___________ are used in an automotive shop and can be harmful to the environment if not disposed of properly.Oil filters
Fluorescent lamps
Mercury-containing lamps
All of the above

Answers

Answer 1
Answer:

Answer: D all above

Explanation:

Jus done it


Related Questions

1. (5 pts) An adiabatic steam turbine operating reversibly in a powerplant receives 5 kg/s steam at 3000 kPa, 500 °C. Twenty percent of the flow is extracted at 1000 kPa to a feedwater heater and the remainder flows out at 200 kPa. Find the two exit temperatures and the turbine power output.
Consider the products you use and the activities you perform on a daily basis. Describe three examples that use both SI units and customary units for measurement.
It is proposed to deposit a 5 μm thick nickel coating uniformly on all surfaces of a ceramic strip measuring 15 cm x 5 cm x 2 cm by employing a vapor-phase deposition (evaporation-condensation) technique. The vapor pressure-temperature relationship for liquid Ni is of the following form: ln p (atm) = -(51,590/T) – 2.01 ln T + 32.40. The normal melting point and boiling point of nickel are 1453°C and 2730°C, respectively, and the density and atomic weights of Ni are 8.91 g.cm^-3 and 58.71 atomic mass units respectively. Calculate the energy in joules needed to evaporate the required quantity of nickel.
Annealing is a process by which steel is reheated and then cooled to make it less brittle. Consider the reheat stage for a 100-mm-thick steel plate ( 7830 kg/m3 , c 550 J/kg K, k 48 W/m K), which is initially at a uniform temperature of Ti 200 C and is to be heated to a minimum temperature of 550 C. Heating is effected in a gas-fired furnace, where products of combustion at T 800 C maintain a convection coefficient of h 250 W/m2 K on both surfaces of the plate. How long should the plate be left in the furnace
The information on a can of pop indicates that the can contains 360 mL. The mass of a full can of pop is 0.369 kg, while an empty can weighs 0.153 N. Determine the specific weight, density, and specific gravity of the pop and compare your results with the corresponding values for water at Express your results in SI units.

Where should the Q point of an amplifier be on a DC load line to have the LARGEST linear output?

Answers

Answer:

At the middle of the DC load

Explanation:

For the Q point of an amplifier to have the Largest linear output. the Q plant has to be biased at the middle of the DC load line, this is because when the input voltage is low the transistor will be in the cutoff region while when the input voltage is very high the transistor will be in the saturation, hence when the Q point is biased at the middle it is will be higher linearly in relation to the active region

In an adiabatic process, the temperature of the system remains constant. a)- True b)- False

Answers

Answer:

A. True

Explanation:

If the specific surface energy for magnesium oxide is 1.0 J/m2 and its modulus of elasticity is (225 GPa), compute the critical stress required for the propagation of an internal crack of length 0.8 mm.

Answers

Answer:

critical stress required is  18.92 MPa

Explanation:

given data

specific surface energy = 1.0 J/m²

modulus of elasticity = 225 GPa

internal crack of length = 0.8 mm

solution

we get here one half length of internal crack that is

2a = 0.8 mm

so a = 0.4 mm = 0.4 × 10^(-3) m

so we get here critical stress that is

\sigma _c = \sqrt{(2E \gamma )/(\pi a)}     ...............1

put here value we get

\sigma _c =   \sqrt{(2* 225* 10^9 * 1 )/(\pi * 0.4* 10^(-3))}

\sigma _c =  18923493.9151 N/m²

\sigma _c =   18.92 MPa

In this type of projection, the angles between the three axes are different:- A) Isometric B) Axonometric C) Trimetric D) Dimetnic

Answers

Answer:

The correct answer is C) Trimetric

Explanation:

The most suitable answer is a trimetric projection because, in this type of projection, we see that the projection of the three angles between the axes are not equal. Therefore, to generate a trimetric projection of an object, it is necessary to have three separate scales.

Given that the debouncing circuit is somewhat expensive in terms of hardware (2 NAND gates, 2 resistors, and a double-pole, single throw switch), describe applications where you would require switch-debouncing circuits as well as applications where you would not need to include the additional hardware for switch debouncing (in other words, applications where you can tolerate switch bouncing). Note, you cannot use the clock and clear inputs of our lab as example applications; instead you need to think of other examples.

