Ask question

Ask question

All categories

1 month ago

13

There are two different size spherical paintballs and the smaller one has a diameter of 5 cm and the larger one is 9 cm in diame

ter. If the smaller paintball holds 28 cm3 of paint, how much does the larger one hold?

1 answer:

Ostrovityanka [3.2K]1 month ago

5 0

Answer:

145.8 cm³ of paint

Explanation:

d₁ = Diameter of the smaller paintball = 5 cm

d₂ = Diameter of the larger paintball = 9 cm

V₂ = Volume of the larger paintball

Volume of the smaller paintball

$V_1=\frac{4}{3}\pi r_1^3\\\Rightarrow V_1=\frac{4}{3}\pi \left(\frac{d_1}{2}\right)^3\\\Rightarrow V_1=\frac{4}{24}\pi d_1^3$

Likewise

$V_2=\frac{4}{24}\pi d_2^3$

By dividing the previous two equations, we derive

$\frac{V_1}{V_2}=\frac{d_1^3}{d_2^3}\\\Rightarrow V_2=\frac{V_1}{\frac{d_1^3}{d_2^3}}\\\Rightarrow V_2=\frac{28}{\frac{125}{729}}\\\Rightarrow V_2=163.296\ cm^3$

∴ The larger one accommodates 163.296 cm³ of paint

You might be interested in

Considering the activity series given for nonmetals, what is the result of the below reaction? Use the activity series provided.

ValentinkaMS [3465]

Response: The result would be no reaction.

Clarification:

7 0

1 month ago

Read 2 more answers

3. A sample of argon of mass 6.56 g occupies 18.5 dm? at 305 K. (a) Calculate the work done when the gas expands isothermally ag

serg [3582]

Response:

(a) $W=-19.25J$

(b) $W=-52.8J$

Clarification:

Greetings.

(a) In this case, since the starting volume is 18.5 dm³ and the ending volume is 21 dm³ (18.5 +2.5), we can calculate the work at constant pressure as shown below:

$W=-P\Delta V=-7.7kPa*\frac{1000Pa}{1kPa} (21dm^3-18.5dm^3)*\frac{1m^3}{1000dm^3}\\ \\W=-19.25J$

This value is negative as it expands against the given pressure.

(b) Furthermore, if the process is conducted reversibly, the pressure might change, hence, we need to calculate the work using:

$W=nRTln(\frac{V_1}{V_2} )$

The moles are calculated based on the provided mass of argon:

$n=6.56g*\frac{1mol}{39.95g}=0.164mol$

Consequently, the work amounts to:

$W=0.164mol*8.314\frac{J}{mol*K} *305Kln(\frac{18.5dm^3}{21dm^3} )\\\\W=-52.8J$

Best regards.

4 0

18 days ago

A composite wall separates combustion gases at 2400°C from a liquid coolant at 100°C, with gas and liquid-side convection coeffi

Ostrovityanka [3204]

Response:

$\text{heat loss} = 24864.05 \ W/m^2$

Clarification:

If

$T_1$ , $T_2$ represent the temperatures of gases and liquids in Kelvins,

$t_1$ and $t_2$ denote the thicknesses of the gas layer and steel slab in meters,
$h_1$ , $h_2$ are the convection coefficients for gas and liquid in $W/m^2 \cdot K$ ,
$R_c$ represents the contact resistance in $m^2 \cdot K/W$ ,

and $k_1, k_2$ signify thermal conductivities of gas and steel in $W/m \cdot K$ ,

then: part(a):

$\text{heat loss } = \frac{T_1 - T_2} { \frac{1}{h_1} + \frac{t_1}{t_2} + R_c + \frac{t_2}{k_2} + \frac{1}{h_2}}$

by employing known values:

$\text {heat loss} = 2486.05 W/m^2$

part(b): Utilizing the rate equation:

$\text {heat loss} = h_1 (T_1 - T_{s1})$

the surface temperature is $T_{s1} = 1678.438 \ K$

and $T_{c1} = T_{s1} - \frac {t_1 (\text{heat loss})}{k_1} = 1664.560 \ K$

Correspondingly

$T_{c2} = T_{c1} - R_c (\text{heat loss}) = 421.357 \ K$

$T_{s2} = T_{c2} - \frac {t_2 (\text{heat loss})}{ k_2} = 397.864 \ K$

The temperature profile is depicted in the image provided

3 0

1 month ago

Find the three longest wavelengths (call them λ1, λ2, and λ3) that "fit" on the string, that is, those that satisfy the boundary

Sav [3153]

For a string that is secured at both ends, the oscillation amplitude must be zero at both extremities of the string. Refer to the attached images for clarity. It is evident that our fundamental harmonic will have the following wavelength: All subsequent harmonics are simply multiples of the fundamental. The three longest wavelengths are as follows:

6 0

21 day ago

A permeability test was run on a compacted sample of dirty sandy gravel. The sample was 175 mm long and the diameter of the mold

Maru [3345]

Answer:

(a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity is 0.0330 cm/s.

(c). The discharge velocity during the test is 0.0187 cm/s.

Explanation:

Given that,

Length = 175 mm

Diameter = 175 mm

Time = 90 sec

Volume= 405 cm³

We need to calculate the discharge

Using formula of discharge

$Q=\dfrac{V}{t}$

Insert the values into the formula

$Q=\dfrac{405}{90}$

$Q=4.5\ cm^3/s$

(a). We will compute the coefficient of permeability

Applying the formula for permeability

$Q=kiA$

$k=\dfrac{Q}{iA}$

$k=\dfrac{Ql}{Ah}$

Where, Q=discharge

l = length

A = cross-sectional area

h=constant head that causes flow

Plugging the value into the formula

$k=\dfrac{4.5\times175\times10^{-1}}{\dfrac{\pi(175\times10^{-1})^2}{4}\times38}$

$k=8.6\times10^{-3}\ cm/s$

The coefficient of permeability measures as $8.6\times10^{-3}\ cm/s$ .

(c). To ascertain the discharge velocity during the testing phase

Utilizing the discharge velocity formula

$v=ki$

$v=\dfrac{kh}{l}$

Substituting the values into the equation

$v=\dfrac{8.6\times10^{-3}\times38}{17.5}$

$v=0.0187\ cm/s$

The discharge velocity during the test measures 0.0187 cm/s.

Thus, (a). The coefficient of permeability is $8.6\times10^{-3}\ cm/s$ .

(b). The seepage velocity computes at 0.0330 cm/s.

(c). The observed discharge velocity during the test equals 0.0187 cm/s.

8 0

1 month ago