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2 years ago

8

For a standard production car, the highest road-tested acceleration ever reported occurred in 1993, when a Ford RS200 Evolution

went from zero to 26.8 m/s (60 mi/h) in 3.323 s. Find the magnitude of the car's acceleration.

1 answer:

Ann [662]2 years ago

3 0

Answer:

a = 8.06 m/s²

Explanation:

The acceleration of this car can be found using the first equation of motion:

$v_f = v_i + at\\\\a = \frac{v_f-v_i}{t}$

where,

a = acceleration = ?

vf = final speed = 26.8 m/s

vi = initial speed = 0 m/s

t = time = 3.323 s

Therefore,

$a = \frac{26.8\ m/s-0\ m/s}{3.323\ s}$

<u>a = 8.06 m/s²</u>

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Moustapha jones drives east at 100km/hr for 3 hours then back west at 80km/hr for 1.5 hours. which pair of answers gives his ave

olya-2409 [2.1K]

The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

The total distance cover in east direction is=100*3=300 km

The total distance cover in the west direction=80*1.5=120 km

The total distance covered is =300+120=420 km

And Total displacement of the car is =300-120=180 km

As we know that the average speed is given as

Avg Speed =Total Distance / Total time

=420/4.5=93.33 km/hr

As we know that the average velocity is given as

Avg Speed =Total Displacement/ Total time

=180/4.5=40 km/hr

Therefore, The average speed of the car is 93.33 km/hr and the average velocity of the car is 40 km/hr.

7 0

3 years ago

A solid sphere of radius R is placed at a height of 30 cm on a15 degree slope. It is released and rolls, without slipping, to th

photoshop1234 [79]

Answer:

The height is $h_c = 42.857$

A circular hoop of different diameter cannot be released from a height 30cm and match the sphere speed because from the conservation relation the speed of the hoop is independent of the radius (Hence also the diameter )

Explanation:

From the question we are told that

The height is $h_s = 30 \ cm$

The angle of the slope is $\theta = 15^o$

According to the law of conservation of energy

The potential energy of the sphere at the top of the slope = Rotational kinetic energy + the linear kinetic energy

$mgh = \frac{1}{2} I w^2 + \frac{1}{2}mv^2$

Where I is the moment of inertia which is mathematically represented as this for a sphere

$I = \frac{2}{5} mr^2$

The angular velocity $w$ is mathematically represented as

$w = \frac{v}{r}$

So the equation for conservation of energy becomes

$mgh_s = \frac{1}{2} [\frac{2}{5} mr^2 ][\frac{v}{r} ]^2 + \frac{1}{2}mv^2$

$mgh_s = \frac{1}{2} mv^2 [\frac{2}{5} +1 ]$

$mgh_s = \frac{1}{2} mv^2 [\frac{7}{5} ]$

$gh_s =[\frac{7}{10} ] v^2$

$v^2 = \frac{10gh_s}{7}$

Considering a circular hoop

The moment of inertial is different for circle and it is mathematically represented as

$I = mr^2$

Substituting this into the conservation equation above

$mgh_c = \frac{1}{2} (mr^2)[\frac{v}{r} ] ^2 + \frac{1}{2} mv^2$

Where $h_c$ is the height where the circular hoop would be released to equal the speed of the sphere at the bottom

$mgh_c = mv^2$

$gh_c = v^2$

$h_c = \frac{v^2}{g}$

Recall that $v^2 = \frac{10gh_s}{7}$

$h_c= \frac{\frac{10gh_s}{7} }{g}$

$= \frac{10h_s}{7}$

Substituting values

$h_c = \frac{10(30)}{7}$

$h_c = 42.86 \ cm$

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Answer:

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