Answers

Explanation:

One of the common application of debouncing g circuit is in microprocessors or microcontrollers or FPGA's where fast processing is required. In such cases, it is extremely important that during the limited processing cycle, the signals remains valid without debouncinng. Because debouncing can complete impact the output of the controller.

A case where debouncing can be compromised where a system is run partially through human intervention or that has different indications for one operation.

For example in a car wash management system, where green and red lights are used to indicate if a car is being washed, green light will be on and then red light means that there no car in washing que

Air enters the first compressor stage of a cold-air standard Brayton cycle with regeneration and intercooling at 100 kPa, 300 K, with a mass flow rate of 6 kg/s. The overall compressor pressure ratio is 10, and the pressure ratios are the same across each compressor stage. The temperature at the inlet to the second compressor stage is 300 K. The turbine inlet temperature is 1400 K. The compressor stages and turbine each have isentropic efficiencies of 80% and the regenerator effectiveness is 80%. For k = 1.4, calculate: a. the thermal efficiency of the cycle
b. the back work ratio
c. the net power developed, in kW
d. the rates of exergy destruction in each compressor stage and the turbine stage as well as the regenerator, in kW, for T 0 = 300 K.

Answers

Answer:

a. \eta _(th) = 77.65%

b. bwr = 6.5%

c. 3538.986 kW

d. -163.169 kJ

Explanation:

a. The given property  are;

P₂/P₁ = 10, P₂ = 10 * 100 kPa = 1000 kPa

p₄/p₁ = 10

P₂/P₁ = p₄/p₃ = √10

p₂ = 100·√10

T_(2s) = T₁×(√10)^(0.4/1.4) = 300 × (√10)^(0.4/1.4) = 416.85 K

T₂ = T₁ + (T_(2s) - T₁)/\eta _c = 300 + (416.85 - 300)/0.8 = 446.0625 K

p₄ = 10×p₁ = 10×100 = 1000 kPa

p₄/p₃ = √10 =

p₃ = 100·√10

T₃ = 300 K

T₃/T_(4s) = (P₂/P₁)^((k - 1)/k) = (√10)^(0.4/1.4)

T_(4s) = T₃/((√10)^(0.4/1.4) ) = 300/((√10)^(0.4/1.4)) = 215.905 K

T₄ = T₃ + (T_(4s) - T₃)/\eta _c = 300 + (215.905- 300)/0.8 = 194.881 K

The efficiency = 1 - (T₄ - T₁)/(T₃ - T₂) = 1 - (194.881 -300)/(300 -446.0625 ) = 0.28

T₄ = 446.0625 K

T₆ = 1400 K

T_(7s)/T₆ = (1/√10)^(0.4/1.4)

T_(7s) = 1400×(1/√10)^(0.4/1.4)  = 1007.6 K

T₇ = T₆ - \eta _t(T₆ - T_(7s)) = 1400 - 0.8*(1400 - 1007.6) = 1086.08 K

T₈ = 1400 K

T₉ = 1086.08 K

T₅ = T₄ + \epsilon _(regen)(T₉ - T₄) = 446.0625 +0.8*(1086.08 - 446.0625) = 958.0765 K

\eta _(th) =(((T₆ - T₇) + (T₈ - T₉)) -((T₂ - T₁) + (T₄ - T₃)))/((T₆ - T₅) + (T₈ - T₇))

(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300)))/((1400 -958.0765 ) + (1400 -1086.08 )) = 0.7765

\eta _(th) = 77.65%

b. Back work ratio, bwr = bwr = (w_(c,in))/(w_(t,out))

((446.0625 - 300)+(194.881 - 300))/((1400 - 1086.08) + (1400 -1086.08 ))

40.9435/627.84 = 6.5%

c. w_(net, out) = c_p[(T_6 -T_7) + (T_8 - T_9)] - [(T_2 - T_1) + (T_4 -T_3)]

Power developed is given by the relation;

\dot m \cdot w_(net, out)

\dot m \cdot w_(net, out)= 6*1.005*(((1400 - 1086.08) + (1400 -1086.08 ))-((446.0625 - 300)+(194.881 - 300))) = 3538.986 kW

d. Exergy destruction = 6*(1.005*(300-446.0625 ) - 300*1.005*(-0.3966766)

-163.169 kJ

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