All categories

4 years ago

Will Mark BRAINLIEST!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

vazorg [7]4 years ago

4 0

r = distance of each mass in each hand from center = 0.6 m

m = mass of each mass in each hand = 2 kg

v = linear speed = 1.1 m/s

L = combined angular momentum of the masses = ?

Combined angular momentum of the masses is given as

L = 2 m v r

L = 2 (2) (1.1) (0.6)

L = 2.64 kg m²/s

v' = linear speed when she pulls her arms = ?

r' = distance of each mass from center after she pulls her arms = 0.15 m

Using conservation of momentum , angular momentum remains same, hence

L = 2 m v' r'

2.64 = 2 (2) (0.15) v'

v' = 4.4 m/s

You might be interested in

The part of a standing wave where there is complete destructive interference is known as a(n)

Alisiya [41]

Crest is the answer

6 0

3 years ago

The position function x (6.0 m) cos[(3p rad/s)t p/3 rad]gives the simple harmonic motionof a body. at t 2.0 s, what are the(a) d

Daniel [21]

The correct formula for this simple harmonic motion is:

x = (6.0) cos [ (3 pi) t + pi/3 ]

A. Calculating for displacement x at t = 2:

x =6.0 cos (19/3 pi)
x = 6 (0.5)

x = 3 m

B. Velocity is equivalent to the 1st derivative of the equation. v = dx / dt

Velocity v = dx/dt = -18 pi sin (3pi t + pi/3)

v = - 18 pi sin (6pi + pi/3)

v = 49.0 m/s

C. Acceleration is the 2nd derivative or the 1st derivative of velocity dv / dt
Acceleration a = dv/dt = -54 pi^2 cos(3pi t + pi/3)

a = -54 pi^2 cos(3pi * 2 + pi/3)

a = 386.34 m/s^2

D. Phase = pi/3

E. frequency f= 3pi/2pi = 1.5Hz

F. period = 1/f = 1/1.5 = .6667sec

5 0

4 years ago

A moving company collects data about couches they move and displays the data in a table. The force along the ramp, the angle of

USPshnik [31]

Answer:

the correct answers are d, e, & f

Explanation:

just took the edge test

7 0

4 years ago

Read 2 more answers

3. Looking at the image below label the areas of highest and lowest kinetic energy of both gravitational potential

MArishka [77]

Explanation:

max potential your welcome i love helping outhers :)

8 0

3 years ago

Dennis throws a volleyball up in the air. It reaches its maximum height 1.1\, \text s1.1s1, point, 1, start text, s, end text la

rewona [7]

Answer:

If max height = 1.1 meters, then initial velocity is 3.28 m/s

If max height is 1.1 feet, then the initial velocity is 5.93 ft/s

Explanation:

Recall the formulas for vertical motion under the acceleration of gravity;

for the vertical velocity of the object we have

$v=v_0-g \,t$

for the object's vertical displacement we have

$y-y_0=v_0\,t - \frac{g}{2} \,t^2$

If the maximum height reached by the object is given in meters, we use the value for g in $m/s^2$ which is: $9.8\,\,m/s^2$

If the maximum height of the object is given in feet, we use the value for g in $ft/s^2$ which is : $32\,\,ft/s^2$

Now, when the ball reaches its maximum height, the ball's velocity is zero, so that allows us to solve for the time (t) the process of reaching the max height takes:

$v=v_0-g \,t\\0=v_0-g \,t\\g\,\,t=v_0\\t=\frac{v_0}{g}$

and now we use this to express the maximum height in the second equation we typed:

$y-y_0=v_0\,t - \frac{g}{2} \,t^2\\max\,height=v_0\,(\frac{v_0}{g}) - \frac{g}{2} \,(\frac{v_0}{g})^2\\max\,height= \frac{v_0^2}{2\,g}$

Then if the max height is 1.1 meters, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,9.8}\\(9.8)\,(1.1)=v_0^2\\v_0=10.78\\v_0=\sqrt{10.78} \\v_0=3.28\,\,m/s$

If the max height is 1.1 feet, we use the following formula to solve for $v_0$ :

$1.1= \frac{v_0^2}{2\,32}\\(32)\,(1.1)=v_0^2\\v_0=35.2\\v_0=\sqrt{35.2} \\v_0=5.93\,\,ft/s$

5 0

3 years ago

Read 2 more answers