A river flows at 2m/s the velociy of ferry relative to the shore is 4m/s

A charge of 4.5 × 10-5 C is placed in an electric field with a strength of 2.0 × 104 . If the charge is 0.030 m from the source

snow_tiger [21]

Answer:

The electrical potential energy is 0.027 Joules.

Explanation:

The values from the question are

charge (q) = $4.5 \times 10^{-5} C$

Electric Field strength (E) = $2.0 \times 10^{4} N/C$

Distance from source (d) = 0.030 m

Now the formula for the electrical potential energy (U) is given by

$U = q \times E \times d$

So now insert the values to find the answer

$U = 4.5 \times 10^{-5} C \times 2.0 \times 10^{4} N/C \times 0.030 m$

On further solving

$U = 0.027 J$

8 0

3 years ago

Help asapp!!!!!!!!!!!!!!

IRINA_888 [86]

Sorry don’t know this one

3 0

3 years ago

What is the difference between current and voltage besides their units of measurement

kondor19780726 [428]

"Voltage" is the "pressure" that makes electrons want to leave where they are
and head in some direction, if there's conducting material in that direction.

"Current" is the rate at which they all migrate in that direction.

8 0

3 years ago

A 39-foot ladder is leaning against a vertical wall. If the bottom of the ladder is being pulled away from the wall at the rate

Viefleur [7K]

Answer:

The rate of change of the area when the bottom of the ladder (denoted by $b$ ) is at 36 ft. from the wall is the following:

$\frac{dA}{dt}|_{b=36}=-571.2\, ft^2/s$

Explanation:

The Area of the triangle is given by $A=h\times b$ where $h=\sqrt{l^2-b^2}$ (by using the Pythagoras' Theorem) and $b$ is the length of the base of the triangle or the distance between the bottom of the ladder and the wall.

The area is then

$A=\sqrt{l^2-b^2}b$

The rate of change of the area is given by its time derivative

$\frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\cdot b\right)$

$\implies \frac{dA}{dt}=\frac{d}{dt}\left(\sqrt{l^2-b^2}\right)\cdot b+\frac{db}{dt}\cdot\sqrt{l^2-b^2}$

$\implies\frac{dA}{dt}=\frac{1}{2\sqrt{l^2-b^2}}\frac{d}{dt}(l^2-b^2)\cdot b+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Product rule

$\implies\frac{dA}{dt}=-\frac{1}{2\sqrt{l^2-b^2}}\cdot 2\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$ Chain rule

$\implies\frac{dA}{dt}=-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2\cdot \frac{db}{dt}+\sqrt{l^2-b^2}}\cdot \frac{db}{dt}$

$\implies\frac{dA}{dt}=\frac{db}{dt}\left(-\frac{1}{\sqrt{l^2-b^2}}\cdot b^2+\sqrt{l^2-b^2}}\right)$

In here we can identify $b=36\, ft$ , $l=39$ and $\frac{db}{dt}=8\,ft/s$ .

The result is then

$\frac{dA}{dt}=8\left(-\frac{1}{\sqrt{39^2-36^2}}\cdot 36^2+\sqrt{39^2-36^2}}\right)=-571.2\, ft^2/s$

3 0

3 years ago

A(n) _____ is an organic compound that changes color in acid or bases.

Romashka-Z-Leto [24]

The answer is indicator.

6 0

3 years ago