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3 years ago

5

Un coche de unos 500 kg viaja a 90 km/h. Percibe un obstáculo y debe frenar a tope. Por las marcas del suelo se sabe que el espa

cio de frenada fue de 125 m. ¿Cuánto valía la fuerza de rozamiento entre el coche y la carretera?

1 answer:

a_sh-v [17]3 years ago

3 0

Answer:

THERE'S A GIRL WHO WILL STEAL YOUR INFO, PLZ DON"T PRESS THE LINK IF SHE ASKS YOU TOO!!!!!

Explanation:

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Both graphs show plotted points forming a curved line. Curved lines have changing slope; they may start with a very small slope and begin curving sharply (either upwards or downwards) towards a large slope. In either case, the curved line of changing slope is a sign of accelerated motion (i.e., changing velocity).

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The generation of a magnetic field by an electric current is___ .

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Answer:

Electromagnetic induction

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2 years ago

Suppose a car travels 106 km at a speed of 28 m/s and uses 1.9 gals of gasoline in the process. Only 30% of the gasoline goes in

USPshnik [31]

Answer:

a) The magnitude of the force is 968 N

b) For a constant speed of 30 m/s, the magnitude of the force is 1,037 N

Explanation:

<em>NOTE: The question b) will be changed in other to give a meaningful answer, because it is the same speed as the original (the gallons would be 1.9, as in the original).</em>

Information given:

d = 106 km = 106,000 m

v1 = 28 m/s

G = 1.9 gal

η = 0.3

Eff = 1.2 x 10^8 J/gal

a) We can express the energy used as the work done. This work has the following expression:

$W=F\cdot d$

Then, we can derive the magnitude of the force as:

$F=\frac{W}{d}=\frac{\eta\cdot (G\cdot Eff)}{d}=\frac{0.3*1.9*(1.8*10^8)}{106*10^3} =968\,N$

b) We will calculate the force for a speed of 30 m/s.

If the force is proportional to the speed, we have:

$F_2=F_1(\frac{v_2}{v_1} )=968(\frac{30}{28} )=968*1.0714=1,037\,N$

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3 years ago

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