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NARA [144]
3 years ago
5

Electric charge that is stationary is​

Physics
1 answer:
Andru [333]3 years ago
8 0

Answer:

is only producing an electric field in the surrounding space

Explanation:

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Mindy places a strip of zinc in a copper sulfate solution, as shown in the
mr Goodwill [35]

Answer:

C. The concentration of the copper sulfate is too low

Explanation:

8 0
3 years ago
Read 2 more answers
Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency 1. 72 x 1015 hz?
postnew [5]

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

The distance between the two positive, two negative, or two minimal points on the waveform is known as the wavelength of the wave. The following formula expresses the relationship between the frequency and wavelength of light:

f = c / λ

where, f = frequency of light

            c = speed of light

            λ = wavelength of light

Given data = f = 1.72×10^{15}Hz

Therefore, λ = 3×10^{8} / 1.72×10^{15}

                  λ = 1.74×10^{-7}m

The wavelength, which represents the size of the smallest detectable detail that uses ultraviolet light  , is calculated as follows: 3×10^{8} / 1.72×10^{15} or approximately 1.74×10^{-7}m.

Learn more about light here;

brainly.com/question/15200315

#SPJ4

7 0
2 years ago
The nonreflective coating on a camera lens with an index of refraction of 1.21 is designed to minimize the reflection of 570-nm
lord [1]

Answer: 117.8 nm

Explanation:

Given,

Nonreflective coating refractive index : n = 1.21

Index of refraction: n_0 = 1.52

Wave length of light = λ = 570 nm = 570\times10^{-9}\ m

\text{ Thickness}=\dfrac{\lambda}{4n}

=\dfrac{570\times10^{-9}\ m}{4\times1.21}\\\\\approx\dfrac{117.8\times 10^{-9}\ m}{1}\\\\=117.8\text{ nm}

Hence, the minimum thickness of the coating that will accomplish= 117.8 nm

5 0
3 years ago
Air bags are designed to deploy in 10 ms. Given that the air bags expand 20 cm as they deploy, estimate the acceleration of the
joja [24]

As it is given that the air bag deploy in time

t = 10 ms = 0.010 s

total distance moved by the front face of the bag

d = 20 cm = 0.20 m

Now we will use kinematics to find the acceleration

d = v_i*t + \frac{1}{2}at^2

0.20 = 0 + \frac{1}{2}a*0.010^2

0.20 = 5 * 10^{-5}* a

a = 4000 m/s^2

now as we know that

g = 10 m/s^2

so we have

a = 400g

so the acceleration is 400g for the front surface of balloon

3 0
3 years ago
PLEASE HELP!
Kobotan [32]

Answer:

No

Explanation:

Cause a monster truck don

3 0
3 years ago
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