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3 years ago

Aluminum has a density of

Physics

2 answers:

Morgarella [4.7K]3 years ago

7 0

Answer:

0.1848

Explanation:

p=m/v

where

p=density

m=mass

v=volume

Substitute the values and solve

the answer is 0.1848

BTW, it is correct for Acellus students

agasfer [191]3 years ago

3 0

<h3><u>Volume is 0.1848 m³</u></h3><h3 />

Explanation:

<h2>Given:</h2>

m = 49.9 kg

ρ = 270 kg/m³

<h2>Required:</h2>

volume

<h2>Equation:</h2>

$ρ \:= \:\frac{m}{v}$

where: ρ - density

m - mass

v - volume

<h2>Solution:</h2>

Substitute the value of ρ and m

$ρ \:= \:\frac{m}{v}$

$270\: kg/m³\:= \:\frac{49.9\:kg}{v}$

$(v)\:270\: kg/m³\:= \:49.9\:kg$

$v\:= \:\frac{49.9\:kg}{270\: kg/m³}$

$v\:= \:0.1848\:m³$

<h2>Final Answer:</h2><h3><u>Volume is 0.1848 m³</u></h3>

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A sound source is moving at 80 m/s toward a stationary listener that is standing in still air (a) Find the wavelength of the sou

Setler [38]

Answer:

a. wavelength of the sound, $\vartheta = 1.315\vartheta_{o}$

b. observed frequecy, $\lambda = 0.7604\lambda_{o}$

Given:

speed of sound source, $v_{s}$ = 80 m/s

speed of sound in air or vacuum, $v_{a}$ = 343 m/s

speed of sound observed, $v_{o}$ = 0 m/s

Solution:

From the relation:

v = $\vartheta \lambda$ (1)

where

v = velocity of sound

$\vartheta$ = observed frequency of sound

$\lambda$ = wavelength

(a) The wavelength of the sound between source and the listener is given by:

$\lambda = \frac{v_{a}}{\vartheta }$ (2)

(b) The observed frequency is given by:

$\vartheta = \frac{v_{a}}{v_{a} - v_{s}}\vartheta_{o}$

$\vartheta = \frac{334}{334 - 80}\vartheta_{o}$

$\vartheta = 1.315\vartheta_{o}$ (3)

Using eqn (2) and (3):

$\lambda = \frac{334}{1.315} = \frac{1}{1.315}\frac{v_{a}}{\vartheta_{o}}$

$\lambda = 0.7604\lambda_{o}$

4 0

3 years ago

Which of the following describes an accelerated motion

Naily [24]

Accelerated motion such as a vehicle (car) or a moving item such as a (football thrown in the air)

3 0

3 years ago

the pressure of a gas is 100.0kpa and its volume is 500.0ml if the volume increase to 1000.0ml what is the new pressure of the g

Ket [755]

50Kpa

Explanation:

P1V1 = P2V2

Where;

P1= 100. P2= ?

V1 = 500. V2 = 1000

100 × 500 = P2 × 1000

50000 = 1000P2

50000/1000 = P2

50 = P2

P2 = 50Kpa

4 0

3 years ago

6) Find the speed a spherical raindrop would attain by falling from 4.00 km. Do this:a) In the absence of air dragb) In the pres

sleet_krkn [62]

We are asked to determine the velocity of a rain drop if it falls from 4 km.

To do that we will use the following formula:

$2ah=v_f^2-v_0^2$

Where:

$\begin{gathered} a=\text{ acceleration} \\ h=\text{ height} \\ v_f,v_0=\text{ final and initial velocity} \end{gathered}$

If we assume the initial velocity to be 0 we get:

$2ah=v_f^2$

The acceleration is the acceleration due to gravity:

$2gh=v_f^2$

Now, we take the square root to both sides:

$\sqrt{2gh}=v_f$

Now, we substitute the values:

$\sqrt{2(9.8\frac{m}{s^2})(4000m)}=v_f$

solving the operations:

$280\frac{m}{s}=v$

Therefore, the velocity without air drag is 280 m/s.

