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3 years ago

What type of material is found in the asthenosphere?

Physics

2 answers:

IceJOKER [234]3 years ago

6 0

Answer:

The correct answer is option C.

Explanation:

The layer below the lithosphere , the upper layer of the mantle of the earth with high heat pressure .It is made up of almost solid rocks but also flows not which indicates that it consist of viscoelastic fluid of molten rocks.This is because in some regions there can be presence of molten rocks.

o-na [289]3 years ago

5 0

Answer: A viscoelastic fluid of molten rock is the type of material found in the asthenosphere.

Hope This Helps

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John has tied his friend, Zory, a stuffed unicorn to a toy car with a massless string whichmakes an angle of 30°. If he rides fo

Westkost [7]

Answer: A

Since the car is moving to the right, the Normal Force is balancing the Weight and the Net Force is 10 N, right.

Explanation:

as the answers says, the only two forces in the y axis are the normal force and the weight, and they balance each other. On the x axis, you have 20N to the right and the friction is a force that opposes the movement, so the 10N are to the left. The net force is 20 - 10 = 10N.

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3 years ago

The measure of the energy of motion of particles of matter?

bekas [8.4K]

Mechanical energy is the top because it has two types.
These two types are:
Kinetic energy - which is the energy in motion and potential energy which is the energy in reserve.
Both energy in motion and reserve is called Joules. Joules is the International System of Measurement unit for energy, this is mainly used for scaling energy in all aspects.

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3 years ago

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Goryan [66]

Bro the picture is too dark I can’t see do it again so I can help you

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3 years ago

1. Una bola de 0.510 kg de masa se mueve al este (dirección +x) con una rapidez de 4.80 m/s y choca frontalmente con una bola de

Grace [21]

Responder:

3,37 m / s, + ve x - dirección

Explicación:

Utilizando la ley de conservación de la cantidad de movimiento expresada por la fórmula;

m1u1 + m2u2 = (m1 + m2) v

m1 y m2 son las masas de los objetos

u1 y u2 son sus velocidades iniciales

v es su velocidad común

Dado

m1 = 0,519 kg

u1 = 4,80 m / s

m2 = 0,220 kg

u2 = 0 m / s (cuerpo en reposo)

Necesario

Velocidad común v

Sustituir en la fórmula los valores dados;

0,519 (4,8) + 0,22 (0) = (0,519 + 0,220) v

2,4912 + 0 = 0,739 v

2,4912 = 0,739v

Dividir ambos lados por 0,739

2,4912 / 0,739 = 0,739 / 0,739

<em>Por lo tanto, la rapidez de ambas bolas después de la colisión es de 3.37 m.s hacia la dirección x positiva, ya que m1> m2 y la velocidad común es positiva.</em>

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3 years ago

A crowbar 2727 in. long is pivoted 66 in. from the end. What force must be applied at the long end in order to lift a 600600 lb

NikAS [45]

To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

$\sum \tau = F*d$

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,

$F*(27-6)= 6*600$

$F = \frac{6*600}{21}$

$F= 171.42 lb$

So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb

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3 years ago