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4 years ago

During most of a cell’s life cycle, DNA is found in a mass of loose strands called?

Physics

1 answer:

Rasek [7]4 years ago

7 0

Chromosomes that's the answer

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3 years ago

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3 years ago

How much work is required to lift a 500 kg block 12 m? (Show work)

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3 years ago

A high-speed drill rotating ccw at 2400 rpm comes to a halt in 2.5 s. What is the magnitude of the drill’s angular acceleration?

WARRIOR [948]

Answer:

The magnitude of the drills angular acceleration is $-32\pi s^{-2}$ .

The drill makes 50 revolutions before it stops.

Explanation:

The revolutions that the drill makes in 1 second is

$\dfrac{2400\:rev}{60s} =40\:rev/s$

And the angular velocity is

$\omega= \dfrac{40 (2\pi )}{1s}$ (one revolution is $2\pi$ radians)

$\boxed{\omega =80\pi\:s^{-1}}$

Now, the drill comes to a halt in 2.5 s, which means the magnitude of it's angular acceleration is

$a= \dfrac{\Delta \omega}{\Delta s}$

$a=\dfrac{80\pi-0 }{0-2.5s}$

$\boxed{a=-32\pi \:s^{-2}}$

In other words, the magnitude of the angular acceleration is $-32\pi\: s^{-2}.$

Now, we find the angular displacement of the drill which is given by the equation

$\theta = \omega t +\dfrac{1}{2} \alpha t^2$

putting in $\omega = 80\pi s^{-1}$ , $\alpha = -32\pi s^{-2}$ , and $t=2.5s$ , we get:

$\theta = (80\pi s^{-1} *2.5s)+\dfrac{1}{2} (-32\pi s^{-2})(2.5s)^2$

$\theta = 100\pi$

which is

$\dfrac{100\pi }{2\pi } =50\:rev$ (one revolution is $2\pi$ radians)

50 revolutions.

In other words ,the drill makes 50 revolutions before it stops.

3 0

3 years ago