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nikitadnepr [17]
4 years ago
9

What is linear speed called when something moves in a circle

Physics
1 answer:
Dmitry [639]4 years ago
4 0

Answer: Tangential Velocity

The tangential velocity V is defined as the angular velocity \omega by the radius r of circular motion. As shown below:

V=\omega. r

Its name is due to the fact that this linear velocity vector is always tangent to the trajectory and is the distance traveled by a body or object in a circular movement in a period of time.

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If the light strikes the plastic (from the water) at an angle θw, at what angle θa does it emerge from the plastic (into the air
Natasha2012 [34]

Answer:

\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

Explanation:

According to Snell's law the angle of incidence and angle of refraction are related by:

\frac{sin(\theta _w)}{sin(\theta _a)}=\frac{n_a}{n_w}\\\\\therefore sin(\theta _a)=\frac{sin(\theta _w)\times n_w}{n_a}\\\\\theta _a=sin^{-1}(\frac{sin(\theta _w)\times n_w}{n_a})\\\\\theta _a=asin(\frac{sin(\theta _w)\times n_w}{n_a})

6 0
3 years ago
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PLEASE HELP!!!!
Aleks [24]
The answer is step by step 65
4 0
3 years ago
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24 POINTS
vesna_86 [32]
80 meters should be it
3 0
3 years ago
Assuming a successful launch, what is the speed of the small CO2 rocket given that it travels about 10 meters in about 3 seconds
antiseptic1488 [7]
Probably 30 im not sure 
4 0
3 years ago
The position of a particle moving along the x axis depends on the time according to the equation x = ct² - bt³, where x is in me
Kamila [148]

Answer:

(a) \frac{m}{s^2}

(b) \frac{m}{s^3}

(c) 1 s

(d) 20 m

(e) 1 m

(f) 0\frac{m}{s}

(g) -12\frac{m}{s}

(h) -36\frac{m}{s}

(i) -72\frac{m}{s}

(j) -6\frac{m}{s^2}

(k) -18\frac{m}{s^2}

(l) -30\frac{m}{s^2}

(m) -42\frac{m}{s^2}

Explanation:

Since <em>x</em> is measured in meters and <em>t</em> in seconds, constants <em>a </em>and <em>b</em> must have units that gives meters when multiplied by square and cubic seconds respectivly, so that would mean \frac{m}{s^2} for <em>a </em>and \frac{m}{s^3} for <em>b</em>.

We can get the velocity <em>v </em>equation by deriving the position with respect to <em>t</em>, which gives:

v=6*t-6*t^2

And the acceleration <em>a</em> equation by deriving again:

a=6-12*t

Now for getting the maximun position between 0 and 4, we must find to points where the positions first derivate is equal to cero and evaluate those points. That is <em>v=0</em>, which gives

6*t-6*t^2=0\\6t*(1-t)=0\\t=0 or t=1

For <em>t = 0</em>,<em> x = 0</em> so the maximun position is archieved at 1 second, which gives <em>x = 1 meter</em>.

For obtaining it's displacement <em>r</em>, we can integrate the velocity from 0 seconds to 4 seconds, which gives the mean value of the position in that interval:

r=\frac{1}{4}*(2*4^3-3*4^2)m\\r= 20m

For the remaining questions, we just replace the values of <em>t</em> on the respective equations.

8 0
4 years ago
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