Choose what colors are absorbed when white light hits a red apple. (Pick all that apply.)

1 answer:

4 0

A red apple absorbs all colors of visible light except red, so red light
is the only light left to bounce off of the apple toward our eyes.
(This is a big part of the reason that we call it a "red" apple.)

Here's how the various items on the list make out when they hit the apple:

Red . . . . . reflected
Orange . . absorbed
Yellow . . . absorbed
Green . . . absorbed
Blue . . . . absorbed
Violet . . . absorbed
Black . . . no light; not a color
White . . . has all colors in it

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A proton moves perpendicular to a uniform magnetic field B at a speed of 2.30 107 m/s and experiences an acceleration of 2.50 10

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Answer:

B = 6.18 10⁻⁶ T

the magnetic field is in the negative direction of the y axis

Explanation:

The magnetic force is given by

F = q v x B

as in the exercise indicate that the velocities perpendicular to the magnetic field,

F = q v B

Newton's second law is

F = m a

let's substitute

q v B = m a

B = m a / q v

let's calculate

B = 9.1 10⁻³¹ 2.50 10¹³ / (1.6 10⁻¹⁹ 2.30 10⁷)

B = 6.18 10⁻⁶ T

The direction of the field can be obtained with the right hand rule, where the thumb points in the direction of the velocity, the fingers extended in the direction of the magnetic field and the palm in the direction of the force for a positive charge.

In the exercise indicate that the velocity is the z axis

the acceleration and therefore the force in the x axis

therefore the magnetic field is in the negative direction of the y axis

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2 years ago

Porque deja de funcionar una estufa electrica cuando la electricidad es excesiva??

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Cho lực F ⃗=6x^3 i ⃗-4yj ⃗ tác dụng lên vật làm vật chuyển động từ A(-2,5) đến B(4,7). Vậy công của lực là:

Natasha2012 [34]

The work done by $\vec F$ along the given path C from A to B is given by the line integral,

$\displaystyle \int_C \mathbf F\cdot\mathrm d\mathbf r$

I assume the path itself is a line segment, which can be parameterized by

$\vec r(t) = (1-t)(-2\,\vec\imath + 5\,\vec\jmath) + t(4\,\vec\imath+7\,\vec\jmath) \\\\ \vec r(t) = (6t-2)\,\vec\imath+(2t+5)\,\vec\jmath \\\\ \vec r(t) = x(t)\,\vec\imath + y(t)\,\vec\jmath$

with 0 ≤ t ≤ 1. Then the work performed by F along C is

$\displaystyle \int_0^1 \left(6x(t)^3\,\vec\imath-4y(t)\,\vec\jmath\right)\cdot\frac{\mathrm d}{\mathrm dt}\left[x(t)\,\vec\imath + y(t)\,\vec\jmath\right]\,\mathrm dt \\\\ = \int_0^1 (288(3t-1)^3-8(2t+5)) \,\mathrm dt = \boxed{312}$