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3 years ago

11

A car travelling at 14.0 m/s approaches a traffic light. The driver applies the brakes and is able to come to halt in 5.6 s. Det

ermine the average acceleration of the car during this time interval.

1 answer:

FromTheMoon [43]3 years ago

3 0

Answer:

$a=2.5\ m/s^2$

Explanation:

Given that,

Initial speed of the car, u = 14 m/s

Finally, it comes to rest, v = 0

Time, t = 5.6 s

We need to find the average acceleration of the car during this time interval. We know that,

$a=\dfrac{v-u}{t}\\\\a=\dfrac{0-14}{5.6}\\\\a=-2.5\ m/s^2$

So, the acceleration of the car is $2.5\ m/s^2$ in the opposite direction of motion.

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To solve this problem, we apply the concepts related to the sum of forces and balance in a diagram that will be attached, in order to identify the behavior, direction and sense of the forces. The objective is to find an expression that is in terms of the mass, the angle, the coefficient of friction and the length that allows us to identify when the ladder begins to slip. For equilibrium of the ladder we have,

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3 years ago

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Ezra is pulling a sled, filled with snow, by pulling on a rope attached to the sled. The rope makes an angle θ with respect to t

svetoff [14.1K]

Answer:

Explanation:

Given

rope makes an angle of $\theta$

Mass of sled and snow is m

Normal Force $=F_N$

applied Force is F

as Force is pulling in nature therefore normal reaction is given by

$F_N=mg-F\sin \theta$

Also $F\cos \theta =f_r$

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$f_r=\mu _k\cdot (mg-F\sin \theta )$

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Squaring 1 & 2 and then adding

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$F=\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}$

Substitute value of F in 1

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$\theta =cos^{-1}(\frac{\mu _KF_N}{\sqrt{(\mu _kF_N)^2+(mg-F_N)^2}})$

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3 years ago