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xxMikexx [17]
4 years ago
11

How does frequency and pitch affect sound

Physics
1 answer:
aleksandr82 [10.1K]4 years ago
7 0
<span>The word that musicians use for </span>frequency<span> is </span>pitch<span>. The shorter the wavelength, the higher the </span>frequency<span>, and the higher the </span>pitch<span>, of the </span>sound<span>. In other words, short waves </span>sound<span> high; long waves </span>sound<span> low.</span>
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Each element has an atomic number state which is meant by atomic number
netineya [11]

Answer: The atomic number is the number of protons in an atom, and isotopes have the same atomic number but differ in the number of neutrons. The number of protons in an atom is called its atomic number. This number is very important because it is unique for atoms of a given element. All atoms of an element have the same number of protons, and every element has a different number of protons in its atoms.

Explanation:

3 0
3 years ago
Up-regulation involves the loss of receptors and prevents the target cells from overreacting to persistently high hormone levels
kotegsom [21]

Answer:

False.

Explanation:

Up regulation involves the increase of receptors and it makes the cells more sensitive to the hormones.It's the down regulation that involves the loss of receptors and prevents the target cells from overreacting to persistently high hormone levels.

5 0
3 years ago
What voltage is required to move 6A through 5Ω?
Scorpion4ik [409]
30 volts

E= IxR

See ohms law calculator image below

7 0
3 years ago
The mass of a hypothetical planet is 1/100 that of Earth and its radius is 1/4 that of Earth. If a person weighs 600 N on Earth,
omeli [17]

To solve this problem we will apply the Newtonian concept of gravitational acceleration produced by a planet. This relationship is given by:

g = \frac{GM}{r^2}

Where,

G = Gravitational Universal Constant

M = Mass of Earth

r = Radius

The values given are based on the constants of the earth, so they can be expressed as

M_p = \frac{1}{100} M_e

r_p = \frac{1}{4} r_e

The relationship of gravity would then be given:

g_e = \frac{GM_e}{r_e^2}

The relationship with the new planet, from the gravity of the earth would be given

g_p = \frac{GM_p}{r_p^2}

g_p = \frac{G(1/100)M_e}{(1/4 r_e)^2}

g_p = \frac{GM_e 16}{100 r_e^2}

g_p = 0.16 \frac{GM_e}{r_e^2}

g_p = 0.16g_e

The relationship with the weight of the earth would be given as:

W_e = m*g_e = 600N

W_p = m*g_p = m(0.16g_p)

W_p = (m*g_p)(0.16)

W_p = 600*0.16

W_p = 96N

Therefore the weigh on this planet would be 96N

3 0
3 years ago
In designing structures in an earthquake-prone region, how should the natural frequencies of oscillation of a structure relate t
Genrish500 [490]

Answer:

Natural frequencies of oscillation and typical earthquake frequencies should be different.

Damping on the structure should be large.

Explanation:

The natural frequency of the structure must be different from the typical earthquake frequency, the more different the better. This is because if both frequencies were the same or similar there is a risk that the building will <u>resonate </u>and collapse.

As for the damping, it must have a high value. This so that the building resists earthquakes better and prevents it from moving dangerously, thus damping on the structure should be large.

7 0
4 years ago
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