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2 years ago

Question in image. Question from OSHA.

Engineering

2 answers:

olganol [36]2 years ago

7 0

The answer would be 2. I hope u have a good day!

marin [14]2 years ago

3 0

Answer:

Explanation:

my sister did this and its the answer

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(40%) A bank wants to store the account number of its customers (an 8-digit number) in encrypted form on magnetic stripe ATM car

9966 [12]

Answer:

Explanation:

Attached is the solution to the question

3 0

3 years ago

Please read and answer each question carefully.

Klio2033 [76]

the answer is (c)

After the vehicle is involved in a car accident or fire

5 0

3 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago

Vector A extends from the origin to a point having polar coordinates (7, 70ᵒ ) and vector B extends from the origin to a point h

yaroslaw [1]

Answer:

13.95

Explanation:

Given :

Vector A polar coordinates = ( 7, 70° )

Vector B polar coordinates = ( 4, 130° )

To find A . B we will

A ( r , ∅ ) = ( 7, 70 )

A = rcos∅ + rsin∅

therefore ; A = 2.394i + 6.57j

B ( r , ∅ ) = ( 4, 130° )

B = rcos∅ + rsin∅

therefore ; B = -2.57i + 3.06j

Hence ; A .B

( 2.394 i + 6.57j ) . ( -2.57 + 3.06j ) = 13.95

8 0

2 years ago

A cylindrical insulation for a steam pipe has an inside radius rt = 6 cm, outside radius r0 = 8 cm, and a thermal conductivity k

goldfiish [28.3K]

Answer:

heat loss per 1-m length of this insulation is 4368.145 W

Explanation:

given data

inside radius r1 = 6 cm

outside radius r2 = 8 cm

thermal conductivity k = 0.5 W/m°C

inside temperature t1 = 430°C

outside temperature t2 = 30°C

to find out

Determine the heat loss per 1-m length of this insulation

solution

we know thermal resistance formula for cylinder that is express as

Rth = $\frac{ln\frac{r2}{r1}}{2 \pi *k * L}$ .................1

here r1 is inside radius and r2 is outside radius L is length and k is thermal conductivity

heat loss is change in temperature divide thermal resistance

Q = $\frac{t1- t2}{\frac{ln\frac{r2}{r1}}{2 \pi *k * L}}$

Q = $\frac{(430-30)*(2 \pi * 0.5 * 1}{ln\frac{8}{6} }$

Q = 4368.145 W

so heat loss per 1-m length of this insulation is 4368.145 W

4 0

3 years ago