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2 years ago

If you wanted to get yellow light , what part of the of the spectrum would the color filter have to absorb ?

Physics

1 answer:

densk [106]2 years ago

4 0

The filter would need to absorb every color EXCEPT yellow. It would have to absorb red orange green blue and violet.

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Average speed can be represented by the mathematical expression

nevsk [136]

Average speed is defined as total distance moved in total interval of time

so it is given as

$v_{avg} = \frac{distance}{time}$

now here is we show distance by "d" and time by"t"

then we will have mathematical expression as follows

$v = \frac{d}{t}$

5 0

3 years ago

A soccer player extends her lower leg in a kicking motion by exerting a force with the muscle above the knee in the front of her

AlekseyPX

Answer:

28.025 Nm

Explanation:

Angular acceleration, α = 29.5 rad/s^2

oment of inertia, I = 0.95 kg m^2

The torque is defined as

τ = I x α

τ = 0.95 x 29.5

τ = 28.025 Nm

Thus, the torque is 28.025 Nm.

Explanation:

7 0

3 years ago

What is the ultimate source of energy for an ecosystem?

Romashka-Z-Leto [24]

Answer:

Obviously the answer is Sun...

4 0

3 years ago

The teachers of Scott, Christopher, Dianne, and Kailee have last names Koeninger, Dannemiller, Briscoe, and Carter. Match the ki

Pani-rosa [81]

Scott-Dannemiller, Koeninger, Briscoe, and Carter

Christopher-Briscoe, Dannemiller, and Koeninger

Dianne-Koeninger, Briscoe, and Carter

Kailee-Koeninger and Carter

4 0

3 years ago

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz. Find the possible range of wavelengths in ai

taurus [48]

Answer:

The possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

Explanation:

Given that,

The notes produced by a tuba range in frequency from approximately 45 Hz to 375 Hz.

The speed of sound in air is 343 m/s.

To find,

The wavelength range for the corresponding frequency.

Solution,

The speed of sound is given by the following relation as :

$v=f_1\lambda_1$

Wavelength for f = 45 Hz is,

$\lambda_1=\dfrac{v}{f_1}$

$\lambda_1=\dfrac{343}{45}=7.62\ m$

Wavelength for f = 375 Hz is,

$\lambda_2=\dfrac{v}{f_2}$

$\lambda_2=\dfrac{343}{375}=0.914\ m/s$

So, the possible range of wavelengths in air produced by the instrument is 7.62 m and 0.914 m respectively.

6 0

3 years ago