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3 years ago

What is necessary to designate a position? A. a reference point B. a direction C. fundamental units D. motion E. all of these

Physics

1 answer:

max2010maxim [7]3 years ago

8 0

Answer:

E. all of these

Explanation:

The designation of a point in space all the points that necessary

- reference point

- a direction

- fundamental units

- a direction

- motion

all are necessary to designate a point in space. Hence option E is correct.

For example in simple harmonic motion we need to specify all the above factors of the object in order to designate the position of the object.

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Once a baseball has been hit into the air, what forces are acting upon it? How can you tell that any forces are acting upon the

Tom [10]

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2 years ago

A lady bug is sitting on the bottom of a can while you twirl it overhead on a string that is 65.0

MA_775_DIABLO [31]

The linear speed of the ladybug is 4.1 m/s

Explanation:

First of all, we need to find the angular speed of the lady bug. This is given by:

$\omega=\frac{2\pi}{T}$

where

T is the period of revolution

The period of revolution is the time taken by the ladybug to complete one revolution: in this case, since it does 1 revolution every second, the period is 1 second:

T = 1 s

Therefore, the angular speed is

$\omega=\frac{2\pi}{1 s}=6.28 rad/s$

Now we can find the linear speed of the ladybug, which is given by

$v=\omega r$

where:

$\omega=6.28 rad/s$ is the angular speed

r = 65.0 cm = 0.65 m is the distance of the ladybug from the axis of rotation

Substituting, we find

$v=(6.28)(0.65)=4.1 m/s$

Learn more about angular speed:

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3 years ago

II Force on a tennis ball. The record speed for a tennis ball that is served is 73.14 m/s. During a serve, the ball typically st

AveGali [126]

Answer:

F=248.5W N

Explanation:

Newton's 2nd Law tells us that F=ma. We will use their averages always. The average acceleration the tennis ball experimented is, by definition:

$a=\frac{\Delta x}{\Delta t}=\frac{v-v_0}{t-t_0}$

Since we start counting at 0s and the ball departs from rest, this is just $a=\frac{v}{t}$

So we can write:

$F=ma=\frac{mv}{t}=\frac{gmv}{gt}$

Where in the last step we have just multiplied and divided by g, the acceleration of gravity. This allows us to introduce the weight of the ball W since W=gm, so we have:

$F=\frac{Wv}{gt}=\frac{v}{gt}W$