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yuradex [85]
3 years ago
9

I just need help on problem B

Engineering
1 answer:
Komok [63]3 years ago
8 0
The problem b is An example of a law being implemented by bureaucratic agency is____
A. Congress passing legislation by major vote
B. OSHA deciding how hazardous substance should be stored
C. the USPS delivering mail throughout the country
D. A watchdog group blowing the whistle on government corruption
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What is the tolerance of number 4?
Kamila [148]

Answer:

Answer: ±0.02 units or 20±0.02 units or 19.98-20.02 units depending on how they prefer its written (typically the first or second one)

Explanation:

says on the sheet. Unless otherwise stated 0.XX = ±0.02 tolerance

(based on image sent in other post)

5 0
3 years ago
A 20cm-long rod with a diameter of 0.250 cm is loaded with a 5000 N weight. If the diameter of the bar is 0.490 at this load, de
Margaret [11]

If the diameter of the bar is 0.490 at this load, determine I. the engineering stress and strain, and [2] II. the true stress and strain is 1561. 84 MPa.

<h3>What is strain?</h3>

Strain is a unitless degree of ways a great deal an item receives larger or smaller from an implemented load. Normal stress happens while the elongation of an item is in reaction to an everyday pressure (i.e. perpendicular to a surface), and is denoted via way of means of the Greek letter epsilon.

  1. L = 20 cm d x 1 = 0.21 cm
  2. dx 2 = 0.25 cmF=5500 a) σ= F/A1= 5000/(π/4x(0.0025)^2)= 1018.5916 MPa lateral stress= Ad/d1= (0.0021-0.0025)/0.0025 = - 0.1 longitudinal stress (ɛ_l)= -lateral stress/v = -(-0.16)/0.3
  3. (assuming a poisson's ration of 0.3) ε_l=0.16/0.3 = 0.5333
  4. b) σ_true= σ(1+ ɛ_I)= 1018.5916(1+0.5333
  5. = 1561.84 MPa.

Read more about the diameter :

brainly.com/question/358744

#SPJ1

4 0
2 years ago
A non-inductive load takes a current of 15A at 125V. An inductor is then connected in series in order that the same current shal
Norma-Jean [14]

Answer:

The inductance of the inductor is 0.051H

Explanation:

From Ohm's law;

  V = IR .................. 1

The inductor has its internal resistance referred to as the inductive reactance, X_{L}, which is the resistance to the flow of current through the inductor.

From equation 1;

V = IX_{L}

X_{L} = \frac{V}{I} ................ 2

Given that; V = 240V, f = 50Hz, \pi = \frac{22}{7}, I = 15A, so that;

From equation 2,

X_{L}= \frac{240}{15}

    = 16Ω

To determine the inductance of the inductor,

X_{L} = 2\pifL

L = \frac{X_{L} }{2 \pi f}

  = \frac{16}{2*\frac{22}{7}*50 }

 = 0.05091

The inductance of the inductor is 0.051H.

4 0
4 years ago
Determine the initial void ratio, the relative density and the unit weight (in pounds per cubic foot) of the specimens for each
Elina [12.6K]

The initial void ratio is the <em>parameter </em>which is used to show the structural foundations for each <em>specimen of sand </em>so that the method and speed of compression would be <em>measured</em>.

Relative density is the mass per unit volume of each specimen of sand which is <em>measured </em>and it has to do with the<em> relative ratio</em> of the density of the sand.

Unit weight is the the exact weight per cubic foot of the sand which is measured.

Please note that your question is incomplete so I gave you a general overview to help you better understand the concept

Read more here:

brainly.com/question/15220801

5 0
3 years ago
Nitrogen (N2) contained in a piston–cylinder arrangement, initially at 10 bar and 405 K, undergoes an expansion to a final tempe
yKpoI14uk [10]

Answer:28.21 kJ/kg

Explanation:

Given

P_1=10\ bar

T_1=405\ K

T_2=300\ K

Process PV^{1.3}=constant

Work done for Polytropic process

W=\dfrac{P_1V_1-P_2V_2}{n-1}

where n=Polytropic index

W=\dfrac{R(T_1-T_2)}{n-1}

W=\dfrac{0.296(405-300)}{1.3-1}\quad [R_{N_2}=\frac{8.314}{28}]

W=103.6\ kJ\kg

Now Calculating change in Internal energy

\Delta U=c_v(T_2-T_1)

\Delta U=0.718\times (300-405)

\Delta U=-75.39\ kJ/kg

Now applying First law concept

\Delta U=Q-W

Q=W+\Delta U

Q=103.6-75.392

Q=28.21\ kJ/kg

6 0
4 years ago
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