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MaRussiya [10]
3 years ago
7

Which of the following is true when an object of mass m moving on a horizontal frictionless surface hits and sticks to an object

of mass M > m, which is initially at rest on the surface?The collision is elastic.A. All of the initial kinetic energy of the less-massive object is lost.B. The momentum of the objects that are stuck together has a smaller magnitude than the initial momentum of the less massive object.C. The speed of the objects that are stuck together will be less than the initial speed of the less-massive object.D. The direction of motion of the objects that are stuck together depends on whether the hit is a head on collision
Physics
1 answer:
zhannawk [14.2K]3 years ago
3 0

Answer:

c)

Explanation:

A collision is said to be elastic when the total kinetic energy is the same after the collision. The speed of objects that are stuck together will always be less than the initial speed of the object that was in motion given that the other particle was at rest. It is because the kinetic energy of the system was due to the moving object. The objects have a greater overall mass when they are stuck. If the kinetic energy is the same and the mass increases, the velocity must decrease.

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A 12 V battery is connected to a 1200 Ω resistor. How much current is flowing through the resistor?
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A jet makes a landing traveling due east with a speed of 120 m/s .
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Average acceleration over a time interval lasting \Delta t is

a_{\rm ave}=\dfrac{\Delta v}{\Delta t}

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Which of the following best describes how heat is transferred by conduction?
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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,
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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
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