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KengaRu [80]
3 years ago
14

Break lines may be drawn as

Engineering
1 answer:
pav-90 [236]3 years ago
6 0

Break lines can be drawn as a way to remove a part of a drawing for precision or shorten objects to fit if they're too long to place anywhere else. Short and long ones are used on flat surfaces, while the cylindrical break lines are used on different forms.

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This might reduce outage probability because multiple base stations are able to receive a given mobile signal at a time which lead to signal outage decrease or weakness.

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The team needs to choose a primary view for the part drawing. Three team members make suggestions:
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<u>Option 1</u>

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As the team has already submitted the plans for the part drawing, the best way to proceed would be how it was given in the plans. Hence, the option to be selected :

  • <u>Team member 1 suggests an orthographic top view because that is how the plans for the part were submitted.</u>
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3 0
4 years ago
transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of
hram777 [196]
<h2>Corrected Question:</h2>

A step-up transformer has 26 turns on the primary coil and 720 turns on the secondary coil. If this transformer is to produce an output of 4100 V with a 16-mA current, what input current and voltage are needed?

<h2>Answer:</h2>

The input current and voltage needed are 443 mA and 148 V respectively.

<h2>Explanation:</h2>

In a step-up transformer, the relationship between the number of turns in its primary coil (N_{p}), the number of turns in its secondary coil (N_{s}), the input voltage (V_{p}), the output voltage (V_{s}), the input current (I_{p}), and the output current (I_{s}) is given by;

\frac{N_{s} }{N_{p} } = \frac{V_{s} }{V_{p} } = \frac{I_{p} }{I_{s} }

This implies that;

\frac{N_{s} }{N_{p} } = \frac{V_{s} }{V_{p} }  ---------------------(i)

\frac{N_{s} }{N_{p} } = \frac{I_{p} }{I_{s} }  ---------------------(ii)

From the question;

N_{p} = 26 turns

N_{s} = 720 turns

V_{s} = 4100V

I_{s} = 16mA = 16 x 10⁻³A

(a) Substitute these values into equation (ii) as follows;

\frac{720}{26} = \frac{I_{p} }{16*10^{-3}}

Solve for I_{p};

I_{p} = 720 x 16 x 10⁻³ / 26

I_{p} = 443 x 10⁻³

I_{p} = 443 mA

Therefore the input current needed is 443mA

(b) Also, substitute those values into equation (i) as follows;

\frac{720}{26} = \frac{4100 }{V_{p} }

Solve for V_{p};

V_{p} = 4100 x 26 / 720

V_{p} = 148 V

Therefore, the input current needed is 148 V

5 0
3 years ago
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