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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°

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Arbeitsauftrag 2

kramer

Explanation:

<em>The height of the pendulum is measured from the lowest point it reaches (point 3). </em>

At 1, the kinetic energy of the pendulum is zero (because it is not moving), and it has maximum potential energy.

At 2, the pendulum has both kinetic and potential energy, and how much of each it has depends on its height—smaller the height greater the kinetic energy and lower the potential energy.

At 3, the height is zero; therefore, the pendulum has no potential energy, and has maximum kinetic energy.

At 4, the pendulum again gains potential energy as it climbs back up, Again how much of each forms of energy it has depends on its height.

At 5, the maximum height is reached again; therefore, the pendulum has maximum potential energy and no kinetic energy.

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3 years ago

enot [183]

Answer: 3 square feet

Explanation: I took the test

6 0

3 years ago

the magnitude of the magnetic field at point p for a certain electromagnetic wave is 2.21. What is the magnitude of the elctic f

vesna_86 [32]

Answer:

$6.63\times 10^8\ N/C$

Explanation:

Given that,

The magnitude of magnetic field, B = 2.21

We need to find the magnitude of the electric field. Let it is E. So,

$\dfrac{E}{B}=c\\\\E=Bc$

Put all the values,

$E=2.21\times 3\times 10^8\\\\=6.63\times 10^8\ N/C$

So, the magnitude of the electric field is equal to $6.63\times 10^8\ N/C$ .

7 0

3 years ago

A double charged ion is accelerated to an energy of 32.0 keV by the electric field between two parallel conducting plates separa

Zanzabum

Answer:

$E=8*10^5\frac{V}{m}$

Explanation:

The magnitude of the electric field between two parallel conducting plates is defined as:

$E=\frac{\Delta V}{d}$

Here $\Delta V$ is the potential difference between the plates and d its separation.

The electric potential energy is defined as the product between the particle's charge and the potential difference:

$U=q\Delta V$

Solving for $\Delta V$ and replacing in the electric field formula:

$\Delta V=\frac{U}{q}\\E=\frac{U}{qd}$

In this case we have a double charged ion, so $q=2e$ :

$E=\frac{32*10^3eV}{(2e)(2*10^{-2}m)}\\E=8*10^5\frac{V}{m}$

6 0

3 years ago

Electromagnetic radiation of 8.12×10¹⁸ Hz frequency is applied on a metal surface and caused electron emission. Determine the wo

klemol [59]

Answer:

The work function ϕ of the metal = 53.4196 x 10⁻¹⁶ J

Explanation:

When light is incident on a photoelectric material like metal, photoelectrons are emitted from the surface of the metal. This process is called photoelectric effect.

The relationship between the maximum kinetic energy ( $E_{k}$ ) of the photoelectrons to the frequency of the absorbed photons (f) and the threshold frequency (f₀) of the photoemissive metal surface is:

$E_{k}$ = h(f − f₀)

$E_{k}$ = hf - hf₀

E is the energy of the absorbed photons: E = hf

ϕ is the work function of the surface: ϕ = hf₀

$E_{k}$ = E - ϕ

Frequency f = 8.12×10¹⁸ Hz

Maximum kinetic energy $E_{k}$ = 4.16×10⁻¹⁷ J

Speed of light c = 3 x 10⁸ m/s

Planck's constant h = 6.63 × 10⁻³⁴ Js

E = hf = 6.63 × 10⁻³⁴ x 8.12×10¹⁸

E = 53.8356 x 10⁻¹⁶ J

from $E_{k}$ = E - ϕ ;

ϕ = E - $E_{k}$

ϕ = 53.8356 x 10⁻¹⁶ - 4.16×10⁻¹⁷

ϕ = 53.4196 x 10⁻¹⁶ J

The work function of the metal ϕ = 53.4196 x 10⁻¹⁶ J

4 0

3 years ago