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densk [106]
3 years ago
11

A 1kg sphere rotates in a circular path of radius 0.2m from rest and it reaches an angular speed of 20rad/sec in 10 second calcu

late the angular acceleration​
Physics
1 answer:
ludmilkaskok [199]3 years ago
3 0

Answer:

2 rad/s²

Explanation:

From the question given above, the following data were obtained:

Mass (m) = 1 Kg

Radius (r) = 0.2 m

Angular speed (w) = 20 rad/s

Time (t) = 10 s

Angular acceleration (a) =?

Angular acceleration is defined as:

Angular acceleration (a) =Angular speed (w) / time

a = w/t

With the above formula, we can obtain the angular acceleration of the sphere as follow:

Angular speed (w) = 20 rad/s

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20 / 10

a = 2 rad/s²

Thus, the angular acceleration of the sphere is 2 rad/s²

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A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the
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1) 1.11\cdot 10^{-7} J

The capacitance of a parallel-plate capacitor is given by:

C=\frac{\epsilon_0 A}{d}

where

\epsilon_0 is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

r=\frac{2.0 cm}{2}=1.0 cm=0.01 m

so the area is

A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2

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So the capacitance is

C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F

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U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J

2) 0.71 J/m^3

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E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m

and the energy density of the electric field is given by

u=\frac{1}{2}\epsilon_0 E^2

and using

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7 0
4 years ago
PLEASE HELP!!!!!!
Dafna11 [192]
VF=0
Vi=60mph
T=4.0s

you need to convert hours to seconds... so you do there is 60 minutes in an hour and there is 60 seconds in a minute so you do 60*3=180
180/4.0=45m/s^2
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