All categories

3 years ago

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°

You might be interested in

Johnson is dragging a bag on a ice surface. He pulls on the strap with a force of 112 N at an angle of 45° to the horizontal to

Nady [450]

Answer with Explanation:

We are given that

Force=F=112 N

$\theta=45^{\circ}$

Distance, $s=84 m$

Time, t=3.33 minutes

We have to find the work done by Johnson on the bag and the power generated by Johnson.

Work done, W= $Fscos\theta$

Using the formula

Work done, W= $112\times 84cos45=6652.46 J$

Power, P= $\frac{W}{t}=\frac{6652.46}{3.33\times 60}=33.3 watt$

Using 1 minute=60 s

Hence, the power generated by Johnson=33.3 watt

5 0

3 years ago

All but one statement applies to electromagnetic radiation

lina2011 [118]

<span>D) Electromagnetic radiation travels in the form of longitudinal waves.</span>

3 0

3 years ago

Give five characteristics of a metal

Kipish [7]

Answer:

are good conductors of heat,are you used to make doors utensils e.t.c.

3 0

3 years ago

A trebuchet was a hurling machine built to attack the walls of a castle under siege. A large stone could be hurled against a wal

Studentka2010 [4]

(a) 18.9 m/s

The motion of the stone consists of two independent motions:

- A horizontal motion at constant speed

- A vertical motion with constant acceleration ( $g=9.8 m/s^2$ ) downward

We can calculate the components of the initial velocity of the stone as it is launched from the ground:

$u_x = v_0 cos \theta = (25.0)(cos 41.0^{\circ})=18.9 m/s\\u_y = v_0 sin \theta = (25.0)(sin 41.0^{\circ})=16.4 m/s$

The horizontal velocity remains constant, while the vertical velocity changes due to the acceleration along the vertical direction.

When the stone reaches the top of its parabolic path, the vertical velocity has became zero (because it is changing direction): so the speed of the stone is simply equal to the horizontal velocity, therefore

$v=18.9 m/s$

(b) 22.2 m/s

We can solve this part by analyzing the vertical motion only first. In fact, the vertical velocity at any height h during the motion is given by

$v_y^2 - u_y^2 = 2ah$ (1)

where

$u_y = 16.4 m/s$ is the initial vertical velocity

$v_y$ is the vertical velocity at height h

$a=g=-9.8 m/s^2$ is the acceleration due to gravity (negative because it is downward)

At the top of the parabolic path, $v_y = 0$ , so we can use the equation to find the maximum height

$h_{max} = \frac{-u_y^2}{2a}=\frac{-(16.4)^2}{2(-9.8)}=13.7 m$

So, at half of the maximum height,

$h = \frac{13.7}{2}=6.9 m$

And so we can use again eq(1) to find the vertical velocity at h = 6.9 m:

$v_y = \sqrt{u_y^2 + 2ah}=\sqrt{(16.4)^2+2(-9.8)(6.9)}=11.6 m/s$

And so, the speed of the stone at half of the maximum height is

$v=\sqrt{v_x^2+v_y^2}=\sqrt{18.9^2+11.6^2}=22.2 m/s$

(c) 17.4% faster

We said that the speed at the top of the trajectory (part a) is

$v_1 = 18.9 m/s$

while the speed at half of the maximum height (part b) is

$v_2 = 22.2 m/s$

So the difference is

$\Delta v = v_2 - v_2 = 22.2 - 18.9 = 3.3 m/s$

And so, in percentage,

$\frac{\Delta v}{v_1} \cdot 100 = \frac{3.3}{18.9}\cdot 100=17.4\%$

So, the stone in part (b) is moving 17.4% faster than in part (a).

4 0

4 years ago

If an athlete completes a cycle around a circular track, his displacement eauals to ZERO.

RSB [31]

The statement is <em>TRUE</em>.

"Displacement" means the distance and direction from the start-point to the end-point, NO MATTER how you got there.

If you go out cycling and go back home when you're done, your displacement is zero. it doesn't matter how far you cycled, or what streets or woods you cycled through, or whether your route was a circle, a square, an ellipse, a triangle, or a jumble, or whether you cycled around and around the same path 27 times. You ended right where you started. The distance from your start-point to your end-point is zero, so your displacement is zero.

3 0

3 years ago