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den301095 [7]
3 years ago
7

the measure of each exterior angle of a regular pentagon is ___ the measure of each exterior angle of a regular nonagon

Physics
1 answer:
Cerrena [4.2K]3 years ago
4 0

Answer:

(a) 72°

(b) 40°

Explanation:

PENTAGON

First, we calculate the total angles in a Pentagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 5.

Hence, the total angle in a polygon is

180(5 - 2) = 180 * 3 = 540°

Therefore, each angle will be:

540°/5 = 108°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular pentagon will be:

180 - 108 = 72°

The exterior angle of a regular Pentagon is 72°

NONAGON

First, we calculate the total angles in a Nonagon using:

180(n - 2)

Where n = number of sides of the polygon, in this case, 9.

Hence, the total angle in a polygon is

180(9 - 2) = 180 * 7 = 1260°

Therefore, each angle will be:

1260°/9 = 140°

Because the interior angle and exterior angle form a straight line (180°), the exterior angle of a regular nonagon will be:

180 - 140 = 40°

The exterior angle of a regular Nonagon is 40°

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An object initially at rest experiences an acceleration of 0.281 m/s2 to the South for a time of 5.44 seconds. It then increases
andre [41]

Answer:

12.0 meters

Explanation:

Given:

v₀ = 0 m/s

a₁ = 0.281 m/s²

t₁ = 5.44 s

a₂ = 1.43 m/s²

t₂ = 2.42 s

Find: x

First, find the velocity reached at the end of the first acceleration.

v = at + v₀

v = (0.281 m/s²) (5.44 s) + 0 m/s

v = 1.53 m/s

Next, find the position reached at the end of the first acceleration.

x = x₀ + v₀ t + ½ at²

x = 0 m + (0 m/s) (5.44 s) + ½ (0.281 m/s²) (5.44 s)²

x = 4.16 m

Finally, find the position reached at the end of the second acceleration.

x = x₀ + v₀ t + ½ at²

x = 4.16 m + (1.53 m/s) (2.42 s) + ½ (1.43 m/s²) (2.42 s)²

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5 0
3 years ago
A 2.00-m rod of negligible mass connects two very small objects at its ends. The mass of one object is 1.00 kg and the mass of t
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Answer:

<h2> 4kg</h2>

Explanation:

Step one:

given

length of rod=2m

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let the unknown mass be x

center of mass<em> c.m</em>= 1.6m

hence 1kg is 1.6m from the <em>c.m</em>

and x is 0.4m from the <em>c.m</em>

Taking moment about the <em>c.m</em>

<em>clockwise moment equals anticlockwise moments</em>

1*1.6=x*0.4

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divide both sides by 0.4 we have

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x=4kg

The mass of the other object is 4kg

3 0
3 years ago
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