Answer:
The correct option is;
A) Steel becomes too hot in the Sun and can burn the children
Explanation:
The properties of steel includes;
Low specific heat capacity, high thermal and electrical toughness, high hardness, high tensile strength, high yield strength, appreciable elongation, high fatigue strength, can easily corrode, high malleability and ability to creep
Therefore, due to the low specific heat capacity, which is 0.511 J/(g·°C) and high conductivity of steel which is about 32 W/(m·k), the temperature of the steel can rapidly rise and the hot steel surface can readily conduct the heat, (due to the temperature difference) to other bodies that come in contact
Answer: The electric field decreases because of the insertion of the Teflon.
Explanation:
If the charge on the capacitor is held fixed, the electric field as a consequence of this charge distribution (directed from the positive charged plate to the negative charged one remains unchanged.
However, as the Teflon is a dielectric material, even though doesn't allow the free movement of the electrons as an answer to an applied electric field, it allows that the electrons be displaced from the equilibrium position, leaving a local negative-charged zone close to the posiitive plate of the capacitor, and an equal but opposite charged layer close to the negative plate.
In this way, a internal electric field is created, that opposes to the external one due to the capacitor, which overall effect is diminishing the total electric field, reducing the voltage between the plates, and increasing the capacitance proportionally to the dielectric constant of the Teflon.
Answer:
Explanation:
A plane wall of thickness 2L=40 mm and thermal conductivity k=5W/m⋅Kk=5W/m⋅K experiences uniform volumetric heat generation at a rateq
˙
q
q
˙
, while convection heat transfer occurs at both of its surfaces (x=-L, +L), each of which is exposed to a fluid of temperature T∞=20∘CT
∞
=20
∘
C. Under steady-state conditions, the temperature distribution in the wall is of the form T(x)=a+bx+cx2T(x)=a+bx+cx
2
where a=82.0∘C,b=−210∘C/m,c=−2×104C/m2a=82.0
∘
C,b=−210
∘
C/m,c=−2×10
4
C/m
2
, and x is in meters. The origin of the x-coordinate is at the midplane of the wall. (a) Sketch the temperature distribution and identify significant physical features. (b) What is the volumetric rate of heat generation q in the wall? (c) Determine the surface heat fluxes, q
′′
x
(−L)q
x
′′
(−L) and q
′′
x
(+L)q
x
′′
(+L). How are these fluxes related to the heat generation rate? (d) What are the convection coefficients for the surfaces at x=-L and x=+L? (e) Obtain an expression for the heat flux distribution q
′′
x
(x)q
x
′′
(x). Is the heat flux zero at any location? Explain any significant features of the distribution. (f) If the source of the heat generation is suddenly deactivated (q=0), what is the rate of change of energy stored in the wall at this instant? (g) What temperature will the wall eventually reach with q=0? How much energy must be removed by the fluid per unit area of the wall (J/m2)(J/m
2
) to reach this state? The density and specific heat of the wall material are 2600kg/m32600kg/m
3
and 800J/kg⋅K800J/kg⋅K, respectively.
Answer:
σ = 391.2 MPa
Explanation:
The relation between true stress and true strain is given as:
σ = k εⁿ
where,
σ = true stress = 365 MPa
k = constant
ε = true strain = Change in Length/Original Length
ε = (61.8 - 54.8)/54.8 = 0.128
n = strain hardening exponent = 0.2
Therefore,
365 MPa = K (0.128)^0.2
K = 365 MPa/(0.128)^0.2
k = 550.62 MPa
Now, we have the following data:
σ = true stress = ?
k = constant = 550.62 MPa
ε = true strain = Change in Length/Original Length
ε = (64.7 - 54.8)/54.8 = 0.181
n = strain hardening exponent = 0.2
Therefore,
σ = (550.62 MPa)(0.181)^0.2
<u>σ = 391.2 MPa</u>
Answer:
(166.0 mm Hg) (273.15 mL) = (760.0 mm Hg) (x)
