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Dahasolnce [82]
3 years ago
10

Why do I need to use the Unit Circle in Physics? And how do I use it?

Physics
1 answer:
stiks02 [169]3 years ago
6 0

Answer:

The unit circle helps in making so many calculations and equations easy.

Explanation:

You cannot separate the knowledge of trigonometry to application in equations to physics. The unit circle is known to have a radius of one. This means that the distance from the centre of the circle, regardless of the unit of measurement, to any point of the edge of the circle is 1. Since the unit circle is very helpful in trigonometry, and trigonometry in turn is the projection of triangles and angles that is very crucial in the calculation of momentum, velocity and other factors of physics, the importance of the unit circle cannot be overemphasized in physics.

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A punter wants to kick a football so that the football has a total flight time of 4.70s and lands 56.0m away (measured along the
Sindrei [870]

Answer:

Explanation:

How long does the football need to rise?

4.70/3 = 2.35 s

What height will the football reach?

h = ½(9.81)2.35² = 27.1 m

With what speed does the punter need to kick the football?

vy = g•t = 9.81(2.35) = 23.1 m/s

vx = d/t = 56.0/4.70 = 11.9 m/s

v = √(vx²+vy²) = 26.0 m/s

At what angle (θ), with the horizontal, does the punter need to kick the football?

θ = arctan(vy/vx) = 62.7°

3 0
2 years ago
The first law of motion applies<br> to what?
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3 years ago
Is a computer an open or closed system
soldier1979 [14.2K]

Answer:

OPEN

Explanation:

7 0
2 years ago
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Answer:b

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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
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