All categories

3 years ago

Ohms law in symbols is

Physics

1 answer:

serious [3.7K]3 years ago

6 0

Answer:

The potential difference across a given given wire of an circuit is directly proportional to the current flowing through it provided it's temperature remains the same. This is known as ohm's law

V is proportional to I

V/I = constant

V/I = R

V = IR where R is the constant of proportionality and resistance of the wire

Hope this helps!

You might be interested in

A trait of an organism that helps it survive in its environment is calles

kompoz [17]

Answer is adaptation. An organism develops a trait over time to help survive in its environment called an adaptation. You could take a giraffe for example. A long time ago giraffes actually had short necks, but now since their food is higher they soon developed a longer neck and this is what we now see in the present. This goes for any artic animal. Polar bears and seals have a white fur adaptation to help them blend in with their environment. A chameleon changes colors in order to hide from predators and sneak up on prey. These are all adaptations

7 0

2 years ago

Read 2 more answers

A charge is divided q1 and (q-q1)what will be the ratio of q/q1 so that force between the two parts placed at a given distance i

Arturiano [62]

Answer:

$q / q_{1} = 2$ , assuming that $q_{1}$ and $(q - q_{1})$ are point charges.

Explanation:

Let $k$ denote the coulomb constant. Let $r$ denote the distance between the two point charges. In this question, neither $k$ and $r$ depend on the value of $q_{1}$ .

By Coulomb's Law, the magnitude of electrostatic force between $q_{1}$ and $(q - q_{1})$ would be:

$\begin{aligned}F &= \frac{k\, q_{1}\, (q - q_{1})}{r^{2}} \\ &= \frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\end{aligned}$ .

Find the first and second derivative of $F$ with respect to $q_{1}$ . (Note that $0 < q_{1} < q$ .)

First derivative:

$\begin{aligned}\frac{d}{d q_{1}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q\, q_{1} - {q_{1}}^{2})\right] \\ &= \frac{k}{r^{2}}\, \left[\frac{d}{d q_{1}} [q\, q_{1}] - \frac{d}{d q_{1}}[{q_{1}}^{2}]\right]\\ &= \frac{k}{r^{2}}\, (q - 2\, q_{1})\end{aligned}$ .

Second derivative:

$\begin{aligned}\frac{d^{2}}{{d q_{1}}^{2}}[F] &= \frac{d}{d q_{1}} \left[\frac{k}{r^{2}}\, (q - 2\, q_{1})\right] \\ &= \frac{(-2)\, k}{r^{2}}\end{aligned}$ .

The value of the coulomb constant $k$ is greater than $0$ . Thus, the value of the second derivative of $F$ with respect to $q_{1}$ would be negative for all real $r$ . $F\!$ would be convex over all $q_{1}$ .

By the convexity of $\! F$ with respect to $\! q_{1} \!$ , there would be a unique $q_{1}$ that globally maximizes $F$ . The first derivative of $F\!$ with respect to $q_{1}\!$ should be $0$ for that particular $\! q_{1}$ . In other words:

$\displaystyle \frac{k}{r^{2}}\, (q - 2\, q_{1}) = 0$ .

$2\, q_{1} = q$ .

$q_{1} = q / 2$ .

In other words, the force between the two point charges would be maximized when the charge is evenly split:

$\begin{aligned} \frac{q}{q_{1}} &= \frac{q}{q / 2} = 2\end{aligned}$ .

3 0

2 years ago

How to find the gradient of a velocity time graph

Sidana [21]

The area-

The area under the line in a velocity-time graph represents the distance travelled. To find the distance travelled in the graph above, we need to find the area of the light-blue triangle and the dark-blue rectangle.

Area of light-blue triangle -
The width of the triangle is 4 seconds and the height is 8 meters per second. To find the area, you use the equation: area of triangle = 1⁄2 × base × height so the area of the light-blue triangle is 1⁄2 × 8 × 4 = 16m. Area of dark-blue rectangle The width of the rectangle is 6 seconds and the height is 8 meters per second. So the area is 8 × 6 = 48m. Area under the whole graph The area of the light-blue triangle plus the area of the dark-blue rectangle is:16 + 48 = 64m.This is the total area under the distance-time graph. This area represents the distance covered.

7 0

3 years ago

Aaron created the chart to list the benefits and limitations of using synthetic polymers . In which row has Aaron made an error

Sever21 [200]

Answer:

Its B trust me

8 0

3 years ago

Read 2 more answers

Which of the following is not a reason fluorescent lamps are advantageous over incandescent lamps?A. Fluorescent lamps are more

mr_godi [17]

Answer:

B. Fluorescent lamps operate at a higher temperature than incandescent

Explanation:

Fluorescent lamps have a number of advantages over incandescent lamps which are given in the options given in A, C and D. The option available in B is a drawback, not an advantage. This is because it can give out and radiate more heat as a result of working at a higher temperature. Hence B option is correct.

8 0

3 years ago