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djverab [1.8K]
3 years ago
14

An object of mass 521 kg, initially having a velocity of 90 m/s, decelerates to a final velocity of 14 m/s. What is the change i

n kinetic energy of the object, in kJ?
Engineering
1 answer:
Harman [31]3 years ago
7 0

Answer:2058.992KJ

Explanation:

Given data

Mass of object\left ( m\right )=521kg

initial velocity\left ( v_0\right )=90m/s

Final velocity\left ( v\right )=14m/s

kinetic energy of body is given by=\frac{1}{2}mv^{2}

change in kinectic energy is given by substracting  final kinetic energy from initial kinetic energy of body.

Change in kinetic energy=\frac{1}{2}\times m\left ( V_0^{2}-V^2\right )

Change in kinetic energy=\frac{1}{2}\times521\left ( 90^{2}-14^2\right )

Change in kinetic energy=2058.992KJ

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Given a two-dimensional steady inviscid air flow field with no body forces described by the velocity field given below. Assuming
kolbaska11 [484]

Answer:

the pressure gradient in the x direction = -15.48Pa/m

Explanation:

  • The concept of partial differentiation was used in the determination of the expression for u and v.
  • each is partially differentiated with respect to x and the appropriate substitution was done to get the value of the pressure gradient as shown in the attached file.

4 0
3 years ago
A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3
labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass co_2 = 1 kg

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of co_2 = 44

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

V_1 = \frac{m RT_1}{P_1}

       =\frac{1*0.189*298}{104}

V_1 = 0.5415 m3

V_2 = \frac{m RT_2}{P_2}

     =\frac{1*0.189*584}{1068}

V_1 = 0.1033 m3

WORK DONE

W =P_{avg}*{V_2-V_1}

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

\Delta U  = m *c_v*{V_2-V_1}

\Delta U  = 1*0.657*(584-298)

\Delta U  =187.902 kJ

HEAT TRANSFER

Q = \Delta U  +W

   = 187.902 +(-256.46)

Q = -68.859 kJ

7 0
3 years ago
Does the Diesel engine have engine knock or detonation problem? Why?
Luda [366]

Explanation:

Yes Diesel engine have problem of knocking.

We know that knocking is phenomenon in which suddenly large amount of power generates this large amount of power will cause the failure of diesel engine.

Actually when one set of fuel inject inside the cylinder to burn with already compressed air (in general up to 10-15 bar) then this fuel does not burn complete and accumulate inside the cylinder.After that second set of fuel inject inside the cylinder then that one set of fuel burns with second set of fuel and produces large amount of sudden power for engine and causes the breaks in the crank or connecting rod of engine.it leads to damage the engine.

6 0
3 years ago
What type of car engine is best for cold weather.
Komok [63]

Answer:Antifreeze/coolant

Explanation: keeps your engine cool in warm weather and keeps it from freezing up in the winter. A 50-50 mix of full strength coolant and water generally protects to around -30 degrees Fahrenheit. Make sure you check with the supplier or your owner's manual for the correct formulation

5 0
2 years ago
The temperature of a system rises by 10 °C during a heating process. Express the rise in temperature of K, R, and °F.
Lorico [155]

Explanation:

Given T = 10 °C

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T = (10 + 273.15) K = 283.15 K

<u>T = 283.15 K </u>

The conversion of T( °C) to T(F) is shown below:

T (°F) = (T (°C) × 9/5) + 32  

So,

T (°F) = (10 × 9/5) + 32 = 50 °F

<u>T = 50 °F</u>

The conversion of T( °C) to T(R) is shown below:

T (R) = (T (°C) × 9/5) + 491.67

So,

T (R) = (10 × 9/5) + 491.67 = 509.67 R

<u>T = 509.67 R</u>

6 0
3 years ago
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