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3 years ago

12

Caleb is swinging Rachel in a circle with a centripetal force of 533 N. If the radius of the circle is 0.75 m and Rachel has a m

ass of 16.4 kg, how fast is Rachel moving? Round to the nearest tenth. m/s

2 answers:

Maslowich3 years ago

7 0

Answer:

4.9

Explanation:

I just did it

Alika [10]3 years ago

6 0

Velocity = √((Centrifugal force * radius)/mass)

Velocity = √(533 * 0.75 / 16.4)

Velocity = √(399.75 / 16.4)

Velocity = √24.375

Velocity = 4.9 m/s

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The parasailing system shown uses a winch to pull the rider in towards the boat, which is traveling with a constant velocity. Du

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Answer:

The magnitude of the force is $F_{net}= 1.837 *10^4N$

the direction is 57.98° from the horizontal plane in a counter clockwise direction

Explanation:

From the question we are told that

At t = 0 , $\theta = 20^o$

The rate at which the angle increases is $w = 2 \ ^o/s$

Converting this to revolution per second $\theta ' = 2 \ ^o/s * \frac{\pi}{180} =0.0349\ rps$

The length of the rope is defined by

$r = 125- \frac{1}{3}t^{\frac{3}{2} }$

At $\theta =30^o$ , The tension on the rope T = 18 kN

Mass of the para-sailor is $M_p = 75kg$

Looking at the question we see that we can also denote the equation by which the length is defied as an an equation that define the linear displacement

Now the derivative of displacement is velocity

So

$r' = -\frac{1}{3} [\frac{3}{2} ] t^{\frac{1}{2} }$

represents the velocity, again the derivative of velocity gives us acceleration

So

$r'' = -\frac{1}{4} t^{-\frac{1}{2} }$

Now to the time when the rope made angle of 30° with the water

generally angular velocity is mathematically represented as

$w = \frac{\Delta \theta}{\Delta t}$

Where $\theta$ is the angular displacement

Now considering the interval between $20^o \ to \ 30^o$ we have

$2 = \frac{30 -20 }{t -0}$

making t the subject

$t = \frac{10}{2}$

$= 5s$

Now at this time the displacement is

$r = 125- \frac{1}{3}(5)^{\frac{3}{2} }$

$= 121.273 m$

The linear velocity is

$r' = -\frac{1}{3} [\frac{3}{2} ] (5)^{\frac{1}{2} }$

$= -1.118 m/s$

The linear acceleration is

$r'' = -\frac{1}{4} (5)^{-\frac{1}{2} }$

$= -0.112m/s^2$

Generally radial acceleration is mathematically represented by

$\alpha _R = r'' -r \theta'^2$

$= -0.112 - (121.273)[0.0349]^2$

$= 0.271 m/s^2$

Generally angular acceleration is mathematically represented by

$\alpha_t = r \theta'' + 2 r' \theta '$

Now $\theta '' = \frac{d (0.0349)}{dt} = 0$

So

$\alpha _t = 121.273 * 0 + 2 * (-1.118)(0.0349)$

$= -0.07805 m/s^2$

The net resultant acceleration is mathematically represented as

$a = \sqrt{\alpha_R^2 + \alpha_t^2 }$

$= \sqrt{(-0.07805)^2 +(-0.027)^2}$

$= 0.272 m/s^2$

Now the direction of the is acceleration is mathematically represented as

$tan \theta_a = \frac{\alpha_R }{\alpha_t }$

$\theta_a = tan^{-1} \frac{-0.271}{-0.07805}$

$= 73.26^o$

The force on the para-sailor along y-axis is mathematically represented as

$F_y = mg + Tsin 30^o + ma sin(90- \theta )$

$= (75 * 9.8) + (18 *10^3) sin 30 + (75 * 0.272)sin(90-73.26)$

$= 9.74*10^3 N$

The force on the para-sailor along x-axis is mathematically represented as

$F_x = mg + Tcos 30^o + ma cos(90- \theta )$

$= (75 * 9.8) + (18 *10^3) cos 30 + (75 * 0.272)cos(90-73.26)$

$= 1.557 *10^4 N$

The net resultant force is mathematically evaluated as

$F_{net} = \sqrt{F_x^2 + F_y^2}$

$=\sqrt{(1.557 *10^4)^2 + (9.74*10^3)^2}$

$F_{net}= 1.837 *10^4N$

The direction of the force is

$tan \theta_f = \frac{F_y}{F_x}$

$\theta_f = tan^{-1} [\frac{1.557*10^4}{9.74*10^3} ]$

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Velocity = Displacement / time

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3 years ago

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While working out, a man performed 375 J of work in 11 seconds. What was his power?

Sidana [21]

Power is determined by work/time.
We have P = W/T as our formula.
There is 375J of work, and 11s of time.
Plug in 375J for Work, and 11s for Time.
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Our power is 34.09.
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8 0

3 years ago

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