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3 months ago

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A car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow dow

n with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? a car is moving with a speed of 32.0 m/s. the driver sees an accident ahead and slams on the brakes, causing the car to slow down with a uniform acceleration of magnitude 3.50 m/s2. how far does the car travel after the driver put on the brakes until it comes to a stop? 112 m 292 m 4.57 m 146 m 9.14 m

1 answer:

kicyunya [3.2K]3 months ago

5 0

The vehicle will travel a distance of 146.28 m, thus the correct response is 146 m.

Given that the car is decelerating with a constant acceleration, we can apply the third motion equation.

v^2 = u^2 - 2aS

where v represents the final speed of the car, which is 0 as the vehicle halts,[

u is the initial speed=32 m/s

a is the uniform acceleration=3.5 m/s^2

and S signifies the distance covered by the car prior to stopping.

Substituting the values into the third equation gives us 0 = 32^2 - (2*3.5*S)

S = 146.28 m

Thus, the vehicle covers a distance of 146 m before coming to a stop.

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For lunch you and your friends decide to stop at the nearest deli and have a sandwich made fresh for you with 0.300 kg of Italia

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Answer:

Part a)

A = 0.0581 m

Part b)

T = 0.37 s

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A slice is dropped onto the plate from a height of 0.250 m,

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$M = 0.400 kg$

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$0.300 (2.21) = (0.300 + 0.400) v_f$

$v_f = 0.95 m/s$

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When the slice rests on the plate, the new mean position can be expressed as

$x_1 = 0.01962 m$

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We also determine that the speed of SHM is represented as

$x_2 = 0.0343 m$

Here, we derive values from

$v = \omega\sqrt{A^2 - x^2}$

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$\omega = \sqrt{\frac{200}{0.300 + 0.400}}$

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$0.95 = 16.9\sqrt{A^2 - 0.0147^2}$

$A = 0.0581 m$

Part b)

The time period for the scale is computed as

$T = \frac{2\pi}{\omega}$

$T = \frac{2\pi}{16.9}$

$T = 0.37 s$

8 0

3 months ago