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LenaWriter [7]
3 years ago
14

What kind of materials does light pass through?

Physics
1 answer:
Marrrta [24]3 years ago
6 0
It can pass through glass, water, clear plastic and transparent materials.
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Why does the frequency of a wave increase as the wavelength decreases
olchik [2.2K]

Explanation:

We know that the number of complete waves formed in 1 sec time is frequency and the distance between two consecutive crests or troughs is wavelength. And we have the formula that

Velocity = wavelength * frequency

or, frequency = velocity / wavelength

Here we can see frequency is directly proportional to velocity and indirectly proportional to wavelength.

So as the wavelength increases frequency decreases and as the wavelength decreases frequency increases.

Hope you understood

3 0
3 years ago
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What is the correct answer?
damaskus [11]

Answer:

96 million of dollars

Explanation:

p=-6x^3+72x

p=-6(2)^3+72(2)

p=-6(8)+144

p=-48+144

p=96

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2 years ago
What is the velocity of a wave with a frequency of 760Hz and a wavelength of 0.45m?
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3 years ago
Samantha is investigating the densities of several liquids. Which action is the result of her creative thinking.
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3 years ago
You pull straight up on the string of a yo-yo with a force 0.35 N, and while your hand is moving up a distance 0.16 m, the yo-yo
jarptica [38.1K]

Answer:

a) 0.138J

b) 3.58m/S

c) (1.52J)(I)

Explanation:

a) to find the increase in the translational kinetic energy you can use the relation

\Delta E_k=W=W_g-W_p

where Wp is the work done by the person and Wg is the work done by the gravitational force

By replacing Wp=Fh1 and Wg=mgh2, being h1 the distance of the motion of the hand and h2 the distance of the yo-yo, m is the mass of the yo-yo, then you obtain:

Wp=(0.35N)(0.16m)=0.056J\\\\Wg=(0.062kg)(9.8\frac{m}{s^2})(0.32m)=0.19J\\\\\Delta E_k=W=0.19J-0.056J=0.138J

the change in the translational kinetic energy is 0.138J

b) the new speed of the yo-yo is obtained by using the previous result and the formula for the kinetic energy of an object:

\Delta E_k=\frac{1}{2}mv_f^2-\frac{1}{2}mv_o^2

where vf is the final speed, vo is the initial speed. By doing vf the subject of the formula and replacing you get:

v_f=\sqrt{\frac{2}{m}}\sqrt{\Delta E_k+(1/2)mv_o^2}\\\\v_f=\sqrt{\frac{2}{0.062kg}}\sqrt{0.138J+1/2(0.062kg)(2.9m/s)^2}=3.58\frac{m}{s}

the new speed is 3.58m/s

c) in this case what you can compute is the quotient between the initial rotational energy and the final rotational energy

\frac{E_{fr}}{E_{fr}}=\frac{1/2I\omega_f^2}{1/2I\omega_o^2}=\frac{\omega_f^2}{\omega_o^2}\\\\\omega_f=\frac{v_f}{r}\\\\\omega_o=\frac{v_o}{r}\\\\\frac{E_{fr}}{E_{fr}}=\frac{v_f^2}{v_o^2}=\frac{(3.58m/s)}{(2.9m/s)^2}=1.52J

hence, the change in Er is about 1.52J times the initial rotational energy

5 0
3 years ago
Read 2 more answers
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