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Nata [24]
3 years ago
6

Describe where each subatomic particle is found in an atom.

Physics
2 answers:
Artemon [7]3 years ago
4 0

Answer:

Protons and neutrons are found in the nucleus, and electrons are found in the circles, or orbits, around the nucleus.

Explanation:

Anestetic [448]3 years ago
3 0

Answer:

In the nucleus

Explanation:

You find it in the nucleus. This is where protons and neutrons are. Don't forget the quarks as well ;)

You might be interested in
in which type of wave are vibrations at right angles to the direction in which the wave is travelling​
Pani-rosa [81]

Answer:

longitudinal waves have those properties

7 0
3 years ago
A flat sheet of paper of area 0.365 m2 is oriented so that the normal to the sheet is at an angle of 60 ∘ to a uniform electric
andriy [413]

Answer:

A.) 3.65 N*m²/C B) No C) 0º D) 90º

Explanation:

A) The electric flux, when the electric field is uniform across a gausssian surface, can be calculated as the dot product of the electric field vector, and the vector representing the area of the surface (normal to the surface and directed outward it by convention), as follows:

Flux = E*A*cos φ

where E = 20 N/C, A = 0.365 m², φ = 60º.

Replacing by the values, we can get the value of the electric flux, as follows:

Flux = 20 N/C* 0.365 m²*0.5 = 3.65 N*m²/C

B) While the area remains constant, and doesn't change orientation, the value of the flux will be the same, regardless the shape of the sheet.

C) When the normal to the sheet and the electric field are parallel each other, the surface will intercept the maximum number of field lines, i.e. the flux will be directly E*A*cos 0º = E*A (maximum value possible).

D) When the electric field is tangent to the surface, this means that no field lines will be intercepted by the sheet, so the flux is zero.

In this case, φ = 90º, cos φ = 0

⇒ E*A*cos 90º = E*A*0 = 0

5 0
3 years ago
A ray of light traveling through air strikes a piece of diamond at an angle of incidence equal to 56 degrees. Calculate the angu
Montano1993 [528]

Answer:

The angle of separation is  \Delta \theta =  0.93 ^o

Explanation:

From the question we are told that

    The angle of incidence is  \theta  _ i  = 56^o

     The refractive index of violet light  in diamond  is  n_v = 2.46

       The refractive index of red light in diamond is n_r = 2.41

      The wavelength of violet light is  \lambda _v = 400nm = 400*10^{-9}m

         The wavelength of red  light is  \lambda _r = 700nm = 700*10^{-9}m

Snell's  Law can be represented mathematically as

         \frac{sin \theta_i}{sin \theta_r} = n

Where \theta_r is the angle of refraction

=>       sin \theta_r  =   \frac{sin \theta_i}{n}

Now considering violet light

               sin \theta_r__{v}}  =   \frac{sin \theta_i}{n_v}

substituting values

                sin \theta_r__{v}}  =   \frac{sin (56)}{2.46}

                 sin \theta_r__{v}}  =  0.337

                 \theta_r__{v}}  =  sin ^{-1} (0.337)

                 \theta_r__{v}}  =  19.69^o

Now considering red light

               sin \theta_r__{R}}  =   \frac{sin \theta_i}{n_r}

substituting values

                sin \theta_r__{R}}  =   \frac{sin (56)}{2.41}

                 sin \theta_r__{R}}  =  0.344

                 \theta_r__{R}}  =  sin ^{-1} (0.344)

                 \theta_r__{R}}  = 20.12^o

The angle of separation between the red light and the violet light is mathematically evaluated as

                  \Delta \theta = \theta_r__{R}} -  \theta_r__{V}}

substituting values

                  \Delta \theta =20.12 - 19.69

                  \Delta \theta =  0.93 ^o

6 0
4 years ago
Jimmy walks 3 miles east then 1 mile west. What is the distance Jimmy traveled?
AleksAgata [21]

Answer:

2 miles

Explanation:

Jimmy walked forward 3 miles east (to the right)

Then, he backtracked and walked 1 mile west (to the left)

This can be written as:

3 - 1 = ?

3 - 1 = 2

Therefore, Jimmy traveled a total of 2 miles.

4 0
3 years ago
Somebody please help me!! Timed test
Olegator [25]

Answer: i dont know sorry

Explanation:

3 0
4 years ago
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