An electromagnetic wave has a frequency of 793 Hz. What is<br> its wavelength?

A uniform electric field with a magnitude of 5750 N/C points in the positive x direction. Find the change in electric potential

castortr0y [4]

Explanation:

Given that,

Electric field = 5750 N/C

Charge $q=+10.5\times10^{-6}\ C$

Distance = 5.50 cm

(a). When the charge is moved in the positive x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot d$

$\Delta U=-q(E\cdot d)$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times5.50\times10^{-2}$

$\Delta U=-3.32\times10^{-3}\ J$

The change in electric potential energy is $-3.32\times10^{-3}\ J$

(b). When the charge is moved in the negative x- direction

We need to calculate the change in electric potential energy

Using formula of electric potential energy

$\Delta U=-W$

$\Delta U=-F\cdot (-d)$

$\Delta U=-q(E\cdot (-d))$

Put the value into the formula

$\Delta U=-10.5\times10^{-6}\times5750\times(-5.50\times10^{-2})$

$\Delta U=3.32\times10^{-3}\ J$

The change in electric potential energy is $3.32\times10^{-3}\ J$

Hence, This is the required solution.

3 0

3 years ago

What is the acceleration of a 1000kg car subject to a 550N net force?

igomit [66]

Answer:

a=550÷1000

a=0.55m/s²

5 0

3 years ago

A 5cm tall object is placed 4cm in front of a converging lens that has a focal length of 8cm. Where is the image located in ____

OverLord2011 [107]

Answer:

a. -8 cm

Explanation:

$d_{o}$ = distance of the object = 4 cm

$d_{i}$ = distance of the image = ?

$f$ = focal length of the converging lens = 8 cm

using the lens equation

$\frac{1}{d_{o}} + \frac{1}{d_{i}} = \frac{1}{f}$

$\frac{1}{4} + \frac{1}{d_{i}} = \frac{1}{8}$

$d_{i}$ = - 8 cm

4 0

3 years ago

I need help on the data section of the circuit design lab on Edg.

Arte-miy333 [17]

I hope it's not too late, but here you go

8 0

3 years ago

Gauss's law: Group of answer choices can always be used to calculate the electric field. relates the electric field throughout s

kvasek [131]

Answer:

relates the electric field at points on a closed surface to the net charge enclosed by that surface.

Explanation:

Gauss Law states that overall electric flux of a closed surface is equivalent right to charge enclosed which is divided by the permittivity. In other words Gauss Law stress that

net electric flux that pass through an hypothetical closed surface is equivalent to overall electric charge present within that closed surface.

The Gauss law can be expressed mathematically as

ϕ = (Q/ϵ0)

Q = total charge within the surface,

ε0 = the electric constant

5 0

3 years ago

An electromagnetic wave has a frequency of 793 Hz. What is its wavelength?