When the in-plane shear stress is at its highest, it is true that the normal stresses are equal.
Their combined value is σn+σt.
<h3>What is in-plane shear stress?</h3>
- 6.3 ksi is the maximum in-plane shear stress. 10.2 ksi is the maximum out-of-plane shear stress.
- By examining Mohr's circle, we may understand why out of plane shear stress is the highest.
- Maximum in-plane shear stress is indicated by the inner blue circle radius (6.3 ksi).
- Maximum shear stress planes are 45 degrees from the major planes.
- The principle stress is the highest stress that may be created in a plane, and the principal stress refers to the principal plane when shear stress is zero.
The complete question is:
What is true about the normal stresses when the in-plane shear stress is maximum? Select all that apply.
a) They are equal.
b) They are zero.
c) Their sum is equal to σn+σt.
d) They are maximum.
e) They are equal to shear stress.
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Answer:
The basic purpose of adjusting the fuel/air mixture at altitude is to. ... increase the amount of fuel in the mixture to compensate for the decrease in pressure and density of the air
Answer:
According to the standard Safe Practices for Motor Vehicle Operations, ANSI/ASSE Z15.1, defensive driving skills involves "driving to save lives, time, and money, in spite of the conditions around you and the actions of others."
Explanation:
Defensive driving is a type of training provided to the learners eligible to drive motor vehicles. It involves the basics of driving and rules involved in driving. The learners learn about the methods of driving safely with the minimum expense of time and money. It also helps in reducing the risks of collision in adverse situations. The methods to rescue oneself from dangerous situations are also trained. Preventing oneself from the mistakes of others is also termed under defensive driving.
Answer:
The pressure in tank will become 541.3 KPa.
Explanation:
From general gas equation we know that:
PV = nRT
PV = (m/M)RT
P/Tm = R/VM
Now, for initial state:
P1/T1 m1 = R/VM ______ eqn (1)
For Final State, as the gas constant (R), volume (V) and molecular mass of air (M) remains constant;
P2/T2 m2 = R/VM ______ eqn (2)
Comparing eqn (1) and eqn (2):
P1/T1 m1 = P2/T2 m2
Now, we have:
P1 = 400 KPa
P2 = ?
m1 = 2 kg
m2 = 0.6 kg + 2 kg = 2.6 kg
T1 = 20°C = 293 K
T2 = 32°C = 305 K
Therefore,
(400 KPa)/(293 K)(2 kg) = P2/(305 K)(2.6 kg)
<u>P2 = 541.3 KPa</u>