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Rus_ich [418]
2 years ago
3

What equation describes the relationship between energy of light and wavelength

Physics
1 answer:
Vesna [10]2 years ago
3 0

Answer:

E = h. c / λ

Explanation:

The energy of a wave is directly proportional to its frequency, but inversely proportional to its wavelength.

In other words, the greater the energy, the larger the frequency and the shorter (smaller) the wavelength.

When the energy increases the wavelength decreases and vice versa. That is, energy in inversely proportional to wavelength.

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Alex goes cruising on his dirt bike. He rides 700m north, 300m east, 400 m north, 600m west, 1200m south, 300m east and finally
Bess [88]

Find Displacement and Distance

displacement ...

north is 700+400+100 =1200m n

south=1200m

1200-1200=0


east is 300+300=600m

west is 600m

600-600=0

back at dtart. displ zero


distance is 700+ 300m + 400 m + 600m + 1200m + 300m + 100m  = 3600m


3 0
2 years ago
A charge Q is placed on the x axis at x = +4.0 m. A second charge q is located at the origin. If Q = +75 nC and q = −8.0 nC, wha
Stells [14]

Answer:

23.1 N/C

Explanation:

OP = 3 m , OQ = 4 m

PQ = \sqrt{4^{2}+3^{2}}=5 m

q = - 8 nC, Q = 75 nC

Electric field at P due to the charge Q is

E_{1}=\frac{KQ}{PQ^{2}}=\frac{9\times 10^{9}\times 75\times 10^{-9}}{25}=27 N/C

Electric field at P due to the charge q is

E_{2}=\frac{Kq}{PO^{2}}=\frac{9\times 10^{9}\times 8\times 10^{-9}}{9}=8 N/C

According to the diagram, tanθ = 3/4

Resolve the components of E1 along x axis and along y axis.

So, Electric field along X axis, Ex = - E1 Cos θ

Ex = - 27 x 4 / 5 = - 21.6 N/C

Electric field along y axis, Ey = E1 Sinθ - E2

Ey = 27 x 3 /5 - 8 = 8.2 N/C

The resultant electric field at P is given by

E=\sqrt{E_{x}^{2}+E_{y}^{2}}=\sqrt{(-21.6)^{2}+(8.2)^{2}}=23.1 N/C

3 0
2 years ago
Two astronauts of identical mass are connected by a taut cable of negligible mass, as shown in the figure above, and are initial
PolarNik [594]

Answer:

The right answer is "The center of mass doesn't move".

Explanation:

  • It generates a voltage throughout the cable while the astronaut falls on either the wire. At other ends of the spectrum or cable, the tension will be similar. As such, with both astronauts, there would be the same energy, although throughout the opposite way.
  • Thus, the net force seems to be essentially negative on the machine. And therefore the mass center stays stationary.
5 0
2 years ago
As shown in the figure below, Justin walks from the house to his truck on a windy day. He walks 20 m toward
juin [17]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The velocity is   v =0.333 \  m/s in positive x -direction

The speed is s = 0.733 \ m/s

Explanation:

From the question we are told that

The distance from the house to truck is  D =  20 m

  The distance traveled back to retrieve  wind-blown hat is  d =  15

  The distance from the wind-blown hat position too the truck is  k =  20  m

  The total time taken is  t  =  75 s

Generally when calculating the displacement the Justin's backward movement to collect his wind - blown hat is taken as negative

Generally Justin's displacement is mathematically represented as

      L  =  20 - 15 + 20

=>    L  =  25 \ m

Generally the average velocity is mathematically represented as

          v  =  \frac{L}{t}

=>      v = \frac{25}{75}

=>      v =0.333 \  m/s

Generally the distance covered by Justin is mathematically represented as  

         R =  D+ d + k

=>      R =  20 + 15 +20

=>     R =  55 \  m

Generally Justin's average speed over a 75 s period is mathematically represented as

            s = \frac{R}{ t}

=>         s = \frac{55}{ 75}

=>        s = 0.733 \ m/s

8 0
2 years ago
A 25cm×25cm horizontal metal electrode is uniformly charged to +50 nC . What is the electric field strength 2.0 mm above the cen
saw5 [17]

Answer:

The electric field strength is 4.5\times 10^{4} N/C

Solution:

As per the question:

Area of the electrode, A_{e} = 25\times 25\times 10^{- 4} m^{2} = 0.0625 m^{2}

Charge, q = 50 nC = 50\times 10^{- 9} C[/etx]Distance, x = 2 mm = [tex]2\times 10^{- 3} m

Now,

To calculate the electric field strength, we first calculate the surface charge density which is given by:

\sigma = \frac{q}{A_{e}} = \frac{50\times 10^{- 9}}{0.0625} = 8\times 10^{- 7}C/m^{2}

Now, the electric field strength of the electrode is:

\vec{E} = \frac{\sigma}{2\epsilon_{o}}

where

\epsilon_{o} = 8.85\times 10^{- 12} F/m

\vec{E} = \frac{8\times 10^{- 7}}{2\times 8.85\times 10^{- 12}}

\vec{E} = 4.5\times 10^{4} N/C

7 0
2 years ago
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