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Maslowich
3 years ago
10

Ana wonders how light can affect a mystery material. The table below shows what happens when different types of light hit the my

stery material.
Physics
1 answer:
vivado [14]3 years ago
3 0

The light can definitely change the mystery material. This can occur through a change in temperature or color (option C).

A material is affected by light mainly if the material absorbs the light. Based on the chart, we know this mystery material can absorb two types of light:

  • Red light.
  • Yellow light.

Moreover, this phenomenon can lead to two main changes:

  • Change in temperature: Light affects materials by increasing their temperature of these. A common example is the way the temperature of an object increases if it is exposed to sunlight.
  • Change in color: Some materials react to light by changing their color.

Based on this, the material can change its color or temperature.

Note: This question is incomplete; here is the missing part:

A. Yes, but the mystery material can change in only one way, such as by getting warm, because all the light that a material absorbs will affect that material in the same way.

B. No. The mystery material can’t change because the light is not a physical thing. Light cannot change physical things like the mystery material.

C. Yes and the mystery material can change in one or two different ways, such as by getting warm and/or changing color, because different types of light can cause different changes to a material when they are absorbed.

D. There is no way to know whether the mystery material will change or not.

Learn more about sunlight in: brainly.com/question/1603783

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A glass prism causes dispersion or deviation but glass plate doesn't.why?​
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7 0
2 years ago
A particle with mass 1.81*10^-3kg and a charge of 1.22*10^-8C has, at a given instant, a velocity v= (3.0*10^4 m/s)j.
charle [14.2K]

Answer:

a = -0.33 m/s² k^

Direction: negative

Explanation:

From Newton's law of motion, we know that;

F = ma

Now, from magnetic fields, we know that;. F = qVB

Thus;

ma = qVB

Where;

m is mass

a is acceleration

q is charge

V is velocity

B is magnetic field

We are given;

m = 1.81 × 10^(−3) kg

q = 1.22 × 10 ^(−8) C

V = (3.00 × 10⁴ m/s) ȷ^.

B = (1.63T) ı^ + (0.980T) ȷ^

Thus, since we are looking for acceleration, from, ma = qVB; let's make a the subject;

a = qVB/m

a = [(1.22 × 10 ^(−8)) × (3.00 × 10⁴)ȷ^ × ((1.63T) ı^ + (0.980T) ȷ^)]/(1.81 × 10^(−3))

From vector multiplication, ȷ^ × ȷ^ = 0 and ȷ^ × i^ = -k^

Thus;

a = -0.33 m/s² k^

7 0
3 years ago
Scientists are working on a new technique to kill cancer cells by zapping them with ultrahigh-energy (in the range of 1012 W) pu
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Explanation:

4 0
3 years ago
A 5 kg wooden block sitson a flat straight-away12 meters fromthe bottom of an infinitely long ramp, which has an angle of 20 deg
saveliy_v [14]

Answer:

(a) 19.71801m/s Velocity just before going up the ramp.

(b) 74.56338m.

Explanation:

We will solve it in two parts, first we will calculate time that 5kg wooden block would take to just reach ramp and with this time we will calculate final velocity that the wooden block would have in this time.

Second, we will calculate the component of velocity vector along inclined plane and the time that it would take for velocity to be 0 meters/s then with this time we will calculate the distance that inclined plane would travel along inclined plane.

Following formulas will be used.

                                  x(t) = \frac{1}{2} t^2 = 12m =16.2m/s^2 t^2

                                 F =ma

                                 V(t) = V_{o} +at

                                 x(t) = x_{0} +v_{0}t+\frac{1}{2}a t^2

(a) Calculating velocity right before going up the ramp.

 Wooden block is going on a straightaway and has net for on it.

         F_{n} =F-F_{s} = F-uF_{n}  = 100N-0.4*9.8m/s^2*5kg =81N

     and this force produces acceleration of

      a = \frac{F}{m}=\frac{81}{5} =16.2m/s^2 .

With this acceleration, wooden block would reach at the foot of ramp in.

          x(t) = 12m = 16.2m/s^2*t^2

         t = 1.217s

and final velocity will be

v(t) = v_{0}+at = 0+16.2m/s^2*1.2171s = 19.7180m/s.

this velocity of wooden box just before going up the ramp.

(b) How far up the ramp will the wooden block go before stopping.

Ramp is at 20° relative to horizontal therefore velocity along the ramp that the wooden block would have will be.

                              V= V_{h}cos(20) = 18.5288m/s

and deceleration along the ramp is

                              a = \frac{F_{s} }{m}

 Where F_{s} force of friction along the inclined plane.

F_s =  uF_n = u*m*a

a = 9.8m/s^2*cos(20) = 9.2089m/s^2

is a component of g along normal of the inclined plane.

                               F_{s} = 0.25*5kg*9.2089m/s^2

                              = 11.5112N

                              a = \frac{11.5112N}{5kg} = 2.3022m/s^2

And with this deceleration time needed to get wooded block to stop is.

                     v(t) = v_o-at = 18.5288m/s-2.3022m/s^2*t = 0

                        t = \frac{18.5288m/s}{2.3022m/s^2} =8.04813s

 and in that time wooden block would travel

   x(8.04813s) = 18.52881m/s *8.04813s-\frac{1}{} 2.3022m/s^2*(8.0481)^2=74.56338m

This is how up wooden box will go before coming to stop.

3 0
3 years ago
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