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3 years ago

Find the area of this semi-circle with diameter, d = 58cm. Give your answer rounded to 2 DP.

Mathematics

1 answer:

Anastasy [175]3 years ago

5 0

Answer:

The area of the semi-circle

A = 1320.37 cm²

Step-by-step explanation:

Explanation:-

Given the diameter of the circle 'd' = 58cm

The radius of the circle 'd' = 2r

$r = \frac{d}{2} = \frac{58}{2} = 29$

The area of the semi-circle

$A = \frac{1}{2} \pi r^{2}$

$A = \frac{1}{2} (3.14) ((29))^{2}$

A = 1320.37 cm²

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segment BC is an are of a circle with radius 30.0 cm, and point P is at the center of curvature of the arc Segment DA is an arc

Elan Coil [88]

Answer:

The magnetic field is $B = 3.87 *10^{-6} \ T$

Direction is inward

Step-by-step explanation:

The diagram for this question is shown on the first uploaded image

From the question we are told that

The radius of BC is $r_{BC} = 30 \ cm = 0.30 \ m$

The radius of DA is $r _{DA} = 20 \ cm = 0.20 \ m$

The length of CD is $r_{CD} = 10 cm = 0.1 \ m$

The length of AB is $r_{AB} = 10 cm = 0.1 \ m$

The current is $I = 11.4A$

The magnetic field is mathematically represented as

$B = B_{BC} + B_{DA} + B_{CD} + B_{AB}$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\theta_{DA}}{r_{DA}} + \frac{\theta_{BC}}{r_{BC}}+ \frac{\theta_{CD}}{r_{CD}}+\frac{\theta_{AB}}{r_{AB}}]$

Where

$\theta_{BC} = \theta_{DA} = \frac{2\pi}{3}$

Where $\frac{2 \pi}{3} = 120^o$

$\theta_{CD} = \theta_{AB} = 0^o$

$B = \frac{\mu_o I}{4 \pi } [ \frac{\frac{2\pi}{3} }{0.20} - \frac{\frac{2\pi}{3}}{0.30}}+ \frac{0}{0.10}+\frac{0}{0.10}]$

$B = 3.87 *10^{-6} \ T$

The direction is into the page

This because the magnitude of the magnetic field due to arc BC whose direction is outward is less than that of DA whose direction is inward

This is because according to Fleming's Left Hand Rule the direction of current is perpendicular to the direction of magnetic field so since current in arc BC and DA are moving in opposite direction their magnetic field will also be moving in opposite direction

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