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Lunna [17]
3 years ago
15

I need help with finding these?

Physics
1 answer:
Alchen [17]3 years ago
6 0
I think 23 is b and 24 a?
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URGENT!!!
madreJ [45]

Answer:

f=6.97\times 10^{14}\ Hz

Explanation:

Given that,

Wavelength, \lambda=430\ nm=430\times 10^{-9}\ m

We need to find the frequency of the violet light.

We know that the relation between frequency and wavelength is given by :

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{430\times 10^{-9}}\\\\f=6.97\times 10^{14}\ Hz

So, the frequency of violet light is 6.97\times 10^{14}\ Hz.

7 0
3 years ago
The work function for magnesium is 3.70 ev. what is its cutoff frequency?
alexandr402 [8]

The cutoff frequency for magnesium is 8.93 x 10¹⁴ Hz.

<h3>What is cutoff frequency?</h3>

The work function is related to the frequency as

W0 = h x fo

where, fo = cutoff frequency and h is the Planck's constant

Given is the work function for magnesium is  3.70 eV.

fo = 3.7 x 1.6 x 10⁻¹⁹ / 6.626 x 10⁻³⁴

fo = 8.93 x 10¹⁴ Hz.

Thus, the cut off frequency is 8.93 x 10¹⁴ Hz.

Learn more about cutoff frequency.

brainly.com/question/14378802

#SPJ1

7 0
1 year ago
Which situation is NOT the result of an unbalanced force acting on an object? A. an object speeds up B. an object maintains spee
7nadin3 [17]
Im pretty sure its b........
5 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
Why is it important to understand forces?
yanalaym [24]
If people never learned forces, there would be a major gap in the world and how it works, let alone in physics...
as much as you don't wanna admit it, force is everywhere and you see it if not use it EVERY day in your life, something as simple as driving a car down the street or too school, your using force of your wheels to move your car, which is moving you
3 0
3 years ago
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