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Firlakuza [10]
3 years ago
11

A wheel of diameter 8.0 cm has a cord of length 6.0 m wound around its periphery. Starting from rest, the wheel is given a const

ant angular acceleration of 3.0 rad/s2. How long will it take the cord to unwind?
Physics
1 answer:
MAVERICK [17]3 years ago
6 0

Answer: time t = 10s

Explanation:

Given;

Diameter = 8.0cm = 0.08m

Radius r = diameter/2 = 0.08/2 = 0.04m

Cord length d = 6.0m

Angular acceleration = 3.0rads/s2

Time = t

Converting the angular acceleration to linear acceleration

a = a* × r = 3.0 × 0.04 = 0.12m/s

d = vt + 1/2 (a × t^2)

Initial velocity v = 0, vt = 0 therefore;

d = 1/2 ( a × t^2)

t = √(2d/a)

t = √ [(2× 6)/0.12]

t = 10s

You might be interested in
if the intensity of a person's voice is 4.6 x 10^-7 w/m^2 at a distance of 2.0 m, how much sound power does that person generate
777dan777 [17]

The sound power the person generated is 2.313 \times 10^{-5} \ W.

<h3>Area of the sound wave</h3>

The area of the sound wave is calculated as follows;

A = 4\pi r^2\\\\A = 4 \pi \times (2)^2\\\\A = 50.272 \ m^2

<h3>Power generated</h3>

The sound power the person generated is calculated as follows;

P = I A\\\\P = 4.6\times 10^{-7} \ W/m^2 \ \ \times \ \ 50.272 \ m^2\\\\P = 2.313 \times 10^{-5} \ W

Learn more about intensity of sound here: brainly.com/question/4431819

4 0
2 years ago
With what minimum speed must you toss a 130 gg ball straight up to just touch the 15-mm-high roof of the gymnasium if you releas
xxTIMURxx [149]

Answer:

The initial velocity is 0.5114 m/s or 511.4 mm/s

Explanation:

Let the initial velocity be 'v'.

Given:

Mass of the ball (m) = 130 g = 0.130 kg   [ 1 g = 0.001 kg]

Initial height of the ball (h₁) = 1.4 mm = 0.0014 m   [ 1 mm = 0.001 m]

Final height of the ball (h₂) = 15 mm = 0.015 m

Now, from conservation of energy principle, energy can neither be created nor be destroyed but converted from one form to another.

Here, the kinetic energy of the ball is converted to gravitational potential energy of the ball after reaching the final height.

Change in kinetic energy is given as:

\Delta KE=\frac{1}{2}m(v_f^2-v_i^2)\\Where\ v_f\to Final\ velocity\\v_i\to Initial\ velocity

As it just touches the 15 mm high roof, the final velocity will be zero. So,

v_f=0\ m/s.

Now, the change in kinetic energy is equal to:

\Delta KE = \frac{1}{2}\times 0.130\times v^2\\\\\Delta KE = 0.065v^2

Change in gravitational potential energy = Final PE - Initial PE

So,

\Delta U=mg(h_f-h_i)\\\\\Delta U=0.130\times 9.8\times (0.015-0.0014)\\\\\Delta U=0.017\ J                    [ g = 9.8 m/s²]

Now, Change in KE = Change in PE

0.065v^2=0.017\\\\v=\sqrt{\frac{0.017}{0.065}}\\\\v=0.5114\ m/s\\\\1\ m=1000\ mm\\\\So,0.5114\ m=511.4\ mm\\\\\therefore v=511.4\ mm/s

Therefore, the initial velocity is 0.5114 m/s or 511.4 mm/s

4 0
3 years ago
. Determine if approximate cylindrical symmetry holds for the following situations. State why or why not. (a) A 300-cm long copp
MA_775_DIABLO [31]

Answer:

a) Yes

b) No

Explanation:

In the first case, part a, yes we can say for certainty that cylinderical symmetry holds. Why so? You may ask. This is because from the question, we are told that the length of the rod is 300 cm. And this said length is longer than the distance to the point from the center of the rod, which is 5 cm.

In the second half of the question, I beg to disagree that cylindrical symmetry holds. Again, you may ask why, this is because the length of the rod in this case, is having the same order of magnitude as the distance to the center of the rod. Thus, it is not symmetrical.

6 0
2 years ago
A slingshot is used to launch a stone horizontally from the top of a 20.0 meter cliff. The stone lands 36.0 meters away.
alisha [4.7K]
A) 

     It is a launch oblique, therefore the initial velocity in the vertical direction is zero. Space Hourly Equation in vertical, we have:

S=S_{o}+v_{o}t+ \frac{at^2}{2} \\ 20= \frac{10t^2}{2} \\ t=2s
 
     Through Definition of Velocity, comes:

\Delta v=  \frac{\Delta S}{\Delta t}  \\ v_x= \frac{36}{2}  \\ \boxed {v_{x}=18m/s}


B)
 
     Using the Velocity Hourly Equation in vertical direction, we have:

v_{y}=v_{y_{o}}+gt \\ v_{y}=10\times2 \\ \boxed {v_{y}=20m/s}
  
     The angle of impact is given by:

cos(\theta) =\frac{v_{x}}{v_{y}}  \\ cos(\theta) = \frac{18}{20}  \\ cos(\theta) =0.9 \\ arccos(0.9)=\theta \\ \boxed {\theta \approx 25.84}


If you notice any mistake in my english, please let me know, because i am not native.

7 0
3 years ago
Read 2 more answers
NO LINKS
zhenek [66]

Answer: A , D & F

Explanation:

6 0
2 years ago
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