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Julli [10]
3 years ago
13

For every particle there is a corresponding ______________.

Physics
1 answer:
Shtirlitz [24]3 years ago
6 0

Answer:

Anti-Particle

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What is the minimum size copper equipment grounding conductor required to serve a multi section motor control center equipped wi
Yakvenalex [24]

The minimum size copper equipment grounding conductor that is required is ; #4 copper

<h3>Equipment grounding conductor </h3>

An Equipment grounding conductor is a conductive part of ground fault current path which connects the non-conducting metallic parts of the euipment together and also connects the system grounded conductor.

For a copper euipment grounding conductor required to serve a control center connected to a 300 amp over current device the minimum size of copper is #4 copper

Hence we can conclude that The minimum size copper equipment grounding conductor that is required is ; #4 copper.

Learn more about Equipment grounding conductor : brainly.com/question/14124204

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2 years ago
I NEED THE ANSWER TO PASS SO I CAN GET MY PHONE BACK HELPPPPPPPPPPPPPP
lianna [129]

Answer:

it would be c.

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steposvetlana [31]

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a bungee jumper momentarily comes to rest at the bottom of the dive before he springs back upward. at that moment, is the bungee
barxatty [35]

Answer:

Explanation:

No, the bungee jumper is not at equilibrium.

This can be explained when we consider a bungee jumper as a mass that is undergoing simple harmonic motion. At extreme points i.e. at the bottom, the velocity of the jumper is zero but not the acceleration because it is acting in the opposite direction that is why the jumper moves upward.

3 0
4 years ago
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Youare on a train travelling northat 80.0m/s relative to the ground. The air is still relativetothegroundwhen you hear the whist
AURORKA [14]

To solve this problem we will apply the concepts related to the Doppler effect. This is understood as the change in apparent frequency of a wave produced by the relative movement of the source with respect to its observer. Mathematically this is given as,

f = \frac{v \pm v_r}{v \pm v_s}(f_0)

Here,

v = Speed of the waves in the middle

v_r = Speed of the receiver in relation to the medium (Positive if the receiver is moving towards the transmitter or vice versa)

v_s = Speed of the source with respect to the medium (Positive if the source moves away from the receiver or vice versa)

Our values are given as,

v = 342m/s

f_0 = 262Hz

v_r = 80m/s

f = 350Hz

Replacing,

350 = \frac{342+80}{342-v} (262)

Solving for the velocity of the source,

v = 26.1m/s

Therefore the speed of the other train is 26.1m/s

3 0
3 years ago
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