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Serjik [45]
3 years ago
9

Student A says hazardous waste can take the form of solid, liquid, or gas. Student B says hazardous waste can only take the form

of a solid. Who is correct?
Group of answer choices

Both A and B

Neither A nor B

Student A

Student B

Engineering
1 answer:
lina2011 [118]3 years ago
4 0

Answer:

Student A

Explanation:

hope this helps have a great day

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A ball A is thrown vertically upward from the top of a 30-m-high building with an initial velocity of 5 m>s. At the same inst
expeople1 [14]

Answer:

s= 20.4 m  

Explanation:

First lets write down equations for each ball:  

s=so+vo*t+1/2a_c*t^2

for ball A:

s_a=30+5*t+1/2*9.81*t^2

for ball B:  

s_b=20*t-1/2*9.81*t^2

to find time deeded to pass we just put that

s_a = s_b  

30+5*t-4.91*t^2=20*t-4.9*t^2

t=2 s  

now we just have to put that time in any of those equations an get distance from the ground:  

s = 30 + 5*2 -1/2*9.81 *2^2  

s= 20.4 m  

6 0
3 years ago
Steam enters a turbine at 8000 kPa, 440oC. At the exit, the pressure and quality are 150 kPa and 0.19, respectively.
levacccp [35]

Answer:

\dot W_{out} = 3863.98\,kW

Explanation:

The turbine at steady-state is modelled after the First Law of Thermodynamics:

-\dot Q_{out} -\dot W_{out} + \dot m \cdot (h_{in}-h_{out}) = 0

The specific enthalpies at inlet and outlet are, respectively:

Inlet (Superheated Steam)

h_{in} = 3353.1\,\frac{kJ}{kg}

Outlet (Liquid-Vapor Mixture)

h_{out} = 890.1\,\frac{kJ}{kg}

The power produced by the turbine is:

\dot W_{out}=-\dot Q_{out} + \dot m \cdot (h_{in}-h_{out})

\dot W_{out} = -2.93\,kW + (1.57\,\frac{kg}{s} )\cdot (3353.1\,\frac{kJ}{kg} - 890.1\,\frac{kJ}{kg} )

\dot W_{out} = 3863.98\,kW

8 0
3 years ago
Refrigerant 134a enters an air conditioner compressor at 4 bar, 20 C, and is compressed at steady state to 12 bar, 80 C. The vol
sleet_krkn [62]

Answer:

Q=15.7Kw

Explanation:

From the question we are told that:

Initial Pressure P_1=4bar

Initial Temperature T_1=20 C

Final Pressure  P_2=12 bar

Final Temperature T_2=80C

Work Output W= 60 kJ/kg

Generally Specific Energy from table is

At initial state

 P_1=4bar \& T_1=20 C

 E_1=262.96KJ/Kg

With

Specific Volume V'=0.05397m^3/kg

At Final state

 P_2=12 bar \& P_2=80C

 E_1=310.24KJ/Kg

Generally the equation for The Process is mathematically given by

 m_1E_1+w=m_2E_2+Q

Assuming Mass to be Equal

 m_1=m_1

Where

 m=\frac{V}{V'}

 m=frac{0.06666}{V'=0.05397m^3/kg}

 m=1.24

Therefore

 1.24*262.96+60)=1.24*310.24+Q

 Q=15.7Kw

4 0
3 years ago
Can i have answer of this question please?
cestrela7 [59]

uh its a tough one mate

3 0
3 years ago
In an ideal gas, specific enthalpy is a function of i. Entropy ii. Temperature iii, Pressure iv. Mass
Mice21 [21]

Answer:

Temperature

Explanation:

In an ideal gas the specific enthalpy  is exclusively a function of Temperature only this can be also written as h = h(T)  

A gas is said be ideal gas if obeys PV= nRT law

And in a ideal gas both internal energy and specific enthalpy are a function of Temperature only. Therefore the constant volume and constant pressure specific heats Cv and Cp are also function of temperature only.

5 0
3 years ago
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