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Shkiper50 [21]
3 years ago
13

Calculate the power required to accelerate a 1150 kg car from rest to a final velocity of 30 m/s up a 100 m long uphill road wit

h a slope of 30 degrees from horizontal over a time of 12 seconds. Disregard friction, air drag, and rolling resistance.
Engineering
1 answer:
DedPeter [7]3 years ago
6 0

Answer:

power required is 90.131 kW

Explanation:

Given data

mass of car (m) = 1150 kg

starting velocity (v1) = 0 m/s

final velocity (v2) = 30 m/s

slope angle = 30 degrees

slope distance = 100 m

to find out

power required

solution

we know the power =  work / time

here time is given we have calculate the work

work is for car climb 100 m distance is total sum of potential energy + kinetic energy

and we know car climb 100m at 30 degree slope so vertical distance (h) will be h = 100 sin (30) = 50 m

so now potential energy = mass × gravity × height

potential energy = 1150 × 9.81 × 50 = 564075 J

kinetic energy = 1/2 ×mass × velocity²

kinetic energy = 1/2 ×1150 × ( v2 - v1 )²

kinetic energy = 1/2 ×1150 × ( 30 - 0 )² = 517500 J

now put all value in power formula

power = work / time

power = potential energy + kinetic energy / 12

power = 564075  +  517500  / 12 = 90131.25 W

power = 90.131 kW

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Answer:

Thomas Elva Edison

He invented light.

4 0
3 years ago
Consider casting a concrete driveway 40 feet long, 12 feet wide and 6 in. thick. The proportions of the mix by weight are given
Akimi4 [234]

Answer:

Weight of cement = 10968 lb

Weight of sand = 18105.9 lb

Weight of gravel = 28203.55 lb

Weight of water = 5484 lb

Explanation:

Given:

Entrained air = 7.5%

Length, L = 40 ft

Width,w = 12 ft

thickness,b= 6 inch, convert to ft = 6/12 = 0.5 ft

Specific gravity of sand = 2.60

Specific gravity of gravel = 2.70

The volume will be:

40 * 12 * 0.5 = 240 ft³

We need to find the dry volume of concrete.

Dry volume = wet volume * 1.54 (concrete)

Dry volume will be = 240 * 1.54 = 360ft³

Due to the 7% entarained air content, the required volume will be:

V = 360 * (1 - 0.07)

V = 334.8 ft³

At a ratio of 1:2:3 for cement, sand, and gravel respectively, we have:

Total of ratio = 1+2+3 = 6

Their respective volume will be =

Volume of cement = \frac{1}{6}*334.8 = 55.8 ft^3

Volume of sand = \frac{2}{6}*334.8 = 111.6 ft^3

Volume of gravel = \frac{3}{6}*334.8 = 167.4 ft^3

To find the pounds needed the driveway, we have:

Weight = volume *specific gravity * density of water

Specific gravity of cement = 3.15

Weight of cement =

55.8 * 3.15 * 62.4 = 10968 pounds

Weight of sand =

111.6 * 2.60 * 62.4 = 18105.9 lb

Weight of gravel =

167.4 * 2.7 * 62.4 = 28203.55 lb

Given water to cement ratio of 0.50

Weight of water = 0.5 of weight of cement

= 1/2 * 10968 = 5484 lb

4 0
4 years ago
Water is being added to a storage tank at the rate of 500 gal/min. Water also flows out of the bottom through a 2.0-in-inside di
melomori [17]

Answer:

From the answer, the water level is falling (since rate of outflow is more than that of inflow), and the rate at which the water level in the storage tank is falling is

(dh/dt) = - 0.000753

Units of m/s

Explanation:

Let the volume of the system at any time be V.

V = Ah

where A = Cross sectional Area of the storage tank, h = height of water level in the tank

Let the rate of flow of water into the tank be Fᵢ.

Take note that Fᵢ is given in the question as 500 gal/min = 0.0315 m³/s

Let the rate of flow of water out of the storage tank be simply F.

F is given in the form of (cross sectional area of outflow × velocity)

Cross sectional Area of outflow = πr²

r = 2 inches/2 = 1 inch = 0.0254 m

Cross sectional Area of outflow = πr² = π(0.0254)² = 0.00203 m²

velocity of outflow = 60 ft/s = 18.288 m/s

Rate of flow of water from the storage tank = 0.0203 × 18.288 = 0.0371 m³/s

We take an overall volumetric balance for the system

The rate of change of the system's volume = (Rate of flow of water into the storage tank) - (Rate of flow of water out of the storage tank)

(dV/dt) = Fᵢ - F

V = Ah (since A is constant)

dV/dt = (d/dt) (Ah) = A (dh/dt)

dV/dt = A (dh/dt) = Fᵢ - F

Divide through by A

dh/dt = (Fᵢ - F)/A

Fi = 0.0315 m³/s

F = 0.0371 m³/s

A = Cross sectional Area of the storage tank = πD²/4

D = 10 ft = 3.048 m

A = π(3.048)²/4 = 7.30 m²

(dh/dt) = (0.0315 - 0.0370)/7.3 = - 0.000753

(dh/dt) = - 0.000753

4 0
3 years ago
Refrigerant 134a enters the evaporator of a refrigeration system operating at steady state at -16oC and a quality of 20% at a ve
Dmitry [639]

Answer:

mass flow rate = 0.0534 kg/sec

velocity at exit = 29.34 m/sec

Explanation:

From the information given:

Inlet:

Temperature T_1 = -16^0\ C

Quality x_1 = 0.2

Outlet:

Temperature T_2 = -16^0 C

Quality  x_2 = 1

The following data were obtained at saturation properties of R134a at the temperature of -16° C

v_f= 0.7428 \times 10^{-3} \ m^3/kg \\ \\  v_g = 0.1247 \ m^3 /kg

v_1 = v_f + x_1 ( vg - ( v_f)) \\ \\ v_1 = 0.7428 \times 10^{-3} + 0.2 (0.1247 -(0.7428 \times 10^{-3})) \\ \\  v_1 = 0.0255 \ m^3/kg \\ \\ \\  v_2 = v_g = 0.1247 \ m^3/kg

m = \rho_1A_1v_1 = \rho_2A_2v_2 \\ \\  m = \dfrac{1}{0.0255} \times \dfrac{\pi}{4}\times (1.7 \times 10^{-2})^2\times 6  \\ \\ \mathbf{m = 0.0534 \ kg/sec}

\rho_1A_1v_1 = \rho_2A_2v_2 \\ \\ A_1 =A_2  \\ \\  \rho_1v_1 = \rho_2v_2   \\ \\ \implies \dfrac{1}{0.0255} \times6 = \dfrac{1}{0.1247}\times (v_2)\\ \\ \\\mathbf{\\ v_2 = 29.34 \ m/sec}

3 0
3 years ago
Which of the following has led to a safer and more prosperous global community within the last century? the Bronze Age composite
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Answer:

The bronze age

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4 years ago
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