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3 years ago

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While racing on a flat track a car rounds a curve of 56m radius and instantaneously experiences a centripetal acceleration of 36

m/s how fast was the car going.
25m/s
45m/s
35m/s
2.0 x 10^3 m/s

1 answer:

Damm [24]3 years ago

7 0

45m/s should be the answer because the acceleration changes into's velocity and goes towards the centre in which is 56/ms that equals v^2/R the speed at the instant.

Hope this helped

Regards, ShaggySnaps

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The position of a particle as it moves along an y axis is given by y = (2.0cm)sin(πt/4), with t in second and y in centimeters.

irina [24]

Part a)

At t = 0 the position of the object is given as

$x = 0$

At t = 2

$x = 2 sin(\pi/2) = 2cm$

so displacement of the object is given as

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so average speed is given as

$v_{avg} = \frac{2}{2} = 1 cm/s$

Part b)

instantaneous speed is given by

$v = \frac{dy}{dt}$

$v = 2cos(\pi t/4 ) * \frac{\pi}{4}$

now at t= 0

$v = \frac{\pi}{2} cm/s$

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$v = 2 cos(\pi/4) * \frac{\pi}{4}$

$v = \frac{\pi}{2\sqrt2}$

at t = 2

$v = 0$

Part c)

Average acceleration is given as

$a_{avg} = \frac{v_f - v_i}{t}$

$a_{avg} = \frac{0 - \frac{\pi}{2}}{2}$

$a = -\frac{\pi}{4} cm/s^2$

Part d)

Now for instantaneous acceleration

As we know that

$a =- \omega^2 y$

at t = 0

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at t = 1

$y = \sqrt2 cm$

now we have

$a = -\frac{\pi^2}{16}*\sqrt2$

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$a = -\frac{\pi^2}{8}$

<em>so above is the instantaneous accelerations</em>

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3 years ago

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Answer:

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<u>Assume:</u>

Q = charge transported as a result of charging
E = energy expended
C = cost of charging

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Hence, 43.2 kC of charge is transported as a result of charging.

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Answer:

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