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2 years ago

12

in the system to the right, the pulleys are frictionless and the system hangs in equilibrium. Determine the values of each of th

e unknown weights.

1 answer:

irga5000 [103]2 years ago

4 0

Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .

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A roller coaster car rapidly picks up speed as it rolls down a slope as it starts down the slope its speed is 4m/s but 3 seconds

Gwar [14]

Answer:

The acceleration is 6 [m/s^2]

Explanation:

We can find the acceleration of the roller coaster using the kinematic equation for uniformly accelerated motion.

$v_{f} =v_{i} + a*t\\where:\\v_{f} = final velocity = 22 [m/s]\\v_{i} = initial velocity = 4 [m/s]\\t = time = 3 [s]\\$

Now replacing the values we have:

$a=\frac{v_{f} - v_{i} }{t} \\a=\frac{22 - 4 }{3}\\a = 6 [m/s^{2} ]$

3 0

2 years ago

A flute plays a note with a frequency of 266 Hz. What is the speed of sound if the wavelength is 1.3 m?

hjlf

Wave speed = (frequency) x (wavelength)

= (266 /sec) x (1.3 meters)

= 345.8 meters/sec

6 0

2 years ago

Read 2 more answers

Match the scienntist with their accomplishment / discoveries

Art [367]

Answer:

newton - motion, gravity

kepler - orbital paths

brahe - the sun goes around the earth

Explanation:

im not sure about brahe but its the only one that makes sense

8 0

2 years ago

A light ray moving at a 54.3 deg

MrMuchimi

Answer:

$45.25^{\circ}$

Explanation:

Applying Snell's law

$n_1sin\theta_1 = n2*sin\theta_2\\1.33*sin(54.3) = 1.52*sin\theta_2\\sin\theta_2 = 1.33*0.81208353/1.52 = 0.71057\\\theta_2=sin^{-1} (0.71057)\\\theta_2 = 45.25^{\circ}$

3 0

2 years ago

A child is riding a merry-go-round that is turning at 7.18 rpm. if the child is standing 4.65 m from the center of the merry-go-

Alisiya [41]

First we need to convert the angular speed from rpm to rad/s. Keeping in mind that
$1 rev= 2 \pi rad$
$1 min = 60 s$
the angular speed is
$\omega = 7.18 \frac{rev}{min} \cdot \frac{2 \pi}{60} = 0.75 rad/s$

And so now we can calculate the tangential speed of the child, which is the angular speed times the distance of the child from the center of the motion:
$v= \omega r = (0.75 rad/s)(4.65 m)=3.50 m/s$

3 0

3 years ago