Charge flows from low potential to high potential. O True or O False

Physics

1 answer:

inn [45]3 years ago

8 0

Answer:

False

Explanation:

Think of the electric potential in terms of potential energy. If you imagine a place with high elevation (A) and another one at sea level (B), a ball will roll from high potential to low potential (A-->B).

Everything in our universe wants to reach a lower state of energy if no external force is acted upon it. Every object tends to slow down (friction), a radioactive element dissipates energy (an unstable element releases energy to get to a stable state), water in the clouds comes down to the ground (rain experiencing difference in potential energy).

Electric potential is exactly the same, you just can't see it! It flows from higher voltage (which is a synonym for electric potential) to lower voltage.

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3 years ago

A maintenance worker wants to torque an engine bolt to 65.0 N m. If the torque wrench is 35cm in length, what is force applied t

harina [27]

Answer:

Force = 186 N

Explanation:

Torque is the rotational equivalent of linear force. It can be easely calculated using the formula :

$Torque = \vec{r} \times \vec{F}$

Where $\vec{r}$ is a vector that from the origin of the coordinate system to the point at which the force is applied (the position vector), $\vec{F}$ is the applied force.

The easiest way of computing the force is by setting the origin of the coordinate system to the lowest point of the torque wrench. By doing this we have that $r$ (the magnitud of the position vector) is 35cm.

Before computing the force we need to set all our values to the international system of units (SI). The torque is already in SI. The one missing is the length of the torque wrench (it is in centimeters and we need it in meters). So :

$35cm * \dfrac{1m}{100cm} = 0.35m$

Now using the torque formula:

$Torque = \vec{r} \times \vec{F}$

$Torque = rFsin(\theta)}$

Where $\theta$ is the smaller angle between the force and the position vector. Because the force is applied perpendiculary to the position vector $\theta = 90°$ , thus :