Physics, 27.07.2021 19:40 ugh788o02

A particle makes 800 revolution in 4 minutes of a circle of 5cm. find i. It's period ii. Angular Velocity iii. Linear Velocity iv. It's acceleration please with the formula stated clearly and the meaning of the symbols with accurate solutions you can snap and send it THANK YOU

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Answer from: manou76

(i) The particle has a period of 0.3 seconds.

(ii) The angular velocity of the particle is a20.944 radians per second.

(iii) The linear velocity of the particle is 1.047 meters per second.

(iv) The linear acceleration of the particle is 21.933 meters per square second.

Explanation:

Statement is incomplete and have mistakes, complete and correct form is presented below:

A particle makes 800 revolutions in 4 minutes of a circle with a radius of 5 centimeters. Find (i) its period, (ii) its angular velocity, (iii) its linear velocity and (iv) its acceleration.

(i) The period of the particle (), in seconds, is the time taken to make a complete revolution:

$T = \frac{4\,min \times \frac{60\,s}{1\,min} }{800\,rev}$

$T = 0.3\,s$

The particle has a period of 0.3 seconds.

(ii) The angular velocity ( $\omega$ ), in radians per second, is determined by the following formula:

$\omega = \frac{2\pi}{T}$ (1)

$\omega = \frac{2\pi}{0.3\,s}$

$\omega \approx 20.944\,\frac{rad}{s}$

The angular velocity of the particle is a20.944 radians per second.

(iii) The linear velocity (), in meters per second, is calculated by the following formula:

$v = R\cdot \omega$ (2)

Where is the radius of the circle, in meters.

If we know that $R = 0.05\,m$ and $\omega \approx 20.944\,\frac{rad}{s}$ , then the linear velocity of the particle is:

$v = (0.05\,m)\cdot \left(20.944\,\frac{rad}{s} \right)$

$v = 1.047\,\frac{m}{s}$

The linear velocity of the particle is 1.047 meters per second.

(iv) Since angular velocity is constant, linear acceleration of the particle (), in meters per square second, is entirely radial. Acceleration can be found by means of this expression:

$a = \omega^{2}\cdot R$ (3)

If we know that $R = 0.05\,m$ and $\omega \approx 20.944\,\frac{rad}{s}$ , then the linear acceleration of the particle is:

$a = \left(20.944\,\frac{rad}{s} \right)^{2}\cdot (0.05\,m)$

$a = 21.933\,\frac{m}{s^{2}}$

The linear acceleration of the particle is 21.933 meters per square second.

Answer from: Quest

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.

Answer from: Quest

terd muffin is the correct answer