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NNADVOKAT [17]
2 years ago
13

Which of the following are the three basic types of static stresses to which a material can be subjected (three correct answers)

: a. yield b. tensile c. hardness d. compression e. shear f. reduction in area g. true stress
Chemistry
1 answer:
Tasya [4]2 years ago
4 0

Answer:

b, d, e

Explanation:

Three type basic static stresses to which a material can be subjected are

Tensile: Stress generated due to elongation force on the body.

Compressive: The stress generated when the body is compressed.

Shear:  The stress generated when a force parallel to cross-section are of the body.

Hardness, reduction in area and true stress are material properties.

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Summarize the process a scientist goes through to come up with a<br> satisfactory solution.
maria [59]
Guess and check, test, trial and error, completion.
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2 years ago
Sort these six events from the earliest to the most recent. Democritus proposes the existence of atoms- Dalton's atomic theory-
Igoryamba
These events are actually sorted right, according to the time they occurred.
1. Democritus proposes the existence of atoms - this happened in the 5th century BC
2. Dalton's atomic theory - it was first presented in 1803
3. J.J. Thomson discovers the electron - happened in 1897
4. Rutherford's gold foil experiment - somewhere between 1908 and 1913
5. Bohr model - it was introduced in 1913
6. Schrodinger's wave - the equation was published in 1925
4 0
2 years ago
How much energy is required to melt a 20.0lb bag of ice a 0°c? A pound (lb.) is equivalent to 0.4536. The Hfusion of ice is +6.0
erma4kov [3.2K]
To compute the energy, Q, needed to melt a certain amount of ice, n, in moles, we have

Q = n \Delta H_{f}

where the latent heat of fusion for ice is equal to 6.009 kJ/mol.

Now, since we have a 20.0 lb ice, we must first convert its mass to grams. Thus mass = (20.0)(1000)(0.4536) = 9072 g.

To find the number of moles present, we must recall that the molar mass of water (ice) is equal to <span>1.00794(2) + </span><span>15.9994 </span>≈ 18.01 g/mol. Hence, we have

ice_{moles} = (\frac{9072 g}{1})(\frac{1 mol}{18.01 g}) =503.72 moles

Now, to compute for the molar heat of fusion, Q,

Q = (503.72)(6.009) = 3026.9 kJ

Therefore, the amount of heat needed to melt the 20-lb bag of ice is equal to 3026.9 kJ.
 
Answer: 3026.9 kJ
6 0
2 years ago
A freezer compartment is covered with a 2-mm-thick layer of frost at the time it malfunctions. If the compartment is in ambient
ruslelena [56]

Answer:

The time required to melt the frost is 3.25 hours.

Explanation:

The time required to melt the frost dependes on the latent heat of the frost and the amount of heat it is transfered by convection to the air .

The heat transferred per unit area can be expressed as:

q=h_c*A*\Delta T\\\\q/A=h_c*\Delta T

being hc the convective heat transfer coefficient (2 Wm^-2K^-1) and ΔT the difference of temperature (20-0=20 °C or K).

q/A=h_c*\Delta T=2\frac{W}{m^2K}*20K=40\frac{W}{m^2}

If we take 1 m^2 of ice, with 2 mm of thickness, we have this volume

V=T*A = 0.002 m * 1 m^2=0.002m^3

The mass of the frost can be estimated as

M=\rho * V=700\frac{kg}{m^3}*0.002m^3= 1.4 kg

Then,  the amount of heat needed to melt this surface (1 m²) of frost is

Q=L*M=334\frac{kJ}{kg}*1.4kg= 467.6kJ

The time needed to melt the frost can be calculated as

t=\frac{Q}{(q/A)}=\frac{467.6kJ/m2}{40W/m2} = 11.69\frac{kJ}{W}*\frac{1W*s}{1J}*\frac{1000J}{1kJ}=   11690s=3.25h

7 0
2 years ago
It took 4.5 minutes for 1.0 l helium to effuse through a porous barrier. how long will it take for 1.0 l cl2 gas to effuse under
serious [3.7K]
When The rate of effusion is inversely proportional to the √molar mass of the substance.
and we have R(He) = 1L / 4.5 min so,
R(He)/R(Cl2) = (molar mass of Cl2/ molar mass of He)^0.5
and when we have the molar mass of Cl2 = 70.9  & the molar mass of He = 4
so by substitution:
(1L/4.5 min)/ R(Cl2) = (70.9 / 4)^0.5
(1L/4.5 min) / R(Cl2) = 4.21
∴R(Cl2) = (1L/4.5 min) / 4.21 = 1L/ (4.5*4.21)min = 1 L / 18.945 min
∴Cl2 will take 18.945 min for 1 L to effuse under identical conditions

4 0
2 years ago
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