What could be used as another word for electrical potential

Physics

2 answers:

tekilochka [14]2 years ago

5 0

Answer:

hey mate

answer is probably voltage as per me

Explanation:

Voltage, electric potential difference, electric pressure or electric tension is the difference in electric potential between two points, which is defined as the work needed per unit of charge to move a test charge between the two points

Makovka662 [10]2 years ago

3 0

Answer is B: voltage

You might be interested in

Urgent! Based on the diagram below, what color will each pigmented paper appear to be to an observer?

Nastasia [14]

Answer:

Example A will appear green, while Example B will appear greenish-blue.

Explanation:

The color of an object depend on which part of the visible light it reflects towards the observer. Visible light is made up of seven colors: Violet, Indigo, Blue, Green, Yellow, Orange, and Red (VIBGYOR). If all the colors will be reflected object will appear white. If all the colors are absorbed the object appears black. In example A, only green color is being reflected so it will appear <em>Green</em>.

In example B, green and blue are being reflected so the object will appear a mix of green and blue. This color is cyan (greenish blue).

7 0

3 years ago

A guitar player tunes the fundamental frequency of a guitar string to 560 Hz. (a) What will be the fundamental frequency if she

lawyer [7]

Answer:

(a) if she increases the tension in the string is increased by 15%, the fundamental frequency will be increased to 740.6 Hz

(b) If she decrease the length of the the string by one-third the fundamental frequency will be increased to 840 Hz

Explanation:

(a) The fundamental, f₁, frequency is given as follows;

$f_1 = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L}$

Where;

T = The tension in the string

μ = The linear density of the string

L = The length of the string

f₁ = The fundamental frequency = 560 Hz

If the tension in the string is increased by 15%, we will have;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T\times 1.15}{\mu}} }{2 \cdot L} = 1.3225 \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = 1.3225 \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 = (1 + 0.3225) \times f_1$

$f_{(1 \, new)} = 1.3225 \times f_1 =\dfrac{132.25}{100} \times 560 \ Hz = 740.6 \ Hz$

Therefore, if the tension in the string is increased by 15%, the fundamental frequency will be increased by a fraction of 0.3225 or 32.25% to 740.6 Hz

(b) When the string length is decreased by one-third, we have;

The new length of the string, $L_{new}$ = 2/3·L

The value of the fundamental frequency will then be given as follows;

$f_{(1 \, new)} = \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \times \dfrac{2 \times L}{3} } =\dfrac{3}{2} \times \dfrac{\sqrt{\dfrac{T}{\mu}} }{2 \cdot L} = \dfrac{3}{2} \times 560 \ Hz = 840 \ Hz$