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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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A single loop of wire with an area of 0.0780 m2 is in a uniform magnetic field that has an initial value of 3.80 T, is perpendic
Katarina [22]

Answer:

The induced current is 26.7 mA

Explanation:

Given;

area of the loop, A = 0.078 m²

initial magnetic field, B₁ = 3.8 T

change in the magnetic field strength, dB/dt = 0.24 T/s

The induced emf is calculated as;

emf = - \frac{d \phi}{dt} \\\\emf = -\frac{dB.A}{dt} \\\\emf = A (\frac{dB}{dt} )\\\\emf = 0.078(0.24)\\\\emf = 0.0187 \ V

The resistance of the loop = 0.7 Ω

The induced current is calculated as;

V = IR\\\\I = \frac{V}{R} = \frac{emf}{R} = \frac{0.0187}{0.7} = 0.0267 \ A = 26.7 \ mA

4 0
2 years ago
An X-Ray tube is an evacuated glass tube, where the electrons are produced at one end and accelerated by a strong electric field
lawyer [7]

Answer:

a) ΔV = 25.59 V, b)  ΔV = 25.59 V,  c)  v = 7 10⁴ m / s,  v/c= 2.33 10⁻⁴ ,

v/c% = 2.33 10⁻²

Explanation:

a) The speed they ask for electrons is much lower than the speed of light, so we don't need relativistic corrections, let's use the concepts of energy

starting point. Where the electrons come out

          Em₀ = U = e DV

final point. Where they hit the target

          Em_f = K = ½ m v2

energy is conserved

          Em₀ = Em_f

         e ΔV = ½ m v²

         ΔV = \frac{1}{2} mv²/e     (1)

If the speed of light is c and this is 100% then 1% is

         v = 1% c = c / 100

         v = 3 10⁸/100 = 3 10⁶6 m/ s

let's calculate

         ΔV = \frac{1}{2}  \frac{9.1 \ 10^{-31} (3 10^6 )^2 }{ 1.6 10^{-19} }

         ΔV = 25.59 V

b) Ask for the potential difference for protons with the same kinetic energy as electrons

             K_e = K_p

              K_p = ½ m v_e²

              K_p = \frac{1}{2}  9.1 10⁻³¹ (3 10⁶)²

              K_p = 40.95 10⁻¹⁹ J

we substitute in equation 1

              ΔV = Kp / M

              ΔV = 40.95 10⁻¹⁹ / 1.6 10⁻¹⁹

              ΔV = 25.59 V

notice that these protons go much slower than electrons because their mass is greater

c) The speed of the protons is

             e ΔV = ½ M v²

             v² = 2 e ΔV / M

             v² = \frac{2 \ 1.6 \ 10^{-19} \ 25.59 }{1.67 \ 10^{-27} }

              v² = 49,035 10⁸

               v = 7 10⁴ m / s

Relation

        v/c = \frac{7 \ 10^4 }{ 3 \ 10^8}

        v/c= 2.33 10⁻⁴

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D: Dogs might. In general we don't.

I would pick C

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ΔT is the change in temperature.

So the law of conservation of heat tells that:

Sensible heat of Z + Sensible heat of container = Sensible heat of X

Since we have no idea what these substances are, there is no way of knowing the Cp. We can't proceed with the calculations. So, we can only assume that in the duration of 15 minutes, the whole system achieves equilibrium. Therefore, the equilibrium temperature of the system is equal to 32°C. The answer is C.
5 0
3 years ago
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