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klemol [59]
3 years ago
13

Why are the meters squared in the formula to calculate acceleration?

Physics
1 answer:
lianna [129]3 years ago
8 0

Answer:

During acceleration, you are moving across a distance over a time, but also increasing how fast we are doing it. Therefore, it means by how many meters per second the velocity changes every second

Explanation:

You might be interested in
Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to
jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x 10^{-8} J.

(b) The wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency of the photon is 2.47 x 10^{25} Hz.

Explanation:

Let;

E_{1} = -13.60 ev

E_{2} = -3.40 ev

(a) Energy of the emitted photon can be determined as;

E_{2} - E_{1} = -3.40 - (-13.60)

           = -3.40 + 13.60

           = 10.20 eV

           = 10.20(1.6 x 10^{-9})

E_{2} - E_{1} = 1.632 x 10^{-8} Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x 10^{-8} Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x 10^{-34} Js), c is the speed of light (3 x 10^{8} m/s) and λ is the wavelength.

10.20(1.6 x 10^{-9}) = (6.6 x 10^{-34} * 3 x 10^{8})/ λ

λ = \frac{1.98*10^{-25} }{1.632*10^{-8} }

  = 1.213 x 10^{-17}

Wavelength of the photon is 1.2 x 10^{-17} m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x 10^{-8}  = 6.6 x 10^{-34} x f

f = \frac{1.632*10^{-8} }{6.6*10^{-34} }

 = 2.47 x 10^{25} Hz

Frequency of the emitted photon is 2.47 x 10^{25} Hz.

6 0
3 years ago
In a region of space where gravitational forces can be neglected, a sphere is accelerated by a uniform light beam of intensity 8
BlackZzzverrR [31]

Answer:

The correct answer is B

Explanation:

To calculate the acceleration we must use Newton's second law

      F = m a

      a = F / m

To calculate the force we use the defined pressure and the radiation pressure for an absorbent surface

       P = I / c        absorbent surface

       P = F / A

       F / A = I / c

       F = I A / c

The area of ​​area of ​​a circle is

      A = π r²

We replace

     F = I π r² / c

Let's calculate

     F = 8.0 10⁻³ π (1.0 10⁻⁶)²/3 10⁸

     F = 8.375 10⁻²³ N

Density is

      ρ = m / V

      m = ρ V

      m = ρ (4/3 π r³)

      m = 4500 (4/3 π (1 10⁻⁶)³)

      m = 1,885 10⁻¹⁴ kg

Let's calculate the acceleration

     a = 8.375 10⁻²³ / 1.885 10⁻¹⁴

     a = 4.44 10⁻⁹ m/s²               absorbent surface

The correct answer is B

4 0
3 years ago
Write two ways to be protected from the energy crisis?​
Arada [10]
Possible Solutions to the Problem of Global Energy Crisis:

1. Move Towards Renewable Resources.
2. Buy Energy-Efficient Products.
5 0
3 years ago
Which statement describes a benefit of a model that uses three balls to model the Sun-Earth-Moon system? A. Every aspect of the
Dovator [93]

Answer:

D. It represents a very large, complex system.​

Explanation:

I just did it on a p e x...

8 0
2 years ago
The intensity of an earthquake wave passing through the earth is measured to be 2.5×106 j/(m2⋅s) at a distance of 43 km from the
vampirchik [111]

r₁ = distance of the point from the source = 43 km = 43000 m

I₁ = intensity of earthquake wave at distance "r₁" = 2.5 x 10⁶ W/m²

r₂ = distance of the point from the source = 1.5 km = 1500 m

I₂ = intensity of earthquake wave at distance "r₂" = ?

we know that , for a constant power , the intensity of wave is inversely proportional to the distance from the source .

I α 1/r²             where I = intensity of wave , r = distance from source

hence we can write

I₁/I₂ = r₂²/r₁²

inserting the values

(2.5 x 10⁶) /I₂ = (1500/43000)²

I₂ = 2.1 x 10⁹ W/m²

4 0
3 years ago
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