All of these are true about using adhesive EXCEPT:

___________ is NOT a common injury that an automotive tech may experience at work.

Degger [83]

Answer:The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

Explanation:

The most common injuries were sprains/strains (39% of the total), lacerations (22%), and contusions (15%). Forty-nine percent of the injuries resulted in one or more lost or restricted workdays; 25% resulted in 7 or more lost or restricted workdays.

7 0

3 years ago

A three-phase line has a impedance of 0.4+j2.7 per phase. The line feeds 2 balanced three-phase loads that are connected in para

mamaluj [8]

Answer:

a) 4160 V

b) 12 kW and 81 kVAR

c) 54 kW and 477 kVAR

Explanation:

1) The phase voltage is given as:

$V_p=\frac{3810.5}{\sqrt{3} }=2200 V$

The complex power S is given as:

$S=560.1(0.707 +j0.707)+132=660\angle 36.87^o \ KVA$

$where\ S^*\ is \ the \ conjugate\ of \ S\\Therefore\ S^*=660\angle -36.87^oKVA$

The line current I is given as:

$I=\frac{S^*}{3V}=\frac{660000\angle -36.87}{3(2200)} =100\angle -36.87^o\ A$

The phase voltage at the sending end is:

$V_s=2200\angle 0+100\angle -36.87(0.4+j2.7)=2401.7\angle 4.58^oV$

The magnitude of the line voltage at the source end of the line ( $V_{sL}=\sqrt{3} |V_s|=\sqrt{3} *2401.7=4160V$

b) The Total real and reactive power loss in the line is:

$S_l=3|I|^2(R+jX)=3|100|^2(0.4+j2.7)=12000+j81000$

The real power loss is 12000 W = 12 kW

The reactive power loss is 81000 kVAR = 81 kVAR

c) The sending power is:

$S_s=3V_sI^*=3(2401.7\angle 4.58)(100\angle 36.87)=54000+j477000$

The Real power delivered by the supply = 54000 W = 54 kW

The Reactive power delivered by the supply = 477000 VAR = 477 kVAR

5 0

4 years ago

The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

skelet666 [1.2K]

Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

4 0

3 years ago

A steel loop ABCD of length 5ft and of 3/8" diameter is placed as shown around a 1" diameter aluminum rod AC. Cables BE and DF e

lorasvet [3.4K]

Answer:

a

Explanation:

ye men thats the answer#######%

4 0

3 years ago

Determine the carburizing time necessary to achieve a carbon concentration of 0.30 wt% at a position 4 mm into an iron–carbon al

Ahat [919]

Answer:

the carburizing time necessary to achieve a carbon concentration is 31.657 hours

Explanation:

Given the data in the question;

To determine the carburizing time necessary to achieve the given carbon concentration, we will be using the following equation:

(Cs - Cx) / (Cs - C0) = ERF( x / 2√Dt)

where Cs is Concentration of carbon at surface = 0.90

Cx is Concentration of carbon at distance x = 0.30 ; x in this case is 4 mm = ( 0.004 m )

C0 is Initial concentration of carbon = 0.10

ERF() = Error function at the given value

D = Diffusion of Carbon into steel

t = Time necessary to achieve given carbon concentration ,

so

(Cs - Cx) / (Cs - C0) = (0.9 - 0.3) / (0.9 - 0.1)

= 0.6 / 0.8

= 0.75

now, ERF(z) = 0.75; using ERF table, we can say;

Z ~ 0.81; which means ( x / 2√Dt) = 0.81

Now, Using the table of diffusion data

D = 5.35 × 10⁻¹¹ m²/sec at (1100°C) or 1373 K

now we calculate the carbonizing time by using the following equation;

z = (x/2√Dt)

t is carbonizing time

so we we substitute in our values

0.81 = ( 0.004 / 2 × √5.35 × 10⁻¹¹ × √t)

0.81 = 0.004 / 1.4628 × 10⁻⁵ × √t

0.81 × 1.4628 × 10⁻⁵ × √t = 0.004

1.184868 × 10⁻⁵ × √t = 0.004

√t = 0.004 / 1.184868 × 10⁻⁵

√t = 337.5903

t = ( 337.5903)²

t = 113967.21 seconds

we convert to hours

t = 113967.21 / 3600

t = 31.657 hours

Therefore, the carburizing time necessary to achieve a carbon concentration is 31.657 hours

7 0

3 years ago