2 years ago

I WILL MARK BRAINLIEST!!ASAP!!! Wet Lab - Coulomb's Law lab from edge!!

Physics

1 answer:

anyanavicka [17]2 years ago

8 0

Answer:

i don't get what I have to do

You might be interested in

The force between two charges, q, and 92, is F. If the distance between the

Sholpan [36]

Answer:

Explanation:

F = kQ₁Q₂/d²

F' = k2Q₁2Q₂/d²

F' = 4(kQ₁Q₂/d²)

F' = 4F

4 0

2 years ago

Why do we balance chemical equations

antiseptic1488 [7]

So it could follow the correct mass for the atom

5 0

2 years ago

Testosterone is used to treat breast cancer ? True or false?

Triss [41]

Answer:

True

Explanation:

Because I had a test on this

7 0

2 years ago

Natural history research involves ____ unobsevable and unrepeatable past events.

Tanzania [10]

B verifying is the answer

4 0

3 years ago

An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

$v=\sqrt{\frac{T}{\mu} }$

where T= 200 N is tension in the string , $\mu$ =1.0 g/m is the linear mass density

$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

Wavelength of the wave in the string is

$\lambda =2L=2\times 0.8=1.6 m$

The frequency is

$f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz$

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

$\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m$

8 0

2 years ago