11. Newton's Third Law of Motion relates to action and reaction. Which of the following scenarios accurately names the

A trip is taken that passes through the following points in order

riadik2000 [5.3K]

Answer:

35, I got you bro, i got you

8 0

3 years ago

Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ

Angelina_Jolie [31]

Answer:

a). $\frac{\dot{W}}{m}= 311$ kJ/kg

b). $\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

$\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0$

$\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})$

$\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})$

$\frac{\dot{W}}{m}= -30+1.1(980-670)$

$\frac{\dot{W}}{m}= 311$ kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

$\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}$

$\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}$

$\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978$

$\frac{\dot{\sigma _{gen}}}{m}=0.9113$ kJ/kg-K

5 0

3 years ago

True or False. In a parallel circuit, the voltage is the same anywhere in the circuit.

FrozenT [24]

True! hope this helps:)

7 0

3 years ago

Read 2 more answers

250 mL 0.1 M HCl solution is mixed with 250 mL

pishuonlain [190]

Answer:

The concentration of OH⁻ in the mixture is 0.05 M

Explanation:

The reaction of neutralization between HCl and NaOH is the following:

H⁺(aq) + OH⁻(aq) ⇄ H₂O(l)

The number of moles of HCl is:

$n_{HCl} = C*V = 0.1 mol/L*0.250L = 0.025 moles$

Similarly, the number of moles of NaOH is:

$n_{NaOH} = C*V = 0.2 mol/L*0.250L = 0.05 moles$

Now, from the reaction of HCl and NaOH we have the following number of moles of NaOH remaining:

$n_{NaOH} = 0.05 moles - 0.025 moles = 0.025 moles$

Finally, the concentration of OH⁻ in the mixture is:

$C =\frac{n_{NaOH}}{V_{T}}=\frac{0.025 moles}{0.250*2 L} = 0.05 moles/L$

Therefore, the concentration of OH⁻ in the mixture is 0.05 M.

I hope it helps you!

8 0

2 years ago

A car accelerates from rest to 90m/s in 3 seconds. What is the acceleration and how far did it travel?

Inessa [10]

Answer:

Explanation:

1.)acceleration= final velocity-initial velocity/time

90m/s-0m/s /3seconds

=30m/s^2

2.) distance= speed*time

90m/s*3seconds

=270m

4 0

3 years ago