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djverab [1.8K]
2 years ago
9

- A particular light source gives off light waves with a measured wavelength of

Physics
1 answer:
skad [1K]2 years ago
8 0

Answer:

f = 1.5 x 10¹⁵ Hz

Explanation:

Light is an electromagnetic wave and we know that the speed of light is a constant which is 

3 x 10⁸ m/s.

Knowing this; v = λ x f

λ = wavelength

f= frequency of light

v = seed of light

f = v/λ

f =  3 x 10⁸m/s / 2.0 x 10-⁷ m

f = 1.5 x 10¹⁵ Hz

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A conducting rod is moving through a magnetic field, as in the drawing. The magnetic field strength is 0.65 T, and the speed of
Contact [7]

Explanation:

The given data is as follows.

 Magnetic field strength (B) = 0.65 T

 Speed (v) = 2.3 m/s

 Induced emf (E) = ?

Formula for emf induced at the ends of the rod of length L which is moving with a speed of v is as follows.

                              E = BvL

Putting the given values into the formula as follows.

          E_{1} = BvL

                      = 0.65 T \times 2.3 m/s \times L

                      = 1.495 L .............. (1)

When magnetic field is changed to B_{2} = 0.48 T

Now, we assume that the speed be v_{2} to get the emf E_{2} = E_{1}.

Then,    0.48 T \times v_{2} \times L = 0.65 T \times 2.3 m/s \times L

                  v_{2} = 3.11 m/s

Therefore, we can conclude that the speed v of the rod be adjusted to reestablish the emf induced between the ends of the rod at its initial value is 3.11 m/s.

6 0
3 years ago
A cannon with a muzzle speed of 1 000 m/s is used to start an avalanche on a mountain slope. The target is 2 000 m from the cann
Nataliya [291]

Answer:

∅ = 89.44°

Explanation:

In situations like this air resistance are usually been neglected thereby making g= 9.81 m/s^{2}

Bring out the given parameters from the question:

Initial Velocity (V_{1}) = 1000 m/s

Target distance (d) = 2000 m

Target height (h) =  800 m

Projection angle ∅ = ?

Horizontal distance = V_{1x}tcos ∅     .......................... Equation 1

where V_{1x} = velocity in the X - direction

           t = Time taken

Vertical Distance = y = V_{1y} t - \frac{1}{2}gt^{2}        ................... Equation 2

Where   V_{1y} = Velocity in the Y- direction

              t  = Time taken

V_{1y} = V_{1}sin∅

Making time (t) subject of the formula in Equation 1

                    t = d/(V_{1x}cos ∅)

                      t = \frac{2000}{1000coso} = \frac{2}{cos0}  =    \frac{d}{cos o}             ...................Equation 3

substituting equation 3 into equation 2

Vertical Distance = d = V_{1y} \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

                                  Vertical Distance = h = sin∅ \frac{d}{cos o} - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Vertical Distance = h = dtan∅   - \frac{1}{2}g\frac{2}{cos0}   ^{2}

  Applying geometry

                              \frac{1}{cos o} = tan^{2} o + 1

  Vertical Distance = h = d tan∅   - 2 g (tan^{2} o + 1)

               substituting the given parameters

               800 = 2000 tan ∅ - 2 (9.81)( tan^{2} o + 1)

              800 = 2000 tan ∅ - 19.6( tan^{2} o + 1)  Equation 4

Replacing tan ∅ = Q     .....................Equation 5

In order to get a quadratic equation that can be easily solve.

            800 = 2000 Q - 19.6Q^{2} + 19.6

Rearranging 19.6Q^{2} - 2000 Q + 780.4 = 0

                    Q_{1} = 101.6291

                      Q_{2} = 0.411

    Inserting the value of Q Into Equation 5

                 tan ∅ = 101.63    or tan ∅ = 0.4114

Taking the Tan inverse of each value of Q

                  ∅ = 89.44°     ∅ = 22.37°

             

4 0
3 years ago
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