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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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Atomic Mass Unit is the answer
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What will happen when a light bulb is removed from the circuit below, leaving a gap?
Charra [1.4K]

Answer:

The remaining light bulbs will go out.

Explanation:

The light bulb that was taken out routed power to the other light bulbs, there for, not giving power to the next light bulbs will make them turn off or, "go out". This may be incorrect, as you did not provide a picture of the circuit.

7 0
3 years ago
Read 2 more answers
A projectile of mass 1.800 kg approaches a stationary target body at 4.800 m/s. The projectile is deflected through an angle of
Artyom0805 [142]

Answer:

P = 5.22 Kg.m/s

Explanation:

given,

mass of the projectile = 1.8 Kg

speed of the target = 4.8 m/s

angle of deflection = 60°

Speed after collision = 2.9 m/s

magnitude of momentum after collision = ?

initial momentum of the body = m x v

                                                  = 1.8 x 4.8 = 8.64 kg.m/s

final momentum after collision

momentum along x-direction

P_x = m v cos θ

P_x = 1.8 x 2.9 x  cos 60°

P_x = 2.61 kg.m/s

momentum along y-direction

P_y = m v sin θ

P_y = 1.8 x 2.9 x  sin 60°

P_y = 4.52 kg.m/s

net momentum of the body

P = \sqrt{P_x^2 + P_y^2}

P = \sqrt{4.52^2 + 2.61^2}

 P = 5.22 Kg.m/s

momentum magnitude after collision is equal to P = 5.22 Kg.m/s

5 0
3 years ago
Help me with this review question please.
QveST [7]

Answer:

K E=( mv²)/2

=(60×3.5²)/2

=367.5J

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Ross training for a half marathon which is 13.1 miles long. Within four months he has progressed to an 11 mile practice run if R
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  • Many people mistakenly believe they lose muscle mass far more quickly than they actually do because their muscles' ability to store water and glycogen is declining.
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  • Ross will experience Increased VO2 Max from exercise. VO2 Max is almost completely lost in people who train at lower intensities.

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