Part B. we are asked to determine the velocity if there is air drag. To do that we will use the following formula:

$F_d=\frac{1}{2}C\rho_{air}Av^2$

Where:

$\begin{gathered} F_d=drag\text{ force} \\ C=\text{ constant} \\ \rho_{air}=\text{ density of air} \\ A=\text{ area} \\ v=\text{ velocity} \end{gathered}$

We need to determine the drag force. To do that we will use the following free-body diagram:

Since the velocity that the raindrop reaches is the terminal velocity and its a constant velocity this means that the acceleration is zero and therefore the forces are balanced:

$F_d=mg$

Now, we determine the mass of the raindrop using the following formula:

$m=\rho_{water}V$

Where:

$\begin{gathered} \rho_{water}=\text{ density of water} \\ V=\text{ volume} \end{gathered}$

The volume is the volume of a sphere, therefore:

$m=\rho_{water}(\frac{4}{3}\pi r^3)$

Since the diameter of the raindrop is 3 millimeters, the radius is 1.5 mm or 0.0015 meters. Substituting we get:

$m=(0.98\times10^3\frac{kg}{m^3})(\frac{4}{3}\pi(0.0015m)^3)$

Solving the operations:

$m=1.39\times10^{-5}kg$

Now, we substitute the values in the formula for the drag force:

$F_d=(1.39\times10^{-5}kg)(9.8\frac{m}{s^2})$

Solving the operations:

$F_d=1.36\times10^{-4}N$

Now, we substitute in the formula:

$1.36\times10^{-4}N=\frac{1}{2}C\rho_{air}Av^2$

Now, we solve for the velocity:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}C\rho_{air}A}=v^2$

Now, we substitute the values. We will use the area of a circle:

$\frac{1.36\times10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^3})(\pi r^2)}=v^2$

Substituting the radius:

$\frac{1.36\cdot10^{-4}N}{\frac{1}{2}(0.45)(1.21\frac{kg}{m^{3}})(\pi(0.0015m)^2)}=v^2$

Solving the operations:

$70.67\frac{m^2}{s^2}=v^2$

Now, we take the square root to both sides:

$\begin{gathered} \sqrt{70.67\frac{m^2}{s^2}}=v \\ \\ 8.4\frac{m}{s}=v \\ \end{gathered}$

Therefore, the velocity is 8.4 m/s

7 0

1 year ago

A vat of nitrogen at its boiling point (-196 degrees C) absorbs 384000 J of heat. How much mass of nitrogen burns off? (Unit = k

masya89 [10]

Answer:

1.92 kg of nitrogen.

Explanation:

The following data were obtained from the question:

Heat absorbed (Q) = 384000 J

Note: Heat of vaporisation (ΔHv) of nitrogen = 5600 J/mol

Next, we shall determine the number of mole of nitrogen that absorbed 384000 J.

This is illustrated below:

Q = mol·ΔHv

384000 = mole of N2 x 5600

Divide both side by 5600

Mole of N2 = 384000/5600

Mole of N2 = 68.57 moles

Next, we shall convert 68.57 moles of nitrogen, N2 to grams.

This can be obtained as follow:

Molar mass of N2 = 2 x 14 = 28 g/mol.

Mole of N2 = 68.57 moles.

Mass of N2 =..?

Mole = mass /molar mass

68.57 = mass of N2 /28

Cross multiply

Mass of N2 = 68.57 x 28

Mass of N2 = 1919.96 g

Finally, we shall convert 1919.96 g to kilograms.

This can be achieved as shown below:

1000g = 1 kg

Therefore,

1919.96 g = 1919.96/1000 = 1.92 kg.

Therefore, 1.92 kg of nitrogen were burned off.

3 0

3 years ago