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siniylev [52]
3 years ago
6

Three different liquids are mixed together in a calorimeter. The masses of the liquids are m1 = 520 g, m2 = 715 g, m3 = 370 g. T

he specific heats are: C1 = 280 J/kg°C, C2 = 535 J/kg°C, C3 = 805 J/kg°C. The initial temperatures of the liquids are: T1 = 24.5 °C, T2 = 59.0 °C, T3 = 84.5 °C. What will be the temperature of the mixture in Celsius? (Enter only a number without a unit.)
Physics
1 answer:
DochEvi [55]3 years ago
6 0

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The forces are the same for part A 
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A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformati
Slav-nsk [51]

The question is incomplete. The complete question is :

A viscoelastic polymer that can be assumed to obey the Boltzmann superposition principle is subjected to the following deformation cycle. At a time, t = 0, a tensile stress of 20 MPa is applied instantaneously and maintained for 100 s. The stress is then removed at a rate of 0.2 MPa s−1 until the polymer is unloaded. If the creep compliance of the material is given by:

J(t) = Jo (1 - exp (-t/to))

Where,

Jo= 3m^2/ GPA

to= 200s

Determine

a) the strain after 100's (before stress is reversed)

b) the residual strain when stress falls to zero.

Answer:

a)-60GPA

b) 0

Explanation:

Given t= 0,

σ = 20Mpa

Change in σ= 0.2Mpas^-1

For creep compliance material,

J(t) = Jo (1 - exp (-t/to))

J(t) = 3 (1 - exp (-0/100))= 3m^2/Gpa

a) t= 100s

E(t)= ΔσJ (t - Jo)

= 0.2 × 3 ( 100 - 200 )

= 0.6 (-100)

= - 60 GPA

Residual strain, σ= 0

E(t)= Jσ (Jo) ∫t (t - Jo) dt

3 × 0 × 200 ∫t (t - Jo) dt

E(t) = 0

5 0
3 years ago
Snorkeling by humans and elephants. When a person snorkels, the lungs are connected directly to the atmosphere through the snork
fgiga [73]

Answer:

2354.4 Pa

40221 Pa

Explanation:

\rho = Density = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

h = Depth

The pressure difference would be

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 0.24\times 9.81\\\Rightarrow \Delta P=2354.4\ Pa

The pressure difference in the first case is 2354.4 Pa

\Delta P=\rho gh\\\Rightarrow \Delta P=1000\times 4.1\times 9.81\\\Rightarrow \Delta P=40221\ Pa

The pressure difference in the second case is 40221 Pa

7 0
3 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

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8 0
3 years ago
The maximum wavelength that an electromagnetic wave can have and still eject electrons from a metal surface is 459 nm. What is t
spin [16.1K]

Answer:

2.7067 eV

Explanation:

h = Planck's constant = 6.626\times 10^{-34}\ m^2kg/s

c = Speed of light = 3\times 10^8\ m/s

\lambda_0 = Threshold wavelength = 459 nm

Work function is given by

W_0=\frac{hc}{\lambda_0}\\\Rightarrow W_0=\frac{6.626\times 10^{-34}\times 3\times 10^8}{459\times 10^{-9}}\\\Rightarrow W_0=4.33072\times 10^{-19}\ J

Converting to eV

1\ J=\frac{1}{1.6\times 10^{-19}}\ eV

4.33072\times 10^{-19}\ J=4.33072\times 10^{-19}\times \frac{1}{1.6\times 10^{-19}}\ eV=2.7067\ eV

The work function W0 of this metal is 2.7067 eV

4 0
3 years ago
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