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Leya [2.2K]
3 years ago
15

Please help meeeee. I don’t have time

Engineering
1 answer:
Sholpan [36]3 years ago
3 0

Answer: precision

Explanation: because accuracy is right on there and precision is getting closer and closer

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Teresa is emotionally volatile, particularly with friends and boyfriends. She is extremely dramatic about even the smallest disa
Ilia_Sergeevich [38]

Answer: Borderline personality Disorder.

Explanation: This is a type of mental disorder which could affects mood, behavior and relationships.

Its symptoms includes unstable emotions, sense of insecurity, worthlessness, and impulsivity.

This condition can not be cured, but treatments such as therapies, medication (in some cases) could help.

4 0
3 years ago
What is the net force acting on a car cruising at a constant velocity of 70 km/h (a) on a level road and (b) on an uphill road?
ElenaW [278]

Answer:

a) zero b) zero

Explanation:

Newton's first law tells us that a body remains at rest or in uniform rectilinear motion, if a net force is not applied on it, that is, if there are no applied forces or If the sum of forces acting is zero. In this case there is a body that moves with uniform rectilinear motion which implies that there is no net force.

4 0
3 years ago
The gas expanding in the combustion space of a reciprocating engine has an initial pressure of 5 MPa and an initial temperature
Anit [1.1K]

Answer:

a). Work transfer = 527.2 kJ

b). Heat Transfer = 197.7 kJ

Explanation:

Given:

P_{1} = 5 Mpa

T_{1} = 1623°C

                       = 1896 K

V_{1} = 0.05 m^{3}

Also given \frac{V_{2}}{V_{1}} = 20

Therefore, V_{2} = 1  m^{3}

R = 0.27 kJ / kg-K

C_{V} = 0.8 kJ / kg-K

Also given : P_{1}V_{1}^{1.25}=C

   Therefore, P_{1}V_{1}^{1.25} = P_{2}V_{2}^{1.25}

                     5\times 0.05^{1.25}=P_{2}\times 1^{1.25}

                     P_{2} = 0.1182 MPa

a). Work transfer, δW = \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

                                  \left [\frac{5\times 0.05-0.1182\times 1}{1.25-1}  \right ]\times 10^{6}

                              = 527200 J

                             = 527.200 kJ

b). From 1st law of thermodynamics,

Heat transfer, δQ = ΔU+δW

   = \frac{mR(T_{2}-T_{1})}{\gamma -1}+ \frac{P_{1}V_{1}-P_{2}V_{2}}{n-1}

  =\left [ \frac{\gamma -n}{\gamma -1} \right ]\times \delta W

  =\left [ \frac{1.4 -1.25}{1.4 -1} \right ]\times 527.200

  = 197.7 kJ

6 0
3 years ago
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
Harman [31]

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

7 0
3 years ago
When _____ ,the lithium ions are removed from the_____ and added into the _____
bezimeni [28]

Answer:

b. Discharging; anode; cathode

Explanation:

When discharging , it means the battery is producing a flow electric current, the lithium ions are released from the  anode to the cathode which generates the flow of electrons from one side to another. When charging Lithium ions are released by the cathode and received by the anode.

8 0
3 years ago
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