Answer:
Sound waves are longitudinal waves
Explanation:
Sound waves are the longitudinal waves. In longitudinal waves, the particles of the wave move parallel to the direction of propagation of waves.
It moves in the form of compression and rarefaction. When the particles are compact in a space the compression occurs while when they far apart form each other rarefaction occurs.
Answer:
Magnitude of displacement = 2.07 km
Magnitude of average velocity = 1.17 kmph
Explanation:
Let east represent positive x axis and north represent positive y axis.
A bird watcher meanders through the woods, walking 1.93 km due east, 1.03 km due south, and 3.84 km in a direction 52.8 ° north of west.
1.93 km due wast
s ₁ = 1.93 i km
1.03 km due south
s₂ = -1.03 j km
3.84 km in a direction 52.8 ° north of west
s₃ = -3.84 cos 52.8 i + 3.84 sin 52.8 j = -2.32 i + 3.06 j km
Total displacement
s = s ₁+ s₂+ s₃ = 1.93 i - 1.03 j -2.32 i + 3.06 j = -0.39 i + 2.03 j
Magnitude of displacement, 
Time taken = 1.771 hour
Magnitude of average velocity, 
Answer:
the answer is 1.35. Have a nice day!!
The distance mirror M2 must be moved so that one wavelength has produced one more new maxima than the other wavelength is;
<u><em>L = 57.88 mm</em></u>
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We are given;
Wavelength 1; λ₁ = 589 nm = 589 × 10⁻⁹ m
Wavelength 2; λ₂ = 589.6 nm = 589.6 × 10⁻⁹ m
We are told that L₁ = L₂. Thus, we will adopt L.
Formula for the number of bright fringe shift is;
m = 2L/λ
Thus;
For Wavelength 1;
m₁ = 2L/(589 × 10⁻⁹)
For wavelength 2;
m₂ = 2L/(589.6)
Now, we are told that one wavelength must have produced one more new maxima than the other wavelength. Thus;
m₁ - m₂ = 2
Plugging in the values of m₁ and m₂ gives;
(2L/589) - (2L/589.6) = 2
divide through by 2 to get;
L[(1/589) - (1/589.6)] = 1
L(1.728 × 10⁻⁶) = 1
L = 1/(1.728 × 10⁻⁶)
L = 578790.67 nm
L = 57.88 mm
Read more at; brainly.com/question/17161594
From the answers provided, I believe the possible answer would be the last option, silicon, oxygen, and one or more metals. Here's my reasoning: the most abundant mineral group found in the Earth's crust is the silicate group. The silicate materials contain both oxygen and silicon. Silicates are the most common minerals in the rock-formation process, and it has, in fact, been estimated that they make up 75 to 90 percent of the Earth's crust. From this piece of evidence, I can guess that the answer will possibly be D, silicon, oxygen, and one or more metals.
It should also be noted that the additional elements that combine with the silicon-oxygen tetrahedron are involved with the other elements commonly found in the Earth's crust and mantle. They are aluminum, calcium, iron, magnesium, potassium and sodium.
