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hjlf
2 years ago
5

Timbre describes the sound’s _____. pitch loudness tone quality

Physics
2 answers:
Bumek [7]2 years ago
7 0

Answer:

A) pitch

Explanation:

I did band and also g*ogled it to check :)

If it isn't A, however, it is C

Hope this helps :D

GrogVix [38]2 years ago
3 0

Answer:

loudness

Explanation:

allow the ear to distinguish sounds which have the same pitch and loudnes

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What could be the possible answer to the question ?<br><br>thankyou ~​
Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

  • The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

T_1 = \mathbf{\dfrac{M \cdot g}{2}}

Which gives;

M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}

T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})}  = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}

F₀ = T₂·sin(37°)

Which gives;

F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2}  \approx  \mathbf{0.377  \cdot M \cdot g}

  • F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0
2 years ago
Read 2 more answers
Calculate the critical angle for light going from Glycerine to air.
Georgia [21]
The refractive index for glycerine is n_g=1.473, while for air it is n_a = 1.00.

When the light travels from a medium with greater refractive index to a medium with lower refractive index, there is a critical angle over which there is no refraction, but all the light is reflected. This critical angle is given by:
\theta_c = \arcsin ( \frac{n_2}{n_1} )
where n1 and n2 are the refractive indices of the two mediums. If we susbtitute the refractive index of glycerine and air in the formula, we find the critical angle for this case:
\theta_c = \arcsin ( \frac{1.00}{1.473} )=42.8^{\circ}
6 0
3 years ago
/Q: DHSMV definition/ i have to write 20 characters for whatever reason lol
cluponka [151]

<span><span>Department of Highway Safety and Motor Vehicles          OR</span></span>

<span><span /><span><span>Division of Highway Safety and Motor Vehicles </span></span></span>

6 0
3 years ago
A point charge Q of mass M is located initially at the center of the ring. When it is displaced slightly, the point charge accel
dolphi86 [110]

Answer:

n musical notation, stems are the, "thin, vertical lines that are directly connected to the [note] head." Stems may point up or down. ... There is an exception to this rule: if a chord contains a second, the stem runs between the two notes with the higher being placed on the right of the stem and the lower on the left.

Explanation:

6 0
4 years ago
In a double-slit experiment, light from two monochromatic light sources passes through the same double slit. The light from the
Dmitrij [34]

Answer:

\lambda_2 = 573.3 nm

Explanation:

As we know that the position of maximum intensity on the screen is given as

y = \frac{N\lambda L}{d}

here we know that

\lambda = wavelength

L = distance of the screen

d = distance between two slits

now we know that the position of 8th maximum intensity is same as that of 9th maximum on the screen

so we have

\frac{N_1\lambda_1 L}{d} = \frac{N_2 \lambda_2 L}{d}

so here we have

8 (645 nm) = 9 \lambda_2

\lambda_2 = 573.3 nm

4 0
3 years ago
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