Calculate the total resustance<br> Circut 2ohmls, 2Ohlms<br> In parrelle

Marine science- question- what is climate change?

Troyanec [42]

Answer:

Defined below

Explanation:

Climate change is simply the long term change in the average weather patterns that are associated with the local, regional and global climates of the earth. Climate change is usually driven by human activities like burning of fossils; natural processes like cyclic ocean patterns; external factors like volcanic eruptions.

8 0

2 years ago

A one-piece cylinder has a core section protruding from the larger drum and is free to rotate around its central axis. A rope wr

PilotLPTM [1.2K]

Answer:

Magnitude the net torque about its axis of rotation is 2.41 Nm

Solution:

As per the question:

The radius of the wrapped rope around the drum, r = 1.33 m

Force applied to the right side of the drum, F = 4.35 N

The radius of the rope wrapped around the core, r' = 0.51 m

Force on the cylinder in the downward direction, F' = 6.62 N

Now, the magnitude of the net torque is given by:

$\tau_{net} = \tau + \tau'$

where

$\tau$ = Torque due to Force, F

$\tau'$ = Torque due to Force, F'

$tau = F\times r$

$tau' = F'\times r'$

Now,

$\tau_{net} = - F\times r + F'\times r'$

$\tau_{net} = - 4.35\times 1.33 + 6.62\times 0.51 = - 2.41\ Nm$

The net torque comes out to be negative, this shows that rotation of cylinder is in the clockwise direction from its stationary position.

Now, the magnitude of the net torque:

$|\tau_{net}| = 2.41\ Nm$

3 0

2 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$