two charges + 2.6 uc and -5.4uc experience an attractive force of 6.5mN. What is the separation between the charges?

The force F⃗ pulling the string is constant; therefore the magnitude of the angular acceleration α of the wheel is constant for

abruzzese [7]

Answer:

The answer is " $\boxed{\boxed{\omega = \sqrt{\frac{2fd}{kmr^2}}}}$ "

Explanation:

$\to d= r \theta \\\\ \to \theta =\frac{d}{r}\\\\\to \omega^{r} - \omega_{0}^{r} = 2 \alpha \theta\\\\\to \omega^{r} = 2 \alpha \theta - \omega_{0}^{r} \\\\\to \omega^{r} = 2 (\frac{F}{Kmr}) \frac{d}{r}\\\\\to \omega = \sqrt{\frac{2fd}{kmr^2}}$

5 0

3 years ago

What is the relationship between temperature and altitude in the stratosphere? (2 points).

stealth61 [152]

As altitude increases, temperature increases. The stratosphere is the part of the atmosphere that starts in the tropopause and ends in the estratopause. In the troposphere, the air is close to the Earth surface. The air surface can absorb more sunlight energy than the air, so the Earth surface heats the air. As you go higher, the distance to the Earth surface is higher, so the temperature is lower. The troposphere ends in the tropopause, where this trend changes. In the estratopause, there is a lot of ozone, which absorbs the dangerous UV radiation and converts into heat. That heat warms the air. So the air which is close to the estratopause is warm because of the heat released by the ozone reactions. The tropopause is far from the Earth surface and far from the ozone layer, that’s why it is cold. So the tropopause is cold and the estratopause is warm, which means: the air becomes warmer <span>as you rise above the tropopause until you get to the estratopause.</span>

8 0

3 years ago

An automobile engine can produce 153 N · m of torque. Calculate the angular acceleration (in rad/s^2) produced if 85.2% of this

galina1969 [7]

Answer:

46.2 rad/s2

Explanation:

Angular acceleration works very similar to linear acceleration, it follows this equation:

$\gamma = \frac{Mt}{J}$