3 years ago

Select the statement that best describes how the forces relate. (2 points)

2 answers:

4 0

"D. Magnetic and electrical forces are similar because they are both related to the interactions between charged particles" best describes how the forces relate.

telo118 [61]3 years ago

3 0

Your answer would be d

You might be interested in

A pushing force acts on a toy car for 3 seconds causing it to accelerate at a rate A’. If the car is replaced with a similar, bu

Alla [95]

Answer:

one-half of A'

Explanation:

3 0

3 years ago

Read 2 more answers

Josie sees lightning off in the distance. A few seconds later she hears thunder. What can Josie conclude?

Naddik [55]

Light wave travel faster than sound waves

8 0

3 years ago

Read 2 more answers

Dianne's teacher has a chunk of dry ice for the class to look at. Dry ice is a kind of frozen gas—it is very cold. The teacher w

Gelneren [198K]

The answer is A. Hope this helps you with your work.

4 0

3 years ago

Read 2 more answers

You guys cant answer a dam question I ask it fricking 6th grade questions

nexus9112 [7]

Answer:

calm down please its not that serious maybe no one saw it yet

Explanation:

7 0

3 years ago

Enter your answer in the provided box. The mathematical equation for studying the photoelectric effect is hν = W + 1 2 meu2 wher

siniylev [52]

Answer:

$v = 4.44 \times 10^5 m/s$

Explanation:

By Einstein's Equation of photoelectric effect we know that

$h\nu = W + \frac{1}{2}mv^2$

here we know that

$h\nu$ = energy of the photons incident on the metal

$W$ = minimum energy required to remove photons from metal

$\frac{1}{2}mv^2$ = kinetic energy of the electrons ejected out of the plate

now we know that it requires 351 nm wavelength of photons to just eject out the electrons

so we can say

$W = \frac{hc}{351 nm}$

here we know that

$hc = 1242 eV-nm$

now we have

$W = \frac{1242}{351} = 3.54 eV$

now by energy equation above when photon of 303 nm incident on the surface

$\frac{1242 eV-nm}{303 nm} = 3.54 eV + \frac{1}{2}(9.1 \times 10^{-31})v^2$

$4.1 eV = 3.54 eV + (4.55 \times 10^{-31}) v^2$

$(4.1 - 3.54)\times 1.6 \times 10^{-19}) = (4.55 \times 10^{-31}) v^2$

$8.96 \times 10^{-20} = (4.55 \times 10^{-31}) v^2$

$v = 4.44 \times 10^5 m/s$

6 0

3 years ago