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3 years ago

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To reduce the drag coefficient and thus improve the fuel efficiency of cars,the design of side rearview mirrors has changed dram

atically in recent decades from asimple circular plate to a streamlined shape. Determine the amount of fuel and moneysaved per year as a result of replacing a 15-cm-diameter flat mirror by one with ahemispherical back. Assume the car is driven 25,000 km a year at an average speedof 95 km/hr. Take the density and price of gasoline to be 0.75 kg/L and $0.90/L,respectively; the heating value of gasoline to be 44,000 kJ/kg and the overall efficiencyof the engine to be 30 percent.

1 answer:

marissa [1.9K]3 years ago

4 0

Answer:

Amount of fuel used per year is supposed to be 34150 KJ/kg

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A fluid at 300 K flows through a long, thin-walled pipe of 0.2-m diameter. The pipe is enclosed in a concrete casing that is of

andrew-mc [135]

Answer:

The correct answer is "1341.288 W/m".

Explanation:

Given that:

T₁ = 300 K

T₂ = 500 K

Diameter,

d = 0.2 m

Length,

l = 1 m

As we know,

The shape factor will be:

⇒ $SF=\frac{2 \pi l}{ln[\frac{1.08 b }{d} ]}$

By putting the value, we get

⇒ $=\frac{2 \pi l}{ln[\frac{1.08\times 1}{0.2} ]}$

⇒ $=3.7258 \ l$

hence,

The heat loss will be:

⇒ $Q=SF\times K(T_2-T_1)$

$=3.7258\times 1\times 1.8\times (500-300)$

$=3.7258\times 1.8\times (200)$

$=1341.288 \ W/m$

3 0

3 years ago

1. What is the maximum value of the linear density in a crystalline solid (linear density defined as the fraction of the line le

insens350 [35]

Number three number three number three I’m not 100% sure though

4 0

3 years ago

A ballistic pendulum consists of a 3.60 kg wooden block on the end of a long string. From the pivot point to the center‐of‐mass

Pavel [41]

Answer:

17.799°

Explanation:

When the bullet hits the block at that time the momentum is conserved

So, initial momentum = final momentum

$P_i=P_f$

So $28\times 10^{-3}\times 210=(3.6+0.028)v_f$

$v_f=1.6207\ m/sec$

Now energy is also conserved

So $\frac{1}{2}\times (3.6+0.028)\times 1.6207^2=(3.6+0.028)\times 9.81\times 2.8(1-cos\Theta )$

$cos\Theta =0.8521So\ \Theta =17.799^{\circ}$

3 0

2 years ago

All brake lights are dimmer than normal. Technician A says that bad bulbs could be the cause. Technician B says that high resist

yarga [219]

Answer:

All Brake lights are dimmer than normal because high resistance in the brake switch could be the cause according to Technician B.

Explanation:

According to Technician A

When the bulb is faulty then no current will flow through bulb and it will be open circuit.So no light will produce in bulb .

According to Technician B

When a high resistance inserted in series circuit the voltage across each resistance is reduced and this cause the light glow dimly.

Formula of resistance in series circuit

Rt=r1+r2+r3......

5 0

3 years ago

A flywheel made of Grade 30 cast iron (UTS = 217 MPa, UCS = 763 MPa, E = 100 GPa, density = 7100 Kg/m, Poisson's ratio = 0.26) h

hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

$A_{1}= \frac{\pi }{4}\times 150^{2}$

= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

7 0

3 years ago