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Sedaia [141]
3 years ago
14

B. Why does the moon revolve around the earth? Givereason.[2]​

Physics
1 answer:
LenKa [72]3 years ago
6 0

Answer:

The moon revolves around Earth because Earth is larger than the moon, so it is heavier, and has a greater gravitational pull. The plane of the moon's orbit is very close to the plane of Earth's orbit around the Sun. This is why planets revolve around the Sun, because it is larger, so therefore it has a greater gravitational pull.

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A technician attaches one lead of a digital voltmeter to the ground terminal of the TP sensor and the other meter lead to the ne
wariber [46]

Answer:

Neither A or B

Explanation:

The 37.3mv is not the signal voltage

sensor ground circuit does not has excessive resistance.

5 0
3 years ago
Materials with resistivities between those of good conductors and those of insulators are called:
bogdanovich [222]

semiconductors ... silicon, germanium etc ...

8 0
3 years ago
Read 2 more answers
2. ¿Puede haber trabajo en un sistema si no hay movimiento
Andru [333]

Answer:

d. No, porque la ecuación de trabajo lo define.

Explanation:

En Física, el trabajo realizado se puede definir como la cantidad de energía transferida cuando un objeto o cuerpo se mueve a lo largo de una distancia debido a la acción de una fuerza externa.

Matemáticamente, el trabajo realizado viene dado por la fórmula;

W = F * d

<u>Dónde;</u>

  • W es el trabajo realizado.
  • F representa la fuerza que actúa sobre un cuerpo.
  • d representa la distancia recorrida por el cuerpo.

Por lo tanto, podemos deducir de la definición de trabajo y su fórmula que el trabajo se realiza cuando un objeto (cuerpo) se mueve una distancia o experimenta cualquier forma de desplazamiento mientras transfiere energía.

7 0
3 years ago
Which of these can produce an electric current?Required to answer. Single choice.
vivado [14]

Answer:

last one which is; All the above.

4 0
3 years ago
One of the most important properties of materials in many applications is strength. Two of the qualitative measures of the stren
katrin2010 [14]

To solve this exercise it is necessary to take into account the concepts related to Tensile Strength and Shear Strenght.

In Materials Mechanics, generally the bodies under certain loads are subject to both Tensile and shear strenghts.

By definition we know that the tensile strength is defined as

\sigma = \frac{F}{A}

Where,

\sigma =Tensile strength

F = Tensile Force

A = Cross-sectional Area

In the other hand we have that the shear strength is defined as

\sigma_y = \frac{F_y}{A}

where,

\sigma_y =Shear strength

F_y = Shear Force

A_0 =Parallel Area

PART A) Replacing with our values in the equation of tensile strenght, then

311*10^6 = \frac{F}{(15*10^{-6})(30*10^{-2})}

Resolving for F,

F= 1399.5N

PART B) We need here to apply the shear strength equation, then

\sigma_y = \frac{F_y}{A}

210*10^6 = \frac{F_y}{15*10^{-6}30*10^{-2}}

F_y = 945N

In such a way that the material is more resistant to tensile strength than shear force.

6 0
3 years ago
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