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Crazy boy [7]
3 years ago
6

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 tangential acceleration ?​
Physics
1 answer:
Len [333]3 years ago
6 0

Answer:

0.4 m/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/sec

Time (t) = 10 s

Tangential acceleration (aₜ) =?

Next, we shall determine the angular acceleration (a) of the sphere. This can be obtained as follow:

Angular speed (w) = 20 rad/sec

Time (t) = 10 s

Angular acceleration (a) =?

a = w/t

a = 20/10

a = 2 rad/s²

Finally, we shall determine the tangential acceleration (aₜ) of the sphere. This can be obtained as follow:

The tangential acceleration (aₜ) and the angular acceleration (a) are related according to the equation:

Tangential acceleration (aₜ) = Angular acceleration (a) × Radius (r)

aₜ = ar

With the above formula, we can obtain the tangential acceleration (aₜ) as follow:

Radius (r) = 0.2 m

Angular acceleration (a) = 2 rad/s²

Tangential acceleration (aₜ) =?

aₜ = ar

aₜ = 2 × 0.2

aₜ = 0.4 m/s²

Therefore, the tangential acceleration is 0.4 m/s²

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2 years ago
What quantity of heat is needed to convert 1 kg of ice at -13 degrees C to steam at 100 degrees C?
Effectus [21]

Answer:

Heat energy needed = 3036.17 kJ

Explanation:

We have

     heat of fusion of water = 334 J/g

     heat of vaporization of water = 2257 J/g

     specific heat of ice = 2.09 J/g·°C

     specific heat of water = 4.18 J/g·°C

     specific heat of steam = 2.09 J/g·°C

Here wee need to convert 1 kg ice from -13°C to vapor at 100°C

First the ice changes to -13°C from 0°C , then it changes to water, then its temperature increases from 0°C to 100°C, then it changes to steam.

Mass of water = 1000 g

Heat energy required to change ice temperature from -13°C to 0°C

          H₁ = mcΔT = 1000 x 2.09 x 13 = 27.17 kJ

Heat energy required to change ice from 0°C to water at 0°C

          H₂ = mL = 1000 x 334 = 334 kJ

Heat energy required to change water temperature from 0°C to 100°C  

          H₃ = mcΔT = 1000 x 4.18 x 100 = 418 kJ    

Heat energy required to change water from 100°C to steam at 100°C  

          H₄ = mL = 1000 x 2257 = 2257 kJ    

Total heat energy required

          H = H₁ +  H₂ + H₃ + H₄ = 27.17 + 334 + 418 +2257 = 3036.17 kJ

Heat energy needed = 3036.17 kJ

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3 years ago
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