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3 years ago

What is the relationship between the wavelength of a wave and its energy

Physics

2 answers:

posledela3 years ago

5 0

Answer:

Inverse relation.

Explanation:

Let E is the energy of the wave and $\lambda$ is its wavelength. The energy of the wave is given by :

$E=\dfrac{hc}{\lambda}$ ..............(1)

Where

h = Planck's constant

c = speed of light

$\lambda$ = wavelength of a wave

It is clear from equation (1) that the energy is inversely proportional to the wavelength of a wave. So, there exists an inverse relationship between the energy and the wavelength of the wave.

Tanzania [10]3 years ago

3 0

Wavelength is the distance between two consecutive points of the wave that are in the same phase. The relationship between wavelength, frequency, and velocity of a wave is given by v = f λ where f is the frequency of the wave and λ is the wavelength.

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A ball of mass M collides with a stick with moment of inertia I = βml2 (relative to its center, which is its center of mass). Th

ZanzabumX [31]

Answer:

Part a)

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

Explanation:

Since the ball and rod is an isolated system and there is no external force on it so by momentum conservation we will have

$Mv_o = M v_1 + m v_2$

here we also use angular momentum conservation

so we have

$M v_o d = M v_1 d + \beta mL^2 \omega$

also we know that the collision is elastic collision so we have

$v_o = (v_2 + d\omega) - v_1$

so we have

$\omega = \frac{v_o + v_1 - v_2}{d}$

also we know

$M v_o d - M v_1 d = \beta mL^2(\frac{v_o + v_1 - v_2}{d})$

also we know

$v_1 = v_o - \frac{m}{M}v_2$

so we have

$M v_o d - M(v_o - \frac{m}{M}v_2)d = \beta mL^2(\frac{v_o + v_o - \frac{m}{M}v_2 - v_2}{d})$

$mv_2 d = \beta mL^2\frac{2v_o}{d} - \beta mL^2(1 + \frac{m}{M})\frac{v_2}{d}$

now we have

$(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})v_2 = \frac{2\beta mL^2v_o}{d}$

$v_2 = \frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})}$

Part b)

Now we know that speed of the ball after collision is given as

$v_1 = v_o - \frac{m}{M}v_2$

so it is given as

$v_1 = v_0 - \frac{m}{M}(\frac{\frac{2\beta mL^2v_o}{d}}{(md + \frac{\beta mL^2}{d}(1 + \frac{m}{M})})$

3 0

3 years ago

Please help! thank you

BlackZzzverrR [31]

Answer:

poor, too precise

good

Explanation:

8 0

3 years ago

1.A body of certain mass is kept at a height of 15m from the ground, taking the value of g as 10m/s find out the mass of the bod

ehidna [41]

Answer:

1.#potential energy = PE, m = mass in kg, g = force of gravity, h= vertical height above the ground. ** means to the power of ie exponent. * means multiply.

PE = mgh

300 = m(10)(15)

m = 300/(10)(15)

m= 2kg

2. KE = 1/2 mv**2

= 1/2(50)(50)**2

= 2500 joules

Explanation

Is as in solution

4 0

3 years ago

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