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4 years ago

As the wave interacts with a wall, which kind of wave interaction is shown?

Physics

2 answers:

MArishka [77]4 years ago

8 0

The correct answer is reflection

I just took the test

kow [346]4 years ago

5 0

The answer would be a reflection. This is because, t<span>he color of an object is actually the wavelengths of the light reflected while all other wavelengths are absorbed. Color, in this case, refers to the different wavelengths of light in the </span>visible light spectrum<span>perceived by our eyes. The physical and chemical composition of matter determines which wavelength (or color) is reflected.</span>

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What provides the energy for the conversion from the open complex to chain elongation?

Marat540 [252]

TP hydrolysis distinct from any incorporation into the chain.

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3 years ago

Whoever answers correctly gets brainlist!

UNO [17]

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3 years ago

Read 2 more answers

The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m

Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

$\theta=\frac{s}{r}$

For the Moon

$\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad$

The angle subtended by the Moon is 0.00903 rad

For the Sun

$\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad$

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

$\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}$

The area ratio is $6.268\times 10^{-6}$

3 0

4 years ago

(b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as meas

aleksley [76]

Complete Question: A charge q1 = 2.2 uC is at a distance d= 1.63m from a second charge q2= -5.67 uC. (b) Find a point between the two charges on the horizontal line where the electric potential is zero. (Enter your answer as measured from q1.)

Answer:

d= 0.46 m

Explanation:

The electric potential is defined as the work needed, per unit charge, to bring a positive test charge from infinity to the point of interest.

For a point charge, the electric potential, at a distance r from it, according to Coulomb´s Law and the definition of potential, can be expressed as follows:

V = $\frac{k*q}{r}$