2.3c. What is the frequency of a wave with a length of 12 m that travels at a speed of 36 m/s?

Which is the MOST accurate statement about the heating curve and energy?

V125BC [204]

The answer is A). Moving from A to C the temperature and the kinetic energy increases.

3 0

3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

leonid [27]

Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

6 0

3 years ago

Un avión da "una vuelta mortal" de radio R = 500 m con una celeridad constante v = 360 km/h. Halla la fuerza ejercida por el asi

Airida [17]

Answer:

1400 N

Explanation:

Verá, durante el salto mortal, el piloto se mueve en una trayectoria circular y la fuerza que actúa sobre él es una fuerza centrípeta.

Sea la fuerza centrípeta F, la masa del piloto (m) = 70 Kg, el radio (r) = 500 my la velocidad (v) = 360 km / hr * 1000/3600 = 100 m / s

F = mv ^ 2 / r

F = 70 * (100) ^ 2/500

F = 1400 N

8 0

3 years ago

Could an experiment similar to young's two-slit experiment be performed with sound? how might this be carried out? does it matte

Arada [10]

Young's double slit experiment(YDSE) can be used for any kind of waves such as electromagnetic waves, sound waves, water waves, gravity waves. YDSE is based on interference. In this experiment, we make two waves interfere in order to obtain bright and dark fringes on the screen(in case of light).
You can carry this out with water, would be great if you try this at pond or water reservoir in order to see perfect ripples.

7 0

3 years ago

There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor

VMariaS [17]

Answer:

F = 7.68 10¹¹ N, θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

cos 45 = F₂₄ₓ / F₂₄

sin 45 = F_{24y) / F₂₄

F₂₄ₓ = F₂₄ cos 45

F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

Fₓ = -F₂₁ + F₂₄ₓ

Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis

F_y = - F₂₃ + F_{24y}

F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

F₂₁ = F₂₃

Let's use Coulomb's law

F₂₁ = k q₁ q₂ / r₁₂²

the distance between the two charges is

r = a

F₂₁ = k q² / a²

we calculate F₂₄

F₂₄ = k q₂ q₄ / r₂₄²

the distance is

r² = a² + a²

r² = 2 a²

we substitute

F₂₄ = k q² / 2 a²

we substitute in the components of the forces

Fx = $- k \frac{q^2}{a^2} + k \frac{q^2}{2 a^2} \ cos 45$

Fx = $k \frac{q^2}{a^2}$ ( -1 + ½ cos 45)

F_y = k \frac{q^2}{a^2} ( -1 + ½ sin 45)

We calculate

F₀ = 9 10⁹ 4.25² / 0.440²

F₀ = 8.40 10¹¹ N

Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

Fₓ = -5.43 10¹¹ N

remember cos 45 = sin 45

F_y = - 5.43 10¹¹ N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

F = -5.43 10¹¹ (i + j) N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

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

F = 5.43 10¹¹ √2

F = 7.68 10¹¹ N

in angle is

θ = 45º

7 0

3 years ago