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Serhud [2]
2 years ago
10

orange light has a wavelength, λ, of 6.0 × 10-7 m. in a vacuum, the speed of light, c, is 3.0 × 108 m/s. what is the frequency,

f, of the orange light? (remember: 1 hz
Physics
1 answer:
PolarNik [594]2 years ago
3 0

The frequency of orange light is 5 x 10¹⁴ Hz or 500 THz.

We need to know about photon frequency to solve this problem. The photon frequency can be determined as

f = c / λ

where f  is frequency, c is speed of light (3x10⁸ m/s) and λ is wavelength.

From the question above, we know that:

c = 3x10⁸ m/s

λ = 6 x 10¯⁷ m

By substituting parameters, we get

f = c / λ

f =  3 x 10⁸ /( 6 x 10¯⁷ )

f = 5 x 10¹⁴ Hz

For more on frequency at: brainly.com/question/10728818

#SPJ4

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A 4.67-g bullet is moving horizontally with a velocity of +357 m/s, where the sign + indicates that it is moving to the right (s
Leni [432]

Answer:

(a)0.531m/s

(b)0.00169

Explanation:

We are given that

Mass of bullet, m=4.67 g=4.67\times 10^{-3} kg

1 kg =1000 g

Speed of bullet, v=357m/s

Mass of block 1,m_1=1177g=1.177kg

Mass of block 2,m_2=1626 g=1.626 kg

Velocity of block 1,v_1=0.681m/s

(a)

Let velocity of the second block  after the bullet imbeds itself=v2

Using conservation of momentum

Initial momentum=Final momentum

mv=m_1v_1+(m+m_2)v_2

4.67\times 10^{-3}\times 357+1.177(0)+1.626(0)=1.177\times 0.681+(4.67\times 10^{-3}+1.626)v_2

1.66719=0.801537+1.63067v_2

1.66719-0.801537=1.63067v_2

0.865653=1.63067v_2

v_2=\frac{0.865653}{1.63067}

v_2=0.531m/s

Hence, the  velocity of the second block after the bullet imbeds itself=0.531m/s

(b)Initial kinetic energy before collision

K_i=\frac{1}{2}mv^2

k_i=\frac{1}{2}(4.67\times 10^{-3}\times (357)^2)

k_i=297.59 J

Final kinetic energy after collision

K_f=\frac{1}{2}m_1v^2_1+\frac{1}{2}(m+m_2)v^2_2

K_f=\frac{1}{2}(1.177)(0.681)^2+\frac{1}{2}(4.67\times 10^{-3}+1.626)(0.531)^2

K_f=0.5028 J

Now, he ratio of the total kinetic energy after the collision to that before the collision

=\frac{k_f}{k_i}=\frac{0.5028}{297.59}

=0.00169

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