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scZoUnD [109]
3 years ago
7

The acceleration of automobiles is often given in terms of the time it takes to go from 0 mi/h to 60 mi/h. One of the fastest st

reet legal cars, the Bugatti Veyron Super Sport, goes from 0 to 60 in 2.70 s. To the nearest integer, what is its acceleration in m/s^2? Please show work.
Physics
1 answer:
Rudiy273 years ago
6 0

Answer:

9.934 m/s²

Explanation:

Given:

Initial speed of the Bugatti Veyron Super Sport = 0 mi/h

Final speed of the Bugatti Veyron Super Sport = 60 mi/h

Now,

1 mi/h = 0.44704 m / s

thus,

60 mi/h = 0.44704 × 60 = 26.8224 m/s

Time = 2.70 m/s

Now,

The acceleration (a) is given as:

a=\frac{\textup{Change in speed}}{\textup{Time}}

thus,

a=\frac{26.8224 - 0 }{2.70}

or

a = 9.934 m/s²

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A Block slides down an incline that makes an angle of 30? with the horizontal direction. The coefficient of kinetic friction bet
astraxan [27]

Answer:

Acceleration = 2.35 m/s^{2}

Speed = 8.67 m/s

Explanation:

The coefficient of friction , u =0.3

The angle of incline = 30°

The two forces acting on block are weight and friction.

weight along the incline = mg cos60° = \frac{mg}{2} = 0.5 mg

Friction along incline = umg cos30° = mg 0.3\times \frac{\sqrt{3}}{2}

Friction along incline  = 0.26 mg

Net force acting on the weight = (0.5 - 0.26) mg = 0.24 mg

Acceleration = \frac{net force}{mass} = 0.24 g = 2.35 m/s^{2}

The height of incline = 8 m

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v^{2}= 2\times 0.24 \times 9.8\times 16

v= 8.67 m/s

5 0
4 years ago
A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with a speed of 20 m/s at an angle 5.0 ∘ above t
goldfiish [28.3K]

Answer:

Explanation:

initial height, yo = 2 m

initial velocity, u = 20 m/s

angle of projection,θ = 5 degree

distance of net = 7 m

height of net = 1 m

Let it covers a vertical distance y in time t .

Use Second equation of motion for vertical motion

y=y_{0}+uSin\theta t-1/2 gt^{2}

y=2+20Sin5 t-4.9t^{2}

As it hits the ground in time t, so put y = 0

0=2+1.74 t-4.9t^{2}

4.9t^{2}-1.74t-2=0

t= \frac{1.74\pm\sqrt{1.74^{2}+4\times\2\times4.9}}{9.8}

Taking positive sign, t = 0.84 s

The ball travels a horizontal distance x in time t

X = 20 Cos5 x t

X =  16.76 m

As this distance is more than the distance of net, so it clears the net.

Let t' be the time taken to travel a horizontal distance equal to the distance of net

7 = 20 cos5 x t'

t' = 0.35 s

Let the vertical distance traveled by the ball in time t' is y'.

So,

y'=y_{0}+uSin\theta t'-1/2 gt'^{2}

y'=2+20Sin5 t-4.9\times0.35^{2}

y' = 2.008 m

So, it clears the net which is 1 m high.

It clears the net by a vertical distance of 2.008 - 1 = 1.008 m and horizontal distance 16.76 - 7 = 9.76 m

3 0
3 years ago
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