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2 years ago

7

A 45 kg merry go round worker stands on the rides platform 6.3 m from the center. If her speed as she goes around the circle is

4.1 m/s, what is the force of friction necessary to keep her from falling off the platform?

1 answer:

grin007 [14]2 years ago

7 0

I’ve got 120 n, hope it helps!

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You attach a 3.10 kg weight to a horizontal spring that is fixed at one end. You pull the weight until the spring is stretched b

dolphi86 [110]

Answer:

0.785 m/s

Explanation:

Hi!

To solve this problem we will use the equation of motion of the harmonic oscillator, <em>i.e.</em>

$x(t) = A cos(\omega t )+B sin(\omega t)$ - (1)

$\frac{dx}{dt}(t) = \omega (B cos(\omega t )- A sin(\omega t)$ - (1)

The problem say us that the spring is released from rest when the spring is stretched by 0.100 m, this condition is given as:

$x(0) = 0.100$

$\frac{dx}{dt}(0) = 0$

Since cos(0)=1 and sin(0) = 0:

$x(0)=A$

$\frac{dx}{dt}(0) = -B\omega$

We get

$A =0.100\\B = 0$

Now it say that after 0.4s the weigth reaches zero speed. This will happen when the sping shrinks by 0.100. This condition is written as:

$x(0.4) = - 0.100$

Since

$x(t) = 0.100 cos(\omega t)\\ -0.100=x(0.4)=0.100cos(\omega 0.4)$

This is the same as:

$-1 = cos(0.4\omega)$

We know that cosine equals to -1 when its argument is equal to:

(2n+1)π

With n an integer

The first time should happen when n=0

Therefore:

π = 0.4ω

or

ω = π/0.4 -- (2)

Now, the maximum speed will be reached when the potential energy is zero, <em>i.e. </em>when the sping is not stretched, that is when x = 0

With this info we will know at what time it happens:

$0 = x(t) = 0.100cos(\omega t)$

The first time that the cosine is equal to zero is when its argument is equal to π/2

<em>i.e.</em>

$t_{maxV}=\pi /(2\omega)$

And the velocity at that time is:

$\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\omega t_{maxV})\\\frac{dx}{dt}(t_{maxV} ) =- 0.100\omega sin(\pi/2)\\$

But sin(π/2) = 1.

Therefore, using eq(2):

$\frac{dx}{dt}(t_{maxV} ) = 0.100*\omega = 0.100\frac{\pi}{0.400} = \pi/4$

And so:

$V_{max} = \pi / 4 =0.785$

6 0

3 years ago

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

Andrews [41]

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

7 0

3 years ago

NEED HELP ASAPPPPPPP!

AlekseyPX

Answer:

Changeable. Answer is c

Explanation:

3 0

3 years ago

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How do you calculate density?

Mashcka [7]

Mass per cubic metre so kg/m3. Temperature may give different results.

4 0

3 years ago

Martina seems to have several different personalities. In one, she is 7 years

melomori [17]

Answer:Dissociative Identity Disorder

Explanation:I don't say you have to mark my ans brainliest but my friend if it has really helped you don't forget to thank me...

3 0

3 years ago