Gray cast iron, with an ultimate tensile strength of 31 ksi and an ultimate compressive strength of 109 ksi, has the following s
1 answer:
Using an appropriate failure theory, find the factor of safety in each case. State the name of the theory that you are using the theory is max stress theory.
<h3>Wat is the max stress theory?</h3>The most shear strain concept states that the failure or yielding of a ductile fabric will arise whilst the most shear strain of the fabric equals or exceeds the shear strain fee at yield factor withinside the uniaxial tensile test.”
Stress states at various critical locations are f= 2.662.
Read more about strain:
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Answer:
hello your question has a missing part below is the missing part
Consider the string length equal to
answer : 2cos(2t) sin(2x) - 10cos(10t)sin(10x)
Explanation:
Given string length =
distorted function f(x) = 2sin(2x) - 10sin(10x)
Determine the wave formed in the string
attached below is a detailed solution of the problem
Answer:
R= 1.25
Explanation:
As given the local heat transfer,
But we know as well that,
Replacing the values
Reynolds number is define as,
Where V is the velocity of the fluid and \upsilon is the Kinematic viscosity
Then replacing we have
<em>*Note that A is just a 'summary' of all of that constat there.</em>
<em>That is </em>
Therefore at x=L the local convection heat transfer coefficient is
Definen that we need to find the average convection heat transfer coefficient in the entire plate lenght, so
The ratio of the average heat transfer coefficient over the entire plate to the local convection heat transfer coefficient is
Answer:
the width of the turning roadway = 15 ft
Explanation:
Given that:
A ramp from an expressway with a design speed(u) = 30 mi/h connects with a local road
Using 0.08 for superelevation(e)
The minimum radius of the curve on the road can be determined by using the expression:
where;
R= radius
= coefficient of friction
From the tables of coefficient of friction for a design speed at 30 mi/h ;
= 0.20
So;
R = 214.29 ft
R ≅ 215 ft
However; given that :
The turning roadway has stabilized shoulders on both sides and will provide for a onelane, one-way operation with no provision for passing a stalled vehicle.
From the tables of "Design widths of pavement for turning roads"
For a One-way operation with no provision for passing a stalled vehicle; this criteria falls under Case 1 operation
Similarly; we are told that the design vehicle is a single-unit truck; so therefore , it falls under traffic condition B.
As such in Case 1 operation that falls under traffic condition B in accordance with the Design widths of pavement for turning roads;
If the radius = 215 ft; the value for the width of the turning roadway for this conditions = 15ft
Hence; the width of the turning roadway = 15 ft
Answer: Inherent width in the emission line: 9.20 × 10⁻¹⁵ m or 9.20 fm
length of the photon emitted: 6.0 m
Explanation:
The emitted wavelength is 589 nm and the transition time is ∆t = 20 ns.
Recall the Heisenberg's uncertainty principle:-
∆t∆E ≈ h ( Planck's Constant)
The transition time ∆t corresponds to the energy that is ∆E
.
The corresponding uncertainty in the emitted frequency ∆v is:
∆v= ∆E/h = (5.273*10^-27 J)/(6.626*10^ J.s)= 7.958 × 10^6 s^-1
To find the corresponding spread in wavelength and hence the line width ∆λ, we can differentiate
λ = c/v
dλ/dv = -c/v² = -λ²/c
Therefore,
∆λ = (λ²/c)*(∆v) = {(589*10⁻⁹ m)²/(3.0*10⁸ m/s)} * (7.958*10⁶ s⁻¹)
= 9.20 × 10⁻¹⁵ m or 9.20 fm
The length of the photon (<em>l)</em> is
l = (light velocity) × (emission duration)
= (3.0 × 10⁸ m/s)(20 × 10⁻⁹ s) = 6.0 m