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kirza4 [7]
3 years ago
6

The angular speed of digital video discs (DVDs) varies with whether the inner or outer part of the disc is being read. (CDs func

tion in the same way.) Over a 133min playing time, the angular speed varies from 570rpm to 1600rpm .Assuming it to be constant, what is the angular acceleration (in/rads sec ) of such a DVD?
Physics
1 answer:
Kitty [74]3 years ago
8 0

Answer:

α = 0.0135 rad/s²

Explanation:

given,

t = 133 min = 133 x 60 = 7980 s

angular speed varies from 570 rpm to 1600 rpm

now,

570 rpm = 570 \times \dfrac{2\pi}{60}

              = 59.69 rad/s

1600 rpm =  = 570 \times \dfrac{2\pi}{60}

              = 167.6 rad/s

using equation of rotational motion

ωf = ωi + αt    

167.6 = 59.7 + α x 7980

α  x 7980 = 107.9

α = 0.0135 rad/s²

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b

Explanation:

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How does the end point differ from the equivalence <br>point of a titration?​
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<u>Answer:</u>

<em>Equivalence point and end point are terminologies in pH titrations and they are not the same. </em>

<u>Explanation:</u>

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At equivalence point the <em>number of moles of the acid</em> will be equal to the number of moles of the base as given in the equation.  The nature of the final solution determines the <em>pH at equivalence point. </em>

<em>A pH less than 7 will be the result if the resultant is acidic and if it is basic the pH will be greater than 7. </em>In a strong base-strong acid and weak base-weak acid titration the pH at the equivalence point will be 7 indicating <em>neutral nature of the solution. </em>

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3 years ago
Physical properties of matter can be observed without changing the identity of the substance . The phases of matter are physical
ZanzabumX [31]

Solids have molecules that do not vibrate.

4 0
3 years ago
When a piece of aluminium foil is taken out of the oven and cools from 1000C to 500C. What is the change in temperature?
zmey [24]

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3 years ago
Calculate the propellant mass required to launch a 2000 kg spacecraft from a 180 km circular orbit on a Hohmann transfer traject
Finger [1]

Answer:

t = 12,105.96 sec

Explanation:

Given data:

weight of spacecraft is 2000 kg

circular orbit distance to saturn = 180 km

specific impulse = 300 sec

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earth orbit around the sun R_1= 149.6 * 10^ 6 km

time required for the mission is given as t

t = \frac{2\pi}{\sqrt{\mu_sun}} [\frac{1}{2}(R_1 + R_2)]^{3/2}

where

\mu_{sun} is gravitational parameter of sun =  1.32712 x 10^20 m^3 s^2.t = \frac{2\pi}{\sqrt{ 1.32712 x 10^{20}}} [\frac{1}{2}(149.6 * 10^ 6 +1.43 *10^9 )]^{3/2}

t = 12,105.96 sec

6 0
3 years ago
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