleration can be found by using Newton's second law:

$a=\frac{F}{m}$

where

m = 150 g = 0.150 kg is the mass of the particle. Substituting into the equation,

$a=\frac{0.850}{0.150}sin (\frac{x}{2.00})=5.67 sin(\frac{x}{2.00})$ [m/s^2]

When x = 3.14 m, the acceleration is:

$a=5.67 sin(\frac{3.14}{2.00})=5.67 m/s^2$

Now we can find the final speed of the particle by using the suvat equation:

$v^2-u^2=2ax$

where

u = 8.00 m/s is the initial velocity

v is the final velocity

$a=5.67 m/s^2$

x = 3.14 m is the displacement

Solving for v,

$v=\sqrt{u^2+2ax}=\sqrt{8.00^2+2(5.67)(3.14)}=9.98 m/s$

And the speed is just the magnitude of the velocity, so 9.98 m/s.

4 0

2 years ago

a steel block has a mass of 40g.it is in the form of a cube. each edge is 1.74cm long. calculate the density

Vinil7 [7]

Answer:

d ≈ 7,6 g/cm³

Explanation:

d = m/V = 40g/5,27cm³ ≈ 7,6 g/cm³

V = l³ = (1.74cm)³ ≈ 5,27 cm³

3 0

2 years ago

Heat engines were first envisioned and built during the industrial revolution. Explain the thermodynamics of a heat engine comme

Artyom0805 [142]

Heat engines were developed during industrial revolution.

Generally a heat engine contains three parts i.e source, sink and working substance.

The source of a heat engine is present at a higher temperature as compared to the sink. Due to the temperature difference, the heat will flow from source to sink through working substance.

Let us consider $T_{1}\ and\ T_{2}$ are the temperature of source and sink.

As the source is at higher temperature as compared to sink, heat will flow from source to sink.

$Let\ Q_{1}\ and\ Q_{2}$ are the heat provided by source and heat rejected to sink.

Hence, the work done by the working substance will be -

$W\ =\ Q_{1}-Q_{2}$

The efficiency of a heat engine is defined as the ratio of output to the input energy.

Here output = workdone [W]

Hence, the efficiency of a heat engine is calculated as -

$Efficiency\ [\eta]=\frac{W}{Q_{1}}$

$\eta\ =\frac{Q_{1}- Q_{2}} {Q_{1}}$

$=\ 1-\frac{Q_{2}} {Q_{1}}$

This is the expression for the efficiency of heat engine.

Here, all the heat absorbed by the working substance can not be converted to desired output. The efficiency of a heat engine can not be 100 percent. Some amount of heat is lost in the form of sound and heat due to the friction which is produced due to the relative motion between various parts of the machine.

6 0

2 years ago

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