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bija089 [108]
2 years ago
6

Why is the heliocentric model not correct

Physics
2 answers:
Svetllana [295]2 years ago
6 0
The heliocentric model is CORRECT because it has been proven to be correct by multiple generations of scientists.
bekas [8.4K]2 years ago
4 0

Answer:

Um, wrong.

Explanation:

The Helioentric model is correct. We orbit the sun. The GEOcentric model is incorrect. The Universe does not orbit the Earth. What are you learning?

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A musical tone sounded on a pi has frequency of 410 Hz and a wavelength of 0.80 m.
spayn [35]
C. I did it bit i forgot how I did it
4 0
3 years ago
A hydrogen atom that has an electron in the n = 2 state absorbs a photon. What wavelength must the photon possess to send the el
Deffense [45]

Answer:

486nm

Explanation:

in order for an electron to transit from one level to another, the wavelength emitted is given by Rydberg Equation which states that

\frac{1}{wavelength}=R.[\frac{1}{n_{f}^{2} } -\frac{1}{n_{i}^{2} }] \\n_{f}=2\\n_{i}=4\\R=Rydberg constant =1.097*10^{7}m^{-1}\\subtitiute \\\frac{1}{wavelength}=1.097*10^{7}[\frac{1}{2^{2} } -\frac{1}{4^{2}}]\\\frac{1}{wavelength}= 1.097*10^{7}*0.1875\\\frac{1}{wavelength}= 2.06*10^{6}\\wavelength=4.86*10{-7}m\\wavelength= 486nm\\

Hence the photon must possess a wavelength of 486nm in order to send the electron to the n=4 state

4 0
3 years ago
An archer shoots an arrow with a mass of 45.0 grams from bow pulled
Sladkaya [172]

Answer:

The force the archer need to pull in order to achieve the height is approximately 101.8 N

Explanation:

By energy conservation principle, puling an elastic bow with a force, for a given distance, performs work which is converted to the potential energy of the arrow at height

The given parameters are;

The mass of the arrow, m = 45.0 grams = 0.045 kg

The distance the elastic bow is pulled, d = 65.0 cm = 0.65 m

The height at which the arrow is reaches, h = 150.0 meters

Let 'F', represent the force the archer need to pull in order to achieve the height

Work done, W = Force × Distance moved in the direction of the force

Therefore;

The work done in pulling the arrow, W = F × d

By energy conservation, we have;

The work done in pulling the arrow, W = The potential energy gained by the arrow, P.E.

W = P.E.

The potential energy gained by the arrow, P.E. = m·g·h

Where;

m = The mass of the arrow

g = The acceleration due to gravity = 9.8 m/s²

h = The height the arrow reaches

∴ by plugging in the values, P.E. = 0.045 kg ×9.8 m/s² × 150 m = 66.15 J

W = F × d = F × 0.065 m

Also, W = P.E. = 66.15 J

∴ W = F × 0.065 m = 66.15 J

F × 0.065 m = 66.15 J

F = 66.15 J/(0.65 m) = 1323/13 N ≈ 101.8 N

The force the archer need to pull in order to achieve the height, F ≈ 101.8 N.

3 0
2 years ago
Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax
kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

1)

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its x-component is

F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its x-component is

F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N

So, the x-component of the resultant force is

F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N

2)

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

180^{\circ}-61^{\circ}

so its y-component is

F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

180^{\circ}+52.8^{\circ}

Therefore its y-component is

F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N

So, the y-component of the resultant force is

F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

6 0
3 years ago
Read 2 more answers
A car starts from rest and accelerates uniformly over a time of 18 seconds for a distance of 390 m. Determine the acceleration o
Sergeeva-Olga [200]

Answer:

a=2.4\ m/s^2

Explanation:

Given that,

The initial speed of a car, u = 0

Time, t = 18 s

Distance, d = 390 m

We need to find the acceleration of the car. Let it is a. Using the second equation of motion to find it.

d=ut+\dfrac{1}{2}at^2

or

d=\dfrac{1}{2}at^2\\\\a=\dfrac{2d}{t^2}\\\\a=\dfrac{2\times 390}{(18)^2}\\\\a=2.4\ m/s^2

So, the acceleration of the car is 2.4\ m/s^2.

5 0
3 years ago
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