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4 years ago

5

A 150-if object takes 1.5 minutes to travel s 2,500-meter straight path. It begins the trip traveling 120 m/s and decelerates to

a velocity of 20 m/s. What was it's acceleration?

1 answer:

aksik [14]4 years ago

4 0

The Answer is it is -1.11 m/s^2.

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3 years ago

The tub of a washer goes into its spin-dry cy-cle, starting from rest and reaching an angularspeed of 3.1 rev/s in 10.7 s.At thi

neonofarm [45]

Answer:

35 revolutions

Explanation:

t = Time taken

$\omega_f$ = Final angular velocity

$\omega_i$ = Initial angular velocity

$\alpha$ = Angular acceleration

$\theta$ = Number of rotation

Equation of rotational motion

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{3-0}{10.7}\\\Rightarrow \alpha=0.28037\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{3.1^2-0^2}{2\times 0.28037}\\\Rightarrow \theta=17.13806\ rev$

Number of revolutions in the 10.7 seconds is 17.13806

$\omega_f=\omega_i+\alpha t\\\Rightarrow \alpha=\frac{\omega_f-\omega_i}{t}\\\Rightarrow \alpha=\frac{0-3.1}{11.2}\\\Rightarrow a=-0.27678\ rev/s^2$

$\omega_f^2-\omega_i^2=2\alpha \theta\\\Rightarrow \theta=\frac{\omega_f^2-\omega_i^2}{2\alpha}\\\Rightarrow \theta=\frac{0^2-3.1^2}{2\times -0.27678}\\\Rightarrow \theta=17.36035\ rev$

Number of revolutions in the 11.2 seconds is 17.36035

Total total number of revolutions in the 21.9 second interval is 17.13806+17.36035 = 34.49841 = 35 revolutions

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4 years ago