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3 years ago

Effieciency of a machine is never 100%

Physics

1 answer:

tatyana61 [14]3 years ago

4 0

Answer:

in real life yes it's true because of energy lost in another way that we do not want. like car engine lost some energy as heat. So the energy from gasoline can't converted 100% to move the tire. Maybe it only 40-60 % energy use for move and the rest lost as heat and other

but sometime it can be 100% of efficiency on physics text book so that we can learn physics in the easiest way.

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Three identical train cars, coupled together, are rolling east at 1.8 m/s . A fourth car traveling east at 4.5 m/s catches up wi

Vedmedyk [2.9K]

Answer:

$v = 1.98\ m/s$

Explanation:

consider the mass of each train car be m

m₁ = m₂ = m₃ = m

speed of the three identical train

u₁ = u₂ = u₃ = 1.8 m/s

m₄ = m u₄ = 4.5 m/s

m₅ = m u₅ = 0 (initial velocity )

final velocity

v₁ = v₂ = v₃ = v₄ = v₅ = v

using conservation of momentum

m₁u₁ + m₂u₂ + m₃u₃ + m₄u₄ + m₅u₅ = m₁v₁ + m₂v₂ + m₃v₃ + m₄v₄ + m₅v₅

m (1.8 + 1.8 + 1.8 +4.5) = 5 m v

$v = \dfrac{9.9}{5}$

$v = 1.98\ m/s$

5 0

2 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

HELP NOW PLEASE ANSWER THE QUESTION THE PICTURE IS ATTACHED BELOW

bagirrra123 [75]

Answer:

Explanation:

6 0

2 years ago

Use the conditions provided in the previous problem. Atmosphere statically stable at the base of the mountain, where pressure =

Ann [662]

Answer:

change in relative vorticity 0.0590

Explanation:

Given data

pressure = 1000 hPa

temperature lapse rate q1 = 3.1◦C per 50 hPa

pressure = 850 hPa

temperature lapse rate q2= -0.61◦C per 50 hPa

to find out

change in relative vorticity

solution

we will apply here formula that is

N = (g / potential temperature ) × (potential vertical temperature) × exp^1/2 ............................1

here we know g = 9.8 m/s

and q1 = potential temperature=3.3 degree celsius

potential vertical temperature gradient = 3.1 - 0.61 / 1000 -850

potential vertical temperature gradient = 0.0166 degree celsius/hpa

N = 9.8 / 2.75 × 0.0166 × exp^1/2

N = 0.0590

8 0

3 years ago

Battery life is what distinguishes one type of mobile computer from another.

BlackZzzverrR [31]

I think the statement is false. It is not only the battery life that would distinguish one type of mobile computer from another. There are other aspects that would characterize every mobile computer. It could be the brand, the size, the operating system, the hardware specifications like the RAM, processor and the video card. Every mobile computer would vary depending on the brand, model and intended use.

8 0

3 years ago