Driving Distraction Brainstorming Session

two cars travel on a straight road from the point. A to point B both cars accelerate to their maximum speed and then continue at

Lady bird [3.3K]

Answer:

Part A

The acceleration of car 2, is larger than the acceleration of car 1

Part B

The average speed of car 2 is larger than the average speed of car 1

Explanation:

The question relates to kinematic motion

Using a similar question for the parameters, we have;

The time it takes car 1 to accelerate to v₁ = 20 m/s is t₁ = 30 s

The time it takes car 1 to reach point B, t₁₂ = 35 s

The time it takes car 2 to accelerate to v₂ = 20 m/s is t₂ = 20 s

The time it takes car 2 to reach point B is t₂₂ = 30 s

Part A

From the kinematic equation of motion, v = u + at, we have;

Acceleration, a = (v - u)/t

Where;

a = The acceleration

v = The final velocity

u = The initial velocity = 0 m/s for both cars as they start from rest

t = The time taken

The acceleration of car 1, a₁ = (v₁ - u₁)/t₁

∴ a₁ = (20 m/s - 0 m/s)/(30 s) = 2/3 m/s²

The acceleration of car 2, a₂ = (v₂ - u₂)/t₂

∴ a₂ = (20 m/s - 0 m/s)/(20 s) = 1 m/s²

The acceleration of car 2, a₂ = 1 m/s² is larger than the acceleration of car 1, a₁ = 2/3 m/s²

Part B

Average speed = Total distance/(Total time taken to reach the distance)

Let 'B' represent the distance to point in meters, we have;

The average speed of car 1 = B/(The time it takes car 1 to reach point B)

∴ The average speed of car 1 = B/35 m/s

The average speed of car 2 = B/(The time it takes car 2 to reach point B)

∴ The average speed of car 2 = B/30 m/s

Noting that the larger velocity is given by the smaller divisor to 'B', we have;

The average speed of car 2 = B/30 m/s is larger than the average speed of car 1 = B/35 m/s

6 0

3 years ago

Problem 1. Network-Flow Programming (25pt) A given merchandise must be transported at a minimum total cost between two origins (

hjlf

Solution :

Cost

Destination Destination Destination Maximum supply

Origin 1 5 7 600

Origin 2 10 10 800

15, for > 200 15, for > 200

Demand 500 700

Variables

Destination 1 2

Origin 1 $X_1$ $$$X_2$

Origin 2 $X_3$ $$$X_4$

Constraints : $X_1$ , $$$X_2$ , $X_3$ , $$$X_4$ ≥ 0

Supply : $X_1$ + $$$X_2$ ≤ 600

$X_3$ + $$$X_4$ ≤ 800

Demand : $X_1$ + $$$X_3$ ≥ 500

$X_2$ + $$$X_4$ ≥ 700

Objective function :

Min z = $5X_1+7X_2+10X_3+10X_4, \ (if \ X_3, X_4 \leq 200)$

$=5X_1+7X_2+(10\times 200)+(X_3-200)15+(10 \times 200)+(X_4-200 )\times 15 , \ \ (\text{else})$

Costs :

Destination 1 Destination 2

Origin 1 5 7

Origin 2 10 10

15 15

Variables :

$X_1$ $$$X_2$

300 300

200 400

$X_3$ $$$X_4$

Objective function : Min z = 10600

Constraints:

Supply 600 ≤ 600

600 ≤ 800

Demand 500 ≥ 500

700 ≥ 500

Therefore, the total cost is 10,600.

4 0

3 years ago

A 5Kw solar system may produce enough energy to power your home. a)-True b)- False

Minchanka [31]

Answer: True

Explanation:

Yes, it is true that 5 Kw solar system may produce enough energy to power your home as, on an average good quality of 5 KW solar system can produced 22 units per day enough to power all home appliances. As, a 5 KW solar system produced energy is basically depends on the three main factor that are:

Quality of the solar panel system.
Location from where the solar system generated its energy.
And also on the positioning of the solar system.

