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My name is Ann [436]
2 years ago
9

What is the name of this piece?

Engineering
1 answer:
Ivanshal [37]2 years ago
7 0

Answer: block o’ metal

Explanation:

You might be interested in
This problem has been solved!
lisov135 [29]

Answer: a) 135642 b) 146253

Explanation:

A)

1- the bankers algorithm tests for safety by simulating the allocation for predetermined maximum possible amounts of all resources, as stated this has the greatest degree of concurrency.

3- reserving all resources in advance helps would happen most likely if the algorithm has been used.

5- Resource ordering comes first before detection of any deadlock

6- Thread action would be rolled back much easily of Resource ordering precedes.

4- restart thread and release all resources if thread needs to wait, this should surely happen before killing the thread

2- only option practicable after thread has been killed.

Bii) ; No. Even if deadlock happens rapidly, the safest sequence have been decided already.

5 0
3 years ago
Which rigid motion maps the solid-line figure onto the dotted-line figure?
Agata [3.3K]
I would love to answer but unfortunately there is no picture.
5 0
3 years ago
Water is pumped steadily through a 0.10-m diameter pipe from one closed pressurized tank to another tank. The pump adds 4.0 kW o
jekas [21]

Complete Question

Complete Question is attached below.

Answer:

V'=5m/s

Explanation:

From the question we are told that:

Diameter d=0.10m

Power P=4.0kW

Head loss \mu=10m

 \frac{P_1}{\rho g}+\frac{V_1^2}{2g}+Z_1+H_m=\frac{P_2}{\rho g}+\frac{V_2^2}{2g}+Z_2+\mu

 \frac{300*10^3}{\rho g}+35+Hm=\frac{500*10^3}{\rho g}+15+10

 H_m=(\frac{200*10^3}{1000*9.8}-10)

 H_m=10.39m

Generally the equation for Power is mathematically given by

 P=\rho gQH_m

Therefore

 Q=\frac{P}{\rho g H_m}

 Q=\frac{4*10^4}{1000*9.81*10.9}

 Q=0.03935m^3/sec

Since

 Q=AV'

Where

 A=\pi r^2\\A=3.142 (0.05)^2

 A=7.85*10^{-3}

Therefore

 V'=\frac{0.03935m^3/sec}{7.85*10^{-3}}

 V'=5m/s

5 0
3 years ago
1. Fatigue equations are based solely on theoretical assumptions. Experimental data is only used to verify the theory. a. True.b
Rainbow [258]

Answer:

1.  b. False

2. b. False

3.  b. False

4.  b. False

5. a. True

6. a. True

7.  b. False

8.  b. False

9. a. True

Explanation:

1. The fatigue properties of a material  are determined by series of test.

2. For most steels there is a level of fatigue limit below which a component will survive an infinite number of cycles, for aluminum and titanium a fatigue limit can not be defined, as failure will eventually occur after enough experienced cycles.

3. Although there is a cyclic stress, there are also stresses complex circumstances involving tensile to compresive and constant stress, where the solution is given into the mean stress and the stress amplitude or stress range, which is double the stress amplitude.

4. Low‐cycle fatigue is defined as few thousand cycles and high cycle fatigue is around more than 10,000 cycles.

5. The number of cycles for failure on brittle materials are less and determined compared with the ductile materials.

6.  The bending fatigue could be handled with specific load requirements  for uniform bending or axial fatigue of the same section size where the material near the surface is subjected to the  maximum stress, as in torsional fatigue, which can be performed on  axial-type specially designed machines also, using the proper fixtures if  the maximum twist required is small, in which linear motion is changed to rotational motion.

7.  A SN-Curve for a given material, is a plot displayed on logarithmic scales of the magnitude of an alternating stress in relation to the number of cycles to failure

8. The strain life method measures the strain resistance of local stresses and strains around stress concentration that controls the fatigue life of the material. It is more accurate than determining fatigue performance as the stress-life method is for long life millions of cycles in elastic stresses, but an it gets an effective stress concentration in fatigue loading.

9. Linear Elastic Fracture Mechanics (LEFM) states that the material is isotropic and linear elastic so, when the stresses near the crack surpasses the material fracture toughness, the crack grows.

7 0
3 years ago
An engine indicator is used to determine the following: (a) speed (d) m.e.p, and IHP (e) BHP (b) temperature (c) volume of cylin
Greeley [361]

Answer:

Option (d) MEP and IHP

Explanation:

MEP stands for Mean Effective Pressure and IHP stands for Indicated Horse Power

In engines (Internal Combustion), engine indicator is generally to indicate the indicate the changes in pressure inside the cylinder of an Internal Combustion Engine or IC engines. Once, Mean Effective Pressure of the engine is calculated it further helps to calculate the Horse power and both these quantities, i.e., MEP and IHP are displayed on the engine indicator.

8 0
3 years ago
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