Please Hurry!! This is very simple!

Physics

1 answer:

trasher [3.6K]3 years ago

7 0

Answer:

What are your thoughts about the connection between exercise and memory?

I think its important to memorize how much u work out and exersize. make sure u dont do it too much or u might pass out. keep track of ur health by memorizing what u ate and when u ate before doing a lot of movement. you might throw up. Exercising is obviously very important for your body.

How often do you exercise?

5 times a week. (5/7 days)

Do you exercise more or less now that you are no longer on campus? Is that a good or bad thing?

A little less. But i think its a good thing

Why?

I used to go to a gym near campus! But now some are a little farther but during this time of quarantine, i work out at home and do the basics like running and agility. I think its good because it teaches me the exercising i can do at home instead of getting memberships and using my money. now i can spend money wisely and still exersize.

Mark as brainliest!

hope this helps! :)

make any changes if ud like

Explanation:

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Hope it helps :)

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3 years ago

A rock is sitting at the edge of a flat merry-go-round at a distance of 1.6 meters from the center. The coefficient of static fr

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Answer:

ω = 2.1 rad/sec

Explanation:

As the rock is moving along with the merry-go-round, in a circular trajectory, there must be an external force, keeping it on track.
This force, that changes the direction of the rock but not its speed, is the centripetal force, and aims always towards the center of the circle.
Now, we need to ask ourselves: what supplies this force?
In this case, the only force acting on the rock that could do it, is the friction force, more precisely, the static friction force.
We know that this force can be expressed as follows:

$f_{frs} = \mu_{s} * F_{n} (1)$

where μs = coefficient of static friction between the rock and the merry-

go-round surface = 0.7, and Fn = normal force.

In this case, as the surface is horizontal, and the rock is not accelerated in the vertical direction, this force in magnitude must be equal to the weight of the rock, as follows:
Fn = m*g (2)
This static friction force is just the same as the centripetal force.
The centripetal force depends on the square of the angular velocity and the radius of the trajectory, as follows:

$F_{c} = m* \omega^{2}*r (3)$

Since (1) is equal to (3), replacing (2) in (1) and solving for ω, we get:

$\omega = \sqrt{\frac{\mu_{s} * g}{r} } = \sqrt{\frac{0.7*9.8m/s2}{1.6m}} = 2.1 rad/sec$

This is the minimum angular velocity that would cause the rock to begin sliding off, due to that if it is larger than this value , the centripetal force will be larger that the static friction force, which will become a kinetic friction force, causing the rock to slide off.

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