4 0

3 years ago

Derive the probability that a receptor is occupied by a ligand using a model that treats the L ligands in solution as distinguis

love history [14]

that is the same thing as you are not going through this week or something

7 0

3 years ago

A commercial refrigerator with refrigerant-134a as the working fluid is used to keep the refrigerated space at -30C by rejecting

chubhunter [2.5K]

Answer:

a) x = 0.4795

b) QL = 5.85 KW

c) COP = 2.33

d) QL_max = 12.72 KW

Explanation:

Solution:-

- Assuming the steady state flow conditions for both fluids R-134a and water.

- The thermodynamic properties remain constant for respective independent intensive properties.

- We will first evaluate the state properties of the R-134a and water.

- Compressor Inlet, ( Saturated Vapor ) - Ideal R-134a vapor cycle

P1 = 60 KPa, Tsat = -36.5°C

T1 = -34°C , h1 = hg = 230.03 KJ/kg

Qin = 450 W - surrounding heat

- Condenser Inlet, ( Super-heated R-134a vapor ):

P2 = 1.2 MPa , Tsat = 46.32°C

T2 = 65°C , h2 = 295.16 KJ/kg

- Condenser Outlet, ( Saturation R-134a point ):

P3 = P2 = 1.2 MPa , Tsat = 46.32°C

T3 = 42°C , h3 = hf = 111.23 KJ/kg

- R-134a is throttled to the pressure of P4 = compressor pressure = P1 = 60 KPa by an "isenthalpic - constant enthalpy pressure reduction" expansion valve.

- Inlet of Evaporator - ( liquid-vapor state )

P4 = P1 = 60 KPa, hf = 3.9 KJ/kg , hfg = 223.9 KJ/kg

h4 = h3 = 111.23 KJ/kg

- The quality ( x ) of the liquid-vapor R-134a at evaporator inlet can be determined:

x4 = ( h4 - hf ) / hfg

x4 = ( 111.23 - 3.9 ) / 223.9

x4 = 0.4795 Answer ( a )

- Water stream at a flow rate flow ( mw ) = 0.25 kg/s is used to take away heat from the R-134a.

- Condenser Inlet, ( Saturated liquid water ):

Ti = 18°C , h = hf = 75.47 KJ/kg

- Condenser Outlet, ( Saturated liquid water ):

To = 26°C , h = hf = 108.94 KJ/kg

- Since the heat of R-134a was exchanged with water in the condenser. The amount of heat added to water (Qh) is equal to amount of heat lost from refrigerant R-134a.

- Apply thermodynamic balance on the R-134a refrigerant in the condenser:

Qh = flow (mr) * [ h2 - h3 ]

Where,

flow ( mr ) : The flow rate of R-134a gas in the refrigeration cycle

flow ( mr ) = Qh / [ h2 - h3 ]

flow ( mr ) = 8.3675 / [ 295.16 - 111.23 ]

flow ( mr ) = 0.0455 kg/s

- The cooling load of the refrigeration cycle ( QL ) is determined from energy balance of the cycle net work input ( Compressor work input ) - "Win" and the amount of heat lost from R-134a in condenser ( Qh ).

- Apply the thermodynamic balance for the compressor:

Win = flow ( mr )*[ h2 - h1 ] - Qin

Win = 0.0455*[ 295.16 - 230.03] KW - 0.45 KW

Win = 2.513 KW

- The cooling load ( QL ) for the refrigeration cycle can now be calculated. Apply thermodynamic balance for the refrigeration cycle:

QL = Qh - Win

QL = 8.3675 - 2.513

QL= 5.85 KW .... Refrigeration Load, Answer ( b )

- The COP of the refrigeration cycle is calculated as the ratio of useful work and total work input required:

COP = QL / Win

COP = 5.85 / 2.513

COP = 2.33 Answer ( c )

- For a compressor to be working at 100% efficiency or ideal then the maximum COP for the refrigeration cycle would be:

COP_max = [ TL ] / [ Th - TL ]

Where,

TL : The absolute temperature of heat sink, refrigerated space

TH : The absolute temperature of heat source, water inlet

COP_max = [ -30+273 ] / [ (18+273) - (-30+273) ]

COP_max = 5.063

- The theoretical ideal refrigeration load ( QL max ) would be:

COP_max = QL_max / Win

QL_max = Win*COP_max

QL_max = 2.513*5.063

QL_max = 12.72 KW Answer ( d )

5 0

4 years ago