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sweet-ann [11.9K]
3 years ago
6

At what rate is soda being sucked out of a cylindrical glass that is 6 in tall and has radius of 2 in? The depth of the soda dec

reases at a constant rate of 0.25 in/sec.
Physics
1 answer:
evablogger [386]3 years ago
4 0

Answer:

The soda is being sucket out at a rate of 3.14 cubic inches/second.

Explanation:

R= 2in

S= π*R²= 12.56 inch²

rate= 0.25 in/sec

rate of soda sucked out= rate* S

rate of soda sucked out=  3.14 inch³/sec

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A tensile test specimen has a gage length = 50 mm and its cross-sectional area = 100 mm2. The specimen yields at 48,000 N, and t
Shalnov [3]

Answer:

a) yield strength

   \sigma_y = \dfrac{F_y}{A} = =\dfrac{48000}{100} = 480 MPa

b) modulus of elasticity

strain calculation

\varepsilon_0=\dfrac{L-L_0}{L_0}=\dfrac{50.23-50}{50} = 0.0046

strain for offset yield point

\varepsilon_{new} = \varepsilon_0 -0.002

                              =0.0046-0.002 = 0.0026

now, modulus of elasticity

 E = \dfrac{\sigma_y}{\varepsilon_{new}}=\dfrac{480}{0.0026}

    = 184615.28 MPa = 184.615 GPa

c) tensile strength

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d) percentage elongation

\% Elongation = \dfrac{L-L_0}{L_0}\times 100 = \dfrac{67.3-50}{50}\times 100 = 34.6\%

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\% Area\ reduction = \dfrac{A-A_f}{A}\times 100=\dfrac{100-53}{100}= 47 \%                            

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3 years ago
What is the voltage across a semiconductor bar if the current through it is 0.17 A? The electron concentration in the bar is 2.7
Anastaziya [24]

Answer:

The voltage across a semiconductor bar is 0.068 V.

Explanation:

Given that,

Current = 0.17 A

Electron concentration n= 2.7\times10^{18}\ cm^{-3}

Electron mobility \mu=1000 cm^2/Vs

Length = 0.1 mm

Area = 500 μm²

We need to calculate the resistivity

Using formula of resistivity

\sigma=n\times q\times \mu

\rho=\dfrac{1}{\sigma}

Put the value into the formula

\rho=\dfrac{1}{2.7\times10^{18}\times10^{6}\times1.6\times10^{-19}\times1000\times10^{-4}}

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Using formula of resistance

R=\dfrac{\rho l}{A}

R=\dfrac{2\times10^{-6}\times0.1\times10^{-3}}{500\times(10^{-6})^2}

R=0.4\ \Omega

We need to calculate the voltage

Using formula of voltage

V= IR

Put the value into the formula

V=0.17\times0.4

V=0.068\ V

Hence, The voltage across a semiconductor bar is 0.068 V.

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