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3 years ago

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An automobile has a mass of 1200 kg. What is its kinetic energy, in ki, relative to the road when traveling at a velocity of 50

km/h? If the vehicle accelerates to 100 km/h in 15 seconds, what is the change in power required?

1 answer:

netineya [11]3 years ago

3 0

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

$Ek=0.5mv^{2}$

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

$Ek=0.5(1200)(13.89)^2$

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

$Ek=0.5(1200)(27.78)^2=462963J\\$

taking into account all of the above the following equation is inferred

ΔE= $\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15} =23146.916W=23.15kW$

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The closed tank of a fire engine is partly filled with water, the air space above being under pressure. A 6 cm bore connected to

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Answer:

The air pressure in the tank is 53.9 $kN/m^{2}$

Solution:

As per the question:

Discharge rate, Q = 20 litres/ sec = $0.02\ m^{3}/s$

(Since, 1 litre = $10^{-3} m^{3}$ )

Diameter of the bore, d = 6 cm = 0.06 m

Head loss due to friction, $H_{loss} = 45 cm = 0.45\ m$

Height, $h_{roof} = 2.5\ m$

Now,

The velocity in the bore is given by:

$v = \frac{Q}{\pi (\frac{d}{2})^{2}}$

$v = \frac{0.02}{\pi (\frac{0.06}{2})^{2}} = 7.07\ m/s$

Now, using Bernoulli's eqn:

$\frac{P}{\rho g} + \frac{v^{2}}{2g} + h = k$ (1)

The velocity head is given by:

$\frac{v_{roof}^{2}}{2g} = \frac{7.07^{2}}{2\times 9.8} = 2.553$

Now, by using energy conservation on the surface of water on the roof and that in the tank :

$\frac{P_{tank}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{tank} = \frac{P_{roof}}{\rho g} + \frac{v_{tank}^{2}}{2g} + h_{roof} + H_{loss}$

$\frac{P_{tank}}{\rho g} + 0 + 0 = \0 + 2.553 + 2.5 + 0.45$

$P_{tank} = 5.5\times \rho \times g$

$P_{tank} = 5.5\times 1\times 9.8 = 53.9\ kN/m^{2}$

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3 years ago

PLSSSSS Help !!!!!!!!!!!. It's due today. <br> I will give brainliest to correct answer

4vir4ik [10]

Answer:

b

Explanation:

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3 years ago

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A horizontal curve on a two-lane road is designed with a 2,300-ft radius, 12-ft lanes, and a 65-mph design speed. Determine the

Ierofanga [76]

Answer:

distance = 22.57 ft

superelevation rate = 2%

Explanation:

given data

radius = 2,300-ft

lanes width = 12-ft

no of lane = 2

design speed = 65-mph

solution

we get here sufficient sight distance SSD that is express as

SSD = 1.47 ut + $\frac{u^2}{30(\frac{a}{g}\pm G)}$ ..............1

here u is speed and t is reaction time i.e 2.5 second and a is here deceleration rate i.e 11.2 ft/s² and g is gravitational force i.e 32.2 ft/s² and G is gradient i.e 0 here

so put here value and we get

SSD = 1.47 × 65 ×2.5 + $\frac{65^2}{30(\frac{11.2}{32.2}\pm 0)}$

solve it we get

SSD = 644 ft

so here minimum distance clear from the inside edge of the inside lane is

Ms = Rv ( 1 - $cos (\frac{28.65 SSD}{Rv})$ ) .....................2

here Rv is = R - one lane width

Rv = 2300 - 6 = 2294 ft

put value in equation 2 we get

Ms = 2294 ( 1 - $cos (\frac{28.65 \times 664}{2294})$ )

solve it we get

Ms = 22.57 ft

and

superelevation rate for the curve will be here as

R = $\frac{u^2}{15(e+f)}$ ..................3

here f is coefficient of friction that is 0.10

put here value and we get e

2300 = $\frac{65^2}{15(e+0.10)}$

solve it we get

e = 2%

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Answer:

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Explanation:

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