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sergiy2304 [10]
3 years ago
5

When light is reflected, the incident rays are bent and change direction. True False

Physics
2 answers:
Helga [31]3 years ago
8 0

Answer: True

Explanation: When light is reflected off lets say a mirror it is bent and changes direction to bounce off of another wall or object. For example if you take a flash light and shine it into a mirror the light reflects into a different direction your welcome

Mars2501 [29]3 years ago
5 0
Answer: True



Step by step explanation:
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Si el periodo de oscilación de resorte es de 0,44 segundos cuando oscila atado a una masa de 2 Kg. ¿Cuál será el valor de la con
boyakko [2]

Answer:

i d k h b u lol I wish I knew it sorry

7 0
2 years ago
a flashlight has a resistance of 2.4 ohms. what voltage is applied by the batteries if the current in the circuit is 2.5A?
Mashcka [7]

E = I R

That means 

        Voltage = (current) x (resistance)

                     =   (2.5 A)  x  (2.4 ohms)

                     =       6 volts .


8 0
3 years ago
Read 2 more answers
Correct me if im wrong
White raven [17]
Your answer is correct. No problem and Have a nice day
6 0
3 years ago
Jane, looking for tarzan, is running at top speed (6.8 m/s) and grabs a vine hanging vertically from a tall tree in the jungle.
erica [24]
Jane's mechanical energy at any time is
E=U+K
where U=mgh is the potential energy, while K= \frac{1}{2} mv^2 is the kinetic energy.

Initially, Jane is on the ground, so the altitude is h=0 and the potential energy is zero: U=0. She's running with speed v, so she has kinetic energy only:
E=K= \frac{1}{2} mv^2
Then she grabs the vine, and when she reaches the maximum height h, her speed is zero: v=0, and so the kinetic energy becomes zero: K=0. So now her mechanical energy is just potential energy:
E=U=mgh

But E must be conserved, so the initial kinetic energy must be equal to the final potential energy:
\frac{1}{2}mv^2=mgh
from which we can find h, the maximum height Jane can reach:
h= \frac{v^2}{2g}= \frac{(6.8 m/s)^2}{2\cdot 9.81 m/s^2}=2.36 m
6 0
3 years ago
The 630-nm light from a helium-neon laser irradiates a grating. The light then falls on a screen where the first bright spot is
dimaraw [331]

Answer:

464.8 nm

Explanation:

The second wavelength of light can be calculated using the next equation:

\lambda = \frac{x*d}{L}        

<u>Where:</u>

<em>λ : is the wavelength of light</em>

<em>x: is the distance from the central maximum</em>

<em>d: is the distance between the spots                      </em>

<em>L: is the lenght from the screen to the bright spot</em>

For the first wavelength of light we have:

\lambda_{1} = \frac{x_{1}*d}{L}

630 \cdot 10^{-9} m = \frac{0.61 m*d}{L}

\frac{d}{L} = \frac{630 \cdot 10^{-9} m}{0.61 m} = 1.033 \cdot 10^{-6}  (1)    

For the second wavelength of light we have:

\lambda_{2} = \frac{x_{2}*d}{L}

\lambda_{2} = 0.45 m*\frac{d}{L}   (2)  

By entering equation (1) into equation (2) we have:

\lambda_{2} = 0.45 m* 1.033 \cdot 10^{-6} = 4.648 \cdot 10^{-7} m = 464.8 nm

Therefore, the second wavelength is 464.8 nm

I hope it helps you!          

3 0
3 years ago
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