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Pavel [41]
3 years ago
6

Cocking your head would be most useful for detecting the ______ of a sound.

Physics
2 answers:
kykrilka [37]3 years ago
8 0

Answer:

It would be most useful for detecting WHERE a sound may come from or the location of said sound.

Explanation:

statuscvo [17]3 years ago
5 0
Cocking your head would be most useful for detecting the LOCATION of a sound.
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-Starts to suck on ur neck giving you hickeys n runs my hand down your body biting my lip against your neck
Hoochie [10]

Answer:

I would shout fore help if I was being raped or try to make him or her stop

6 0
3 years ago
Two toy cars (m1 = 0.200 kg and m2 = 0.250 kg) are held together rear to rear with a compressed spring between them. When they a
timurjin [86]

Answer:

Acceleration of the second particle at that moment is given as

a_2 = 2.12 m/s^2

Explanation:

As we know that both cars are connected by same spring

So on this system of two cars there is no external force

So we will have

F = 0 = m_1a_1 + m_2a_2

now we have

m_1 = 0.200 kg

m_2 = 0.250 kg

a_1 = 2.65 m/s^2

now we have

0.200(2.65) + 0.250a_2 = 0

so we have

a_2 = 2.12 m/s^2

8 0
3 years ago
Which types of geometrical symmetry does a sphere have?.
Helen [10]
A sphere has reflection symmetry across any plane through its center.
8 0
3 years ago
An incident ray through the focal point of a spherical mirror will:
Delvig [45]
Reflect parallel of the principal axis
3 0
3 years ago
Read 2 more answers
Un cuerpo se encuentra en reposo sobre una mesa horizontal. Entonces se puede afirmar que:
vazorg [7]

Answer:

C) solo III

Explanation:

Para solucionar este problema debemos analizar cada una de las opciones hasta llegar a la opcion valida.

I) el cuerpo pesa igual que su masa.

Esta opcion no puede ser ya que el peso de un cuerpo se define como el producto de la masa por la aceleracion gravitacion.

w=m*g

donde:

w = peso [N]

m = masa [kg]

g = aceleracion gravitacional = 9.81 [m/s²]

Como podemos ver el peso siempre sera mayar que la masa, ya que el peso es resultado de la multiplicacion de la masa por la gravedad.

II) Por medio de un analisis de fuerzas en el eje-y, la fuerza del peso se dirige hacia abajo mientras que la fuerza normal tiene igual magnitud, pero se dirige hacia arriba. Por esto la segunda opcion no puede ser.

III) El cuerpo se encuentra en equilibrio, es decir las unicas fuerzas que actuan sobre el cuerpo son el peso y la fuerza normal. Pero estas fuerzas son iguales y opuestas en direccion, por la tanto se cancelan y estan en equilibrio.

Esta es la opcion valida, la fuerza neta es nula.

5 0
3 years ago
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