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KengaRu [80]
3 years ago
8

1.

Physics
1 answer:
Mashutka [201]3 years ago
4 0

Answer:

a) acceleration

Explanation:

Acceleration is, by definition, the change of an object's velocity.

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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
A rowboat has a momentum of 241.5 kg*m/s. if the rowboat has a mass of 115kg, what its it's speed?
Lera25 [3.4K]
Momentum = mass*velocity(or speed)

241.5 kg*m/s = 115 kg * v
v = 241.5 kg*m/s / 115 kg = 2.10 m/s
6 0
3 years ago
Which term best describes a test used to answer a question?
Anastasy [175]

A) experiment. Is the answer.

hypothesis is the educated guess about what the result of the experiment is before conducting the experiment.

Observation is what you see and record during the experiment.

3 0
3 years ago
What is the car's speed at the bottom of the dip?The passengers in a roller coaster car feel 50% heavier thantheir true weight a
Rashid [163]

Answer:

v = 14 m/s

Explanation:

given,

radius of dip = 40 m

The passengers in a roller coaster car feel 50% heavier than their true weight.

Apparent weight

A = W + \dfrac{W}{2}

A =\dfrac{3W}{2}

A =\dfrac{3mg}{2}

When the car is at the bottom,  the weight will be acting downwards and the centripetal force will also be acting downward where as Normal force which is apparent weight will be acting in upward direction.

now,

N = m g + \dfrac{mv^2}{r}

\dfrac{3mg}{2} = m g + \dfrac{mv^2}{r}

\dfrac{mg}{2} = \dfrac{mv^2}{r}

v = \sqrt{\dfrac{rg}{2}}

v = \sqrt{\dfrac{40\times 9.8}{2}}

v = 14 m/s

8 0
3 years ago
A ball is dropped from rest from a height h above the ground. another ball is thrown vertically upwards from the ground at the i
Darya [45]

The position of the first ball is

y_1=h-\dfrac g2t^2

while the position of the second ball, thrown with initial velocity v, is

y_2=vt-\dfrac g2t^2

The time it takes for the first ball to reach the halfway point satisfies

\dfrac h2=h-\dfrac g2t^2

\implies\dfrac h2=\dfrac g2t^2

\implies t=\sqrt{\dfrac hg}

We want the second ball to reach the same height at the same time, so that

\dfrac h2=v\sqrt{\dfrac hg}-\dfrac g2\left(\sqrt{\dfrac hg}\right)^2

\implies h=2v\sqrt{\dfrac hg}-g\left(\dfrac hg\right)

\implies h=v\sqrt{\dfrac hg}

\implies v=\sqrt{hg}

8 0
3 years ago
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