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Doss [256]
3 years ago
13

You push a heavy crate out of a carpeted room and down a hallway with a waxed linoleum floor. While pushing the crate 2.5 m out

of the room you exert a force of 385 N; while pushing it 10 m down the hallway you exert a force of 130 N. To slow it down you exert a force of 350 N through a distance of 2.0 m, opposite to the motion. How much work do you do in all?
Physics
1 answer:
Alinara [238K]3 years ago
3 0

Answer:

W = 1562.5 J

Explanation:

Path 1:

W₁ = F₁*d₁ = 385 N * 2.5 m = 962.5 J

Path 2:

W₂ = F₂*d₂ = 130 N * 10 m = 1300 J

Path 3:

W₃ = F₃*d₃ = (-350 N) * 2 m = - 700 J  (opposite to the motion)

We get

W = W₁ + W₂ + W₃ = 962.5 J + 1300 J + (- 700 J) = 1562.5 J

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An air-standard Diesel cycle has a compression ratio of 16 and a cutoff ratio of 2. At the beginning of the compression process,
Sedbober [7]

Answer:

a.T_3=1723.8kPa\\b.n=0.563\\c.MEP=674.95kPa

Explanation:

a. Internal energy and the relative specific volume at s_1 are determined  from A-17:u_1=214.07kJ/kg, \ \alpha_r_1=621.2.

The relative specific volume at s_2 is calculated from the compression ratio:

\alpha_r_2=\frac{\alpha_r_1}{r}\\=\frac{621.2}{16}\\=38.825

#from this, the temperature and enthalpy at state 2,s_2 can be determined using interpolations T_2=862K and h_2=890.9kJ/kg. The specific volume at s_1 can then be determined as:

\alpha_1=\frac{RT_1}{P_1}\\\\=\frac{0.287\times 300}{95} m^3/kg\\0.906316m^3/kg

Specific volume,s_2:

\alpha_2=\frac{\alpha_1}{r}\\=\frac{0.906316}{16}m^3/kg\\=0.05664m^3/kg

The pressures at s_2 \ and\  s_3 is:

P_2=P_3=\frac{RT_2}{\alpha_2}\\\\=\frac{0.287\times862}{0.05664}\\=4367.06kPa

.The thermal efficiency=> maximum temperature at s_3 can be obtained from the expansion work at constant pressure during s_2-s_3

\bigtriangleup \omega_2_-_3=P(\alpha_3-\alpha_2)\\R(T_3-T_2)=P\alpha(r_c-1)\\T_3=T_2+\frac{P\alpha_2}{R}(r_c-1)\\\\=(862+\frac{4367\times 0.05664}{0.287}(2-1))K\\=1723.84K

b.Relative SV and enthalpy  at s_3 are obtained for the given temperature with interpolation with data from A-17 :a_r_3=4.553 \ and\  h_3=1909.62kJ/kg

Relative SV at s_4 is

a_r_4=\frac{r}{r_c}\alpha _r_3

==\frac{16}{2}\times4.533\\=36.424

Thermal efficiency occurs when the heat loss is equal to the internal energy decrease and heat gain equal to enthalpy increase;

n=1-\frac{q_o}{q_i}\\=1-\frac{u_4-u_1}{h_3-h_2}\\=1-\frac{65903-214.07}{1909.62-890.9}\\=0.563

Hence, the thermal efficiency is 0.563

c. The mean relative pressure is calculated from its standard definition:

MEP=\frac{\omega}{\alpa_1-\alpa_2}\\=\frac{q_i-q_o}{\alpha_1(1-1/r)}\\=\frac{1909.62-890.9-(65903-214.7)}{0.90632(1-1/16)}\\=674.95kPa

Hence, the mean effective relative pressure is 674.95kPa

3 0
3 years ago
I need help with question 4 please​
Yuki888 [10]

Answer:

V = 10 km / 1 hr = 10 km/hr

V = -10 j km / hr   if one were to use i, j, k as unit vectors with the usual orientation

4 0
2 years ago
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