Answer:
the three part are mass, spring, damping
Explanation:
vibrating system consist of three elementary system namely
1) Mass - it is a rigid body due to which system experience vibration and kinetic energy due to vibration is directly proportional to velocity of the body.
2) Spring - the part that has elasticity and help to hold mass
3) Damping - this part considered to have zero mass and zero elasticity.
Answer:
The correct option is;
B) Metamorphic Rocks
Explanation:
Zoisite, which is also referred to saualpite, is a metamorphic rock which is a hydroxy sorosilicate mineral formed from other types of rocks such as sedimentary, metamorphic and ingenious rocks in the process of their metamorphism under the presence high temperatures and pressures and mineral fluids which are hot
Zoiste is named after Sigmund Zois by Abraham Gottlob Werner in 1805 when Sigmund Zois sent Abraham Gottlob Werner the mineral specimen from Saualpe in 1805
Answer:
(A) Because the angle of twist of a material is often used to predict its shear toughness
Explanation:
In engineering, torsion is the solicitation that occurs when a moment is applied on the longitudinal axis of a construction element or mechanical prism, such as axes or, in general, elements where one dimension predominates over the other two, although it is possible to find it in diverse situations.
The torsion is characterized geometrically because any curve parallel to the axis of the piece is no longer contained in the plane initially formed by the two curves. Instead, a curve parallel to the axis is twisted around it.
The general study of torsion is complicated because under that type of solicitation the cross section of a piece in general is characterized by two phenomena:
1- Tangential tensions appear parallel to the cross section.
2- When the previous tensions are not properly distributed, which always happens unless the section has circular symmetry, sectional warps appear that make the deformed cross sections not flat.
Answer:
The flexural strength of a specimen is = 78.3 M pa
Explanation:
Given data
Height = depth = 5 mm
Width = 10 mm
Length L = 45 mm
Load = 290 N
The flexural strength of a specimen is given by
78.3 M pa
Therefore the flexural strength of a specimen is = 78.3 M pa
Answer:
T1 = 625.54 K
Explanation:
We are given;
T_α = Tsur = 25°C = 298K
h = 20 W/m².K,
L = 0.15 m
K = 1.2 W/m.K
ε = 0.8
Ts = T2 = 100°C = 373K
T1 = ?
Assumption:
-Steady- state condition
-One- dimensional conduction
-No uniform heat generation
-Constant properties
From Energy balance equation;
E°in - E°out = 0
Thus,
q"cond – q"conv – q"rad = 0
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴)
Where σ is Stephan-Boltzmann constant and has a value of 5.67 x 10^(-8)
Thus;
K[(T1 - T2)/L] - h(Ts-T_α) - εσ (Ts⁴ – Tsur⁴) = 1.2[(T1 - 373)/0.15] - 20(373 - 298] - 0.8x5.67x10^(-8)[373⁴ - 298⁴] = 0
This gives;
(8T1 - 2984) - (1500) - 520.31 = 0
8T1 = 2984 + 1500 + 520.31
8T1 = 5004.31
T1 = 5004.31/8
T1 = 625.54 K
