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3 years ago

12

If a wind is blowing from a siren toward your ear, does the frequency of the siren change? Why?

2 answers:

bonufazy [111]3 years ago

7 0

Answer:
The frequency of the wave stays the same.

Explanation:
As the wave is moving in an air medium throughout its whole journey and the time is constant, the frequency stays the same

Anna007 [38]3 years ago

5 0

Answer: NO

The frequency of the siren doesn't change because there is no motion of the siren toward your ear

Explanation:

frequency is the rate at which a vibration occurs that constitutes a wave, either in a material (as in sound waves), or in an electromagnetic field (as in radio waves and light), usually measured per second.

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this is energy that is transferred by the movement of electrons through a conductor the electrons create circuit

Paha777 [63]

The correct answer for this question is this one: "Potential Energy."
Potential energy is the energy that is transferred by the movement of electrons through a conductor the electrons create circuit. It is energy at moving. Hope this helps answer your question and have a nice day ahead.

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4 years ago

To rid the body of wastes, including excess water and salts" is the purpose of what human body system? A) Digestive B) Excretory

Verizon [17]

Digestive I believe.

5 0

4 years ago

Read 2 more answers

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

A standard biological microscope is required to have a magnification of 200Ã. what is the focal length of the objective?

NISA [10]

Answer:

The focal length of the objective is L/20.

Explanation:

Given that,

Magnification of the standard biological microscope, m = 200

Let the magnification of the eyepiece is, $m_e=10$

Let $m_o$ is the magnification of objective lens. The formula of the magnification is given by :

$M=m_o\times m_e\\\\200=m_o\times 10\\\\m_o=20$

The formula of the magnification is given by :

$M=\dfrac{L}{f_o}m_e$

L is length of microscope

$f_o$ is focal length of the objective

$f_o=\dfrac{Lm_e}{M}\\\\f_o=\dfrac{10L}{200}\\\\f_o=\dfrac{L}{20}$

So, the focal length of the objective is L/20.

7 0

3 years ago

A truck passes point A on a straight level road with a constant velocity of 12 m/s. At the same instant, an automobile starts fr

REY [17]

Answer:

$t = 13.3 s$

Explanation:

Let the car will overtake the truck after time "t" seconds

so here we can say that after time "t" seconds the distance covered by the car will be same as that of distance covered by truck

So we have

$d_{car} = d_{truck}$

so we have

$d_{truck} = 12 t$

$d_{car} = \frac{1}{2}(1.8 m/s^2) t^2$

so now we have

$12 t = 0.9 t^2$

$t = 13.3 s$

7 0

3 years ago