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Maru [420]
2 years ago
11

The single-pole, single-throw switch is normally wired ? between the source and the load to turn devices on and off.

Physics
1 answer:
zzz [600]2 years ago
7 0

In series.

Single-pole and single-throw switch:

A switch with only one input and one output is referred to as a Single Pole Single Throw (SPST) switch. This indicates that it has a single output terminal and a single input terminal.

A single pole, one throw switch functions as an on/off switch in circuits. The circuit is turned on when the switch is closed. The circuit is shut off when the switch is open.

Thus, SPST switches are relatively basic in design.

Circuit for a single-pole, single-throw (SPST) switch

Types:

According to the application, it can be divided into three categories, including:

  • Simple SPST
  • (ON)-OFF, Push-to-close, SPST Momentary
  • ON-(OFF), Push-to-Open, SPST Momentary
  • Inches Switch SPST

Learn more about terminal here:

brainly.com/question/14236970

#SPJ4

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what is the rotational kinetic energy of the earth? use the moment of inertia you calculated in part a rather than the actual mo
Ivenika [448]

The Earth's rotational kinetic energy is the kinetic Energy that the Earth

has due to rotation.

The rotational kinetic energy of the Earth is approximately <u>3.331 × 10³⁶ J</u>

Reasons:

<em>The parameters required for the question are;  </em>

<em>Mass of the Earth, M = </em><em>5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = </em><em>6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = </em><em>24.0 hrs</em><em>.</em>

The \ moment  \ of \  inertia \  of \  uniform \  sphere \  is \ I =   \mathbf{\dfrac{2}{5} \cdot M \cdot R^2}

Which gives;

\mathbf{I_{Earth}} =   \dfrac{2}{5} \times 5.97 \times 10 ^{24} \cdot \left(6.38 \times 10^6 \right)^2 = 9.7202107 \times 10^{37}

\mathrm{The \ rotational \  kinetic  \ energy \  is} \   E_{rotational} = \mathbf{\dfrac{1}{2} \cdot I \cdot \omega^2}

\mathrm{The \ angular \ speed, \ \omega} = \mathbf{\dfrac{2 \dcdot \pi}{T}}

Therefore;

\omega = \dfrac{2 \cdot \pi}{24}  = \dfrac{\pi}{24}

Which gives;

\mathbf{E_{rotational}} = \dfrac{1}{2} \times  9.7202107 \times 10^{37} \times  \left(  \dfrac{\pi}{12} \right)^2 = 3.331 \times 10^{36}

The rotational kinetic energy of the Earth, E_{rotational} = <u>3.331 × 10³⁶ Joules</u>

Learn more here:

brainly.com/question/13623190

<em>The moment of inertia from part A  of the question (obtained online) is that of the Earth approximated to a perfect sphere</em>.

<em>Mass of the Earth, M = 5.97 × 10²⁴ kg</em>

<em>Radius of the Earth, R = 6.38 × 10⁶ m</em>

<em>The rotational period of the Earth, T = 24.0 hrs</em>

3 0
3 years ago
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