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4 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

Physics

1 answer:

MA_775_DIABLO [31]4 years ago

7 0

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:

brainly.com/question/9575487

brainly.com/question/9329700

brainly.com/question/2506028

#LearnwithBrainly

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17. Ms. Lopez is being chased by a Komodo dragon. She has an initial velocity of 7 m/s

exis [7]

Answer:

Cant say for sure, its been a while sense ive done this, but im almost certain its 70

Explanation:

Take the zero off of your 10 s

Then go 7 m/s x 1 s

Which equals 7

Add the zero back to the end of your answer

10 -- 0 = 1 x 7 = 7 ++ 0 = 70

(PS, two of the same sign is just adding a number to the end of your original answer, that is not what it actually stands for in mathematical terms but that is what i'm using to make it clearer as to whats happening)

I'm not too good at explaining and formulas but i hope this helped

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3 years ago

Which of the following is not one of the emotional expressions that are universally recognized?

inn [45]

Answer:

Explanation:

The answer should be B.

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3 years ago

Read 2 more answers

An object weigh 40N in air ,weigh 20N when submerged in water,and 30N when submerged in a liquid of unknown liquid density.what

muminat

Answer:

The density is $\rho_u =500 kg /m^3$

Explanation:

From the question we are told that

The weight in air is $W_a = 40 \ N$

The weight in water is $W_w = 20 \ N$

The weight in a unknown liquid is $W_u = 30 \ N$

Now according to Archimedes principle the weight of the object in water is mathematically represented as

$W_w = W_a -m _w g$

Where $m_w$ is he mass of the water displaced

substituting value

$m_w g = 40 -20$

$m_w g = 20 \ N --- (1)$

Now according to Archimedes principle the weight of the object in unknown is mathematically represented as

$W_u = W_a -m _u g$

Where $m_u$ is he mass of the unknown liquid displaced

substituting value

$m_u g = 40 -30$

$m_u g = 10 \ N ---(2)$

dividing equation 2 by equation 1

$\frac{m_ug}{m_wg} = \frac{10}{20}$

$\frac{m_u}{m_w} = \frac{1}{2}$

=> $m_u = 0.5 m_w$

Now since the volume of water and liquid displaced are the same then

$\rho _u = 0.5 \rho_w$

This because

$density = \frac{mass}{volume}$

So if volume is constant

mass = constant * density

Where $\rho_u$ is the density of the liquid

and $\rho_ w$ is the density of water which is a constant with a value $\rho_w = 1000 kg/m^3$

$\rho_u = 1000*0.5$

$\rho_u =500 kg /m^3$

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3 years ago

What is the best description of the relationship between emerging scientific ideas and open-mindedness?

Vilka [71]

The third one looks correct to me

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Assuming the bar has no weight where does the fulcrum (the top point of the tringle) need to be positioned for the two sides to

Inessa05 [86]

Fulcrum need to be positioned balanced with weight on both the sides following law of lever.

What is the physical law of the lever?

It is the foundation for issues with weight and balance. According to this rule, a lever is balanced when the weight multiplied by the arm on one side of the fulcrum, which serves as the pivot point for the device, equals the weight multiplied by the arm on the opposing side.
The lever is balanced, in other words, when the sum of the moments about the fulcrum is zero.
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Moving the weights closer to or away from the fulcrum, as well as raising or lowering the weights, can alter the balance point, or CG, of the lever.

Learn more about the Fulcrum with the help of the given link:

brainly.com/question/16422662

#SPJ4

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