Answer: D) All of the above
Explanation:
Burn rate can be affected by all of the above reasons as, variation in chamber pressure because the pressure are dependence on the burn rate and temperature variation in initial gain can affect the rate of the chemical reactions and initial gain in the temperature increased the burning rate. As, gas flow velocity also influenced to increasing the burn rate as it flowing parallel to the surface burning. Burn rate is also known as erosive burning because of the variation in flow velocity and chamber pressure.
Answer:
It will not experience fracture when it is exposed to a stress of 1030 MPa.
Explanation:
Given
Klc = 54.8 MPa √m
a = 0.5 mm = 0.5*10⁻³m
Y = 1.0
This problem asks us to determine whether or not the 4340 steel alloy specimen will fracture when exposed to a stress of 1030 MPa, given the values of <em>KIc</em>, <em>Y</em>, and the largest value of <em>a</em> in the material. This requires that we solve for <em>σc</em> from the following equation:
<em>σc = KIc / (Y*√(π*a))</em>
Thus
σc = 54.8 MPa √m / (1.0*√(π*0.5*10⁻³m))
⇒ σc = 1382.67 MPa > 1030 MPa
Therefore, the fracture will not occur because this specimen can handle a stress of 1382.67 MPa before experience fracture.
Answer:
For expr1 = index 5, length 5
For expr2 = index 0, length 4 and index 21, length 5
string quote = "Four score and seven years ago";
string expr1 = quote.Substring(5, 5).ToUpper(); // "SCORE"
string expr2 = quote.Substring(0, 4) + quote.Substring(21, 5).ToLower(); // "fouryears"
Console.WriteLine(expr1);
Console.WriteLine(expr2);
Explanation:
Then code is written in c# and it produces SCORE and f
ouryears
Substring takes 2 arguments, the start of the specific character and the length
Answer:
a) v = +/- 0.515 m/s
b) x = -0.098164 m
c) v = +/- 1.005 m/s
Explanation:
Given:
The relationship for the acceleration is given as follows:
a = - (0.1 + sin(x/b))
Where, b = 0.98
- IVP is v = 1 m/s @ x = 0
Find:
(a) the velocity of the particle when x = -1 m
(b) the position where the velocity is maximum
(c) the maximum velocity.
Solution:
- We will compute the velocity by integrating a by dt.
a = v*dv / dx = - (0.1 + sin(x/0.98))
- Separate variables:
v*dv = - (0.1 + sin(x/0.98)) . dx
-Integrate from v = 1 m/s to v and @ x = 0 to x:
0.5*(v^2) = - (0.1*x - 0.98*cos(x/0.8)) - 0.98 + 0.5
0.5*v^2 = 0.98*cos(x/0.98) - 0.1*x - 0.48
- Evaluate at, x = -1
0.5*v^2 = 0.98 cos(-1/0.98) + 0.1 -0.48
v = sqrt (0.2651155)
v = +/- 0.515 m/s
- v = v_max when a = 0. Set the given expression to zero and solve for x:
-0.1 = sin(x/0.98)
x = -0.98*0.1002
x = -0.098164 m
- Hence now evaluate velocity through the derived expression:
v^2 = 1.96 cos(-0.098164/0.98) -0.96 -0.2*-0.098164
v = sqrt (1.0098)
v = +/- 1.005 m/s
Answer:
Advantages
The main advantage in the use of pulleys is that the effort becomes less as compared to the normal lifting of the weights. In other words, it reduces the amount of actual force required to lift heavy objects. It also changes the direction of the force applied. These two advantages in the use pulleys make them an important tool for heavy lifting. It also provides a mechanical advantage.
The other advantage in the use of pulleys is that the distance between the operator and weight. There is a safe distance between them which avoids any disaster. Pulleys are easy to assemble and cost-effective. The combination of different directional pulleys can change the position of the load with little effort. Though there are moving parts in the pulley system they require less or no lubrication after installation.
Disadvantages
Apart from the above-said advantages while using pulley systems, there are several disadvantages in their use. The main disadvantage in the use of the pulley system is that it requires large space to install and operate. The mechanical advantage of pulleys can go to higher values but need more space to install them.
In some cases, the ropes/belts move over the wheel with no grooves, the chances of the slip of ropes/belts from the wheel are inevitable. If the system is installed to use for a long time, they require maintenance and regular check-up of ropes/cables as the friction between the wheels and cables/ropes occur causing wear and tear to them. Continuous use of the system makes the ropes weak. The rope may break while using the system causing damages to the operator, surrounding place and the load which is being lifted.
