Does a more efficient car have a larger or smaller LPK number? Explain how you know..

Physics

2 answers:

Sindrei [870]3 years ago

7 0

Answer: smaller

LPK (Litre per kilometer) refers to amount of fuel in litres required per unit km by a car. In Europe, LPK number is used to specify the efficiency of a car. An efficient car would be the one which would consume less amount of fuel per kilometer run. Hence, a more efficient car would have smaller LPK number.

Softa [21]3 years ago

5 0

A more efficient car will have smaller LPK number.

Explanation

LPK number is the measure of liters of fuel consumed by the vehicle to travel one kilometer distance.

The LPK number that is liters per kilometer number is used in Europe to determine the efficiency of cars.

A car will be termed as efficient if it uses less amount of fuel for travelling a larger distance.

This indicates that car which has more mileage or in other word which consumes less quantity of fuel to travel longer distance is termed as efficient cars.

Thus , a more efficient car will have smaller LPK number.

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A football punter wants to kick the ball so that it is in the air for 4.2 s and lands 55 m from where it was kicked. Assume that

earnstyle [38]

Answer:

$57.24^{\circ}$

24.21 m/s

Explanation:

x = Displacement in x direction = 55 m

y = Displacement in x direction = 0

$y_0$ = Height of the ball when the ball is kicked = 1 m

t = Time taken = 4.2 s

$a_y=g$ = Acceleration due to gravity = $-9.81\ \text{m/s}^2$

u = Initial velocity of ball

Displacement in x direction is given by

$x=u_xt+\dfrac{1}{2}a_xt^2\\\Rightarrow 55=4.2u_x+0\\\Rightarrow u_x=\dfrac{55}{4.2}\\\Rightarrow u\cos\theta=13.1\ \text{m/s}\\\Rightarrow u=\dfrac{13.1}{\cos\theta}$

Displacement in y direction is given by

$y=y_0+u_yt+\dfrac{1}{2}a_yt^2\\\Rightarrow 0=1+4.2u\sin\theta+\dfrac{1}{2}\times (-9.81)\times 4.2^2\\\Rightarrow 0=4.2u\sin\theta-85.5242\\\Rightarrow u\sin\theta=20.36\ \text{m/s}\\\Rightarrow u=\dfrac{20.36}{\sin\theta}$

$\dfrac{13.1}{\cos\theta}=\dfrac{20.36}{\sin\theta}\\\Rightarrow \dfrac{20.36}{13.1}=\dfrac{\sin\theta}{\cos\theta}\\\Rightarrow \theta=\tan^{-1}\dfrac{20.36}{13.1}\\\Rightarrow \theta=57.24^{\circ}$