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kaheart [24]
2 years ago
14

A flywheel with radius of 0.300 m starts from rest and accelerates with a constant angular acceleration of 0.400 rad/s2.For a po

int on the rim of the flywheel, what is the magnitude of the tangential acceleration after 2.00 s of acceleration
Physics
1 answer:
oksian1 [2.3K]2 years ago
6 0

Answer:

0.12\ m/s^2

Explanation:

Given that,

The radius of a flywheel, r = 0.3 m

Angular acceleration of a flywheel, \alpha =0.4\ rad/s^2

We need to find the magnitude of the tangential acceleration after 2.00 s of acceleration.

The relation between the tangential and angular acceleration is given by :

a_t=r\alpha \\\\a_t=0.3\times 0.4\\\\a_t=0.12\ m/s^2

So, the required magnitude of tangential acceleration is 0.12\ m/s^2.

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What are the three longest wavelengths for standing waves on a 264- cm -long string that is fixed at both ends
miv72 [106K]

The three longest wavelengths for the standing waves on a 264-cm long string that is fixed at both ends are:

  1. 5.2 meters.
  2. 2.6 meters.
  3. 1.7meters.

Given data:

Length of the fixed string = 264cms = 2.64 meters

The wavelength for standing waves is given by:

λ = 2L/n

where,

  • λ is the wavelength
  • L is the length of the string

For n = 1,

  • λ = 2×2.6/1

= 5.2 meters

For n = 2,

  • λ = 2×2.6/2

= 2.6 meters

For n = 3,

  • λ = 2×2.6/3

= 1.7 meters

To learn more about standing waves: brainly.com/question/14151246

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WILL MARK BRAINLIEST IF RIGHT & HAS EXPLINATION
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Answer:

s= 64m I'm not 100% sure

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What is a Geographic test
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What can be found by counting the number of troughs per second in a wave diagram? amplitude direction frequency speed wavelength
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The period T is time it takes for one complete cycle or from "trough to trough" so the reverse is trough per sec = 1/T = frequency
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While entering a freeway, a car accelerates from rest at a rate of 2.40 m/s2 for 12.0 s. (a) Draw a sketch of the situation. (b)
ArbitrLikvidat [17]

Answer:

a) See attached picture, b) We know the initial velocity = 0, initial position=0, time=12.0s, acceleration=2.40m/s^{2}, c) the car travels 172.8m in those 12 seconds, d) The car's final velocity is 28.8m/s

Explanation:

a) In order to draw a sketch of the situation, I must include the data I know, the data I would like to know and a drawing of the car including the direction of the movement and its acceleration, just like in the attached picture.

b) From the information given by the problem I know:

initial velocity =0

acceleration = 2.40m/s^{2}

time = 12.0 s

initial position = 0

c)

unknown:

displacement.

in order to choose the appropriate equation, I must take the knowns and the unknown and look for a formula I can use to solve for the unknown. I know the initial velocity, initial position, time, acceleration and I want to find out the displacement. The formula that contains all this data is the following:

x=x_{0}+V_{x0}t+\frac{1}{2}a_{x}t^{2}

Once I got the equation I need to find the displacement, I can plug the known values in, like this:

x=0+0(12s)+\frac{1}{2}(2.40\frac{m}{s^{2}} )(12s)^{2}

after cancelling the pertinent units, I get that  my answer will be given in meters. So I get:

x=\frac{1}{2} (2.40\frac{m}{s^{2}} )(12s)^{2}

which solves to:

x=172.8m

So the displacement of the car in 12 seconds is 172.8m, which makes sense taking into account that it will be accelerating for 12 seconds and each second its velocity will increase by 2.4m/s.

d) So, like the previous part of the problem, I know the initial position of the car, the time it travels, the initial velocity and its acceleration. Now I also know what its final position is, so we have more than enough information to find this answer out.

I need to find the final velocity, so I need to use an equation that will use some or all of the known data and the unknown. In order to solve this problem, I can use the following equation:

a=\frac{V_{f}-V_{0} }{t}

Next, since I need to find the final velocity, I can solve the equation just for that, I can start by multiplying both sides by t so I get:

at=V_{f}-V_{0}

and finally I can add V_{0} to both sides so I get:

V_{f}=at+V_{0}

and now I can proceed and substitute the known values:

V_{f}=at+V_{0}

V_{f}=(2.40\frac{m}{s^{2}}} (12s)+0

which solves to:

V_{f}=28.8m/s

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3 years ago
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