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Ket [755]
3 years ago
9

Ben leads a team of a few engineers at a robotics firm. A couple of them would like to improve their skills by taking additional

courses in specialized areas. What are these courses in specialized areas called? A. technical certificate courses B. undergraduate certificate courses C. master’s degree courses D. online courses
Engineering
1 answer:
Anit [1.1K]3 years ago
4 0

Answer:

i dont know

Explanation:

You might be interested in
Based on the concept that it is better to prevent falls happening in the first place, which of the following safety methods meet
alekssr [168]

Answer:fall arrest harness

Explanation:cuz it’s just right

7 0
3 years ago
1 kg of saturated steam at 1000 kPa is in a piston-cylinder and the massless cylinder is held in place by pins. The pins are rem
BARSIC [14]

Answer:

The final specific internal energy of the system is 1509.91 kJ/kg

Explanation:

The parameters given are;

Mass of steam = 1 kg

Initial pressure of saturated steam p₁ = 1000 kPa

Initial volume of steam, = V₁

Final volume of steam = 5 × V₁

Where condition of steam = saturated at 1000 kPa

Initial temperature, T₁  = 179.866 °C = 453.016 K

External pressure = Atmospheric = 60 kPa

Thermodynamic process = Adiabatic expansion

The specific heat ratio for steam = 1.33

Therefore, we have;

\dfrac{p_1}{p_2} = \left (\dfrac{V_2}{V_1} \right )^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

Adding the effect of the atmospheric pressure, we have;

p = 1000 + 60 = 1060

We therefore have;

\dfrac{1060}{p_2} = \left (\dfrac{5\cdot V_1}{V_1} \right )^{1.33}

P_2= \dfrac{1060}{5^{1.33}}  = 124.65 \ kPa

\left [\dfrac{V_2}{V_1} \right ]^k = \left [\dfrac{T_1}{T_2}   \right ]^{\dfrac{k}{k-1}}

\left [\dfrac{V_2}{V_1} \right ]^{k-1} = \left \dfrac{T_1}{T_2}   \right

5^{0.33} = \left \dfrac{T_1}{T_2}   \right

T₁/T₂ = 1.70083

T₁ = 1.70083·T₂

T₂ - T₁ = T₂ - 1.70083·T₂

Whereby the temperature of saturation T₁ = 179.866 °C = 453.016 K, we have;

T₂ = 453.016/1.70083 = 266.35 K

ΔU = 3×c_v×(T₂ - T₁)

c_v = cv for steam at 453.016 K = 1.926 + (453.016 -450)/(500-450)*(1.954-1.926) = 1.93 kJ/(kg·K)

cv for steam at 266.35 K = 1.86  kJ/(kg·K)

We use cv given by  (1.93 + 1.86)/2 = 1.895 kJ/(kg·K)

ΔU = 3×c_v×(T₂ - T₁) = 3*1.895 *(266.35 -453.016) = -1061.2 kJ/kg

The internal energy for steam = U_g = h_g -pV_g

h_g = 2777.12 kJ/kg

V_g = 0.194349 m³/kg

p = 1000 kPa

U_{g1} = 2777.12 - 0.194349 * 1060 = 2571.11 kJ/kg

The final specific internal energy of the system is therefore, U_{g1} + ΔU = 2571.11 - 1061.2 = 1509.91 kJ/kg.

3 0
3 years ago
A particular cloud-to-ground lightning strike lasts 500 µµsec and delivers 30 kA across a potential difference of 100 MV. Assu
sasho [114]

Answer:

a) 15 C charge was delivered by the lightening bolt

b) the lightning delivered 3.0 × 10¹² W of power

c)

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is 416666.67 Wh

d)

the residential retail value of the energy delivered by the strike is $ 40.83

e)

a total of 26 lightening strikes would be required to power an average US home for a year.

Explanation:

Given that;

the lighting strike lasted for t ( time ) = 500 μsecs = 500×10⁻⁶ s

Current I = 30 kA = 30×10³ A

voltage V = 100 mV = 100×10⁶ v

a)

we know that; I = Q/t

so Q = I × t

we substitute

Q =  30×10³ × 500×10⁻⁶

Q = 15 C

Therefore 15 C charge was delivered by the lightening bolt

b)

Power P = V × I

we substitute

Power P = 100×10⁶ × 30×10³

P = 3.0 × 10¹² W

Therefore, the lightning delivered 3.0 × 10¹² W of power

c)

we know that; Power = Energy / Time

Energy = Power × Time

we substitute

E = 3.0 × 10¹²  × 500×10⁻⁶

E = 1500 × 10⁶ J

- total energy delivered by the lightening strike in J is 1500 × 10⁶ J

- total energy delivered by the lightening strike in Wh is;

⇒ 1500×10⁶ / 3600 Wh

= 416666.67 Wh

d)

given that;  1 KWh → $ 0.098

energy delivered by the strike = 416666.67 Wh = 416.66667 KWh

so the residential retail value of the energy delivered by the strike will be;

416.66667 KWh × $ 0.098

= $ 40.83

∴ the residential retail value of the energy delivered by the strike is $ 40.83

e)

Given that; average monthly residential energy consumption is 900 kWh.

for a year; energy consumption = 12 × 900 kWh = 10,800 KWh = 10800000 Wh

Now

1 lightening strike ⇒ 416666.67 Wh

x lightening strike ⇒ 10800000 Wh

x = 10800000 / 416666.67

x = 25.9199 ≈ 26

Therefore; a total of 26 lightening strikes would be required to power an average US home for a year.  

7 0
3 years ago
Can be used to eliminate rubbing friction of wheel touching frame. 1.Traction 2.Thrust washer
Vilka [71]

Answer:

thrust washer

can be used to eliminate rubbing friction of wheel touching frame

5 0
3 years ago
A two-phase mixture of water and steam with a quality of 0.63 and T = 300F expands isothermally until only saturated vapor rema
VMariaS [17]

Answer:

Explanation:

Hello!

To solve this problem you must follow the following steps, which are fully registered in the attached image.

1. Draw the complete outline of the problem.

2. Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties.

3. Use temodynamic tables to find the density of water in state 1, by means of temperature and quality, with this value and volume we can find the mass.

3. Use thermodynamic tables to find the internal energy in state 1 and two using temperature and quality.

4. uses the first law of thermodynamics that states that the energy in a system is always conserved, replaces the previously found values ​​and finds the work done.

5. draw the pV diagram using the 300F isothermal line

5 0
3 years ago
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