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 calculate the tangential acceleration ?

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

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

not moving. pos accel,