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Ainat [17]
3 years ago
5

What is the formula that represents the simplest ratio of the atoms in the components

Physics
1 answer:
Dafna1 [17]3 years ago
5 0

Answer:

Empirical formula

<em>Hope</em><em> </em><em>it'll</em><em> </em><em>help</em><em>!</em>

<em>stay</em><em> </em><em>safe</em><em>:</em><em>)</em>

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Bare free charges do not remain stationary when close together. To illustrate this, calculate the acceleration of two isolated p
Natali5045456 [20]

Answer:

Acceleration, a=3.6\times 10^{16}\ m/s^2

Explanation:

It is given that, two isolated protons are separated by 2 nm. The force due to charged particles is given by :

F_e=\dfrac{kq^2}{d^2}

Force due to mass of proton, F_g=ma

ma=\dfrac{kq^2}{d^2}

a=\dfrac{kq^2}{md^2}

a=\dfrac{9\times 10^9\times (1.6\times 10^{-19})^2}{1.6\times 10^{-27}\times (2\times 10^{-9})^2}

a=3.6\times 10^{16}\ m/s^2

So, the acceleration of two isolated protons is 3.6\times 10^{16}\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
A satellite dish is in the shape of a parabolic surface. Signals coming from a satellite strike the surface of the dish and are
LenKa [72]

Answer: 3.125 ft

Explanation:

If this dish has the form of a concave upward parabola and its vertex p is at the origin, its corresponding equation is:

x^{2}=4py

Where:

x is the radius, which can be found by dividing the diameter d=10 ft by half. Hence x=\frac{d}{2}=\frac{10 ft}{2}=5 ft

y=2 ft is the depth

p is the vertex of the parabola, where its base is

Finding p:

p=\frac{x^{2}}{4y}

p=\frac{(5 ft)^{2}}{4(2 ft)}

Finally:

p=3.125 ft This is where the the receiver should be placed

8 0
3 years ago
Can animal cells be broken down further into a living unit
mezya [45]

Answer: No! Animal cells cannot be broken down further into living cells.

5 0
3 years ago
Read 2 more answers
Dos cargas puntuales están fijas en el eje x: q1 = 6.0µC está en el origen, O, con x1 = 0.0 cm, y q2 = –3.0 µC está situada en e
erik [133]

Answer:

E_total = 1.30 10¹⁰ C / m²

Explanation:

The intensity of the electric field is

     E = k q / r²

on a positive charge proof

The total electric field at the midpoint is

as q₁= 6 10⁻⁶ C the field is outgoing to the right

for charge q₂ = -3 10⁻⁶ C, the field is directed to the right, therefore

E_total = E₁ + E₂

E_total = k q₁ / r₁² + k q₂ / r₂²

r₁ = r₂ = r = 4 10⁻² m

E_total = k/r² (q₁ + q₂)

 we calculate

E_total = 9 10⁹ / (4 10⁻²)²   (6.0 10⁻⁶ +3.0 10⁻⁶)

E_total = 1.30 10¹⁰ C / m²

8 0
3 years ago
An electron and a proton are held on an x axis, with the electron at x = + 1.000 m
mixas84 [53]

Answer:

  r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

Explanation:

For this exercise we must use conservation of energy

the electric potential energy is

          U = k \frac{q_1q_2}{r_{12}}

for the proton at x = -1 m

          U₁ =- k \frac{e^2 }{r+1}

for the electron at x = 1 m

          U₂ = k \frac{e^2 }{r-1}

starting point.

        Em₀ = K + U₁ + U₂

        Em₀ = \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1}

final point

         Em_f = k e^2 ( -\frac{1}{r_2 +1} + \frac{1}{r_2 -1})

   

energy is conserved

        Em₀ = Em_f

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e^2 (- \frac{1}{r_2 +1} + \frac{1}{r_2 -1})              

       

        \frac{1}{2} m v^2 - k \frac{e^2}{r+1} + k \frac{e^2}{r-1} = k e²(  \frac{2}{(r_2+1)(r_2-1)} )

we substitute the values

½ 9.1 10⁻³¹ 450 + 9 10⁹ (1.6 10⁻¹⁹)² [ - \frac{1}{20+1} + \frac{1}{20-1} ) = 9 109 (1.6 10-19) ²( \frac{2}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 2.304 10⁻³⁷ (5.0125 10⁻³) = 4.608 10⁻³⁷ ( \frac{1}{r_2^2 -1} )

          2.0475 10⁻²⁸ + 1.1549 10⁻³⁹ = 4.608 10⁻³⁷     \frac{1}{r_2^2 -1}

          \frac{2.0475 \ 10^{-28} }{1.1549 \ 10^{-37} } = \frac{1}{r_2^2 -1}

          r₂² -1 = (4.443 10⁸)⁻¹

           

          r2 = \sqrt{1 + 2.25 10^{-9}}

          r2 = 1 m

therefore the electron that comes with velocity does not reach the origin, it stops when it reaches the position of the electron at x = 1m

4 0
3 years ago
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