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4 years ago

Which of the following is not a popular surface for a tennis court?

Physics

1 answer:

vovikov84 [41]4 years ago

8 0

The answer is A. wood

because there are three different courts: clay court, grass court, and hard court. wood isn't in there so that would be the answer.

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Graphing data is most used to help you to observe_____ *

kotykmax [81]

Answer: Graphing data is used to display data because it is easier to see trends in the data when it is displayed visually compared to when it is displayed numerically in a table.

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3 years ago

Basketball= What is the half of the court that is farthest from the offensive basket?

klasskru [66]

The answer would be half court I think

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3 years ago

What is a creative name for an element? (like oxygen, phosphorus) try to make it good

Inga [223]

Oxyles i think this suite Oxyles what do you think.

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3 years ago

A fly wheel with a diameter of 1.20m is rotating at an angular speed of 200 Rev/min

gladu [14]

Answer:a) 20.944rad/s (b) 12.57m/s (c) 800rev/min²

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3 years ago

A hoop, a solid cylinder, a spherical shell, and a solid sphere are placed at rest at the top of an inclined plane. All the obje

iogann1982 [59]

Answer:

Solid sphere will reach first

Explanation:

When an object is released from the top of inclined plane

Then in that case we can use energy conservation to find the final speed at the bottom of the inclined plane

initial gravitational potential energy = final total kinetic energy

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}I\omega^2$

now we have

$I = mk^2$

here k = radius of gyration of object

also for pure rolling we have

$v = R\omega$

so now we will have

$mgh = \frac{1}{2}mv^2 + \frac{1}{2}(mk^2)(\frac{v^2}{R^2})$

$mgh = \frac{1}{2}mv^2(1 + \frac{k^2}{R^2})$

$v^2 = \frac{2gh}{1 + \frac{k^2}{R^2}}$

so we will say that more the value of radius of gyration then less velocity of the object at the bottom

So it has less acceleration while moving on inclined plane for object which has more value of k

So it will take more time for the object to reach the bottom which will have more radius of gyration

Now we know that for hoop

$mk^2 = mR^2$

k = R

For spherical shell

$mk^2 = \frac{2}{3}mR^2$

$k = \sqrt{\frac{2}{3}} R$

For solid sphere

$mk^2 = \frac{2}{5}mR^2$

$k = \sqrt{\frac{2}{5}} R$

So maximum value of radius of gyration is for hoop and minimum value is for solid sphere

so solid sphere will reach the bottom at first

7 0

3 years ago