Which do waves transfer

A certain atom has atomic number Z = 25 and atomic mass number A = 52. a. What is the approximate radius of the nucleus of this

wlad13 [49]

Answer:

a)The approximate radius of the nucleus of this atom is 4.656 fermi.

b) The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

Explanation:

$r=r_o\times A^{\frac{1}{3}}$

$r_o=1.25 \times 10^{-15} m$ = Constant for all nuclei

r = Radius of the nucleus

A = Number of nucleons

a) Given atomic number of an element = 25

Atomic mass or nucleon number = 52

$r=1.25 \times 10^{-15} m\times (52)^{\frac{1}{3}}$

$r=4.6656\times 10^{-15} m=4.6656 fm$

The approximate radius of the nucleus of this atom is 4.656 fermi.

b) $F=k\times \frac{q_1q_2}{a^2}$

k= $9\times 10^9 N m^2/C^2$ = Coulombs constant

$q_1,q_2$ = charges kept at distance 'a' from each other

F = electrostatic force between charges

$q_1=+1.602\times 10^{-19} C$

$q_2=+1.602\times 10^{-19} C$

Force of repulsion between two protons on opposite sides of the diameter

$a=2\times r=2\times 4.6656\times 10^{-15} m=9.3312\times 10^{-15} m$

$F=9\times 10^9 N m^2/C^2\times \frac{(+1.602\times 10^{-19} C)\times (+1.602\times 10^{-19} C)}{(9.3312\times 10^{-15} m)^2}$

$F=2.6527 N$

The electrostatic force of repulsion between two protons on opposite sides of the diameter of the nucleus is 2.6527

6 0

3 years ago

Particle 1 has mass 4.6 kg and is on the x-axis at x = 5.7 m. Particle 2 has mass 7.2 kg and is on the y-axis at y = 4.2 m. Part

Iteru [2.4K]

To solve this problem it is necessary to apply the concepts related to the Gravitational Force, for this purpose it is understood that the gravitational force is described as

$F_g = \frac{Gm_1m_2}{r^2}$

Where,

G = Gravitational Universal Force

$m_i =$ Mass of each object

To solve this problem it is necessary to divide the gravitational force (x, y) into the required components and then use the tangent to find the angle generated between both components.

Our values are given as,

$m_1 =4.6 kg\\m_2 = 7.2 kg\\m_3 = 2.6 kg\\r_1 = 5.7 m\\r_2 = 4.2 m$

Applying the previous equation at X-Axis,

$F_x = \frac{Gm_1m_3}{R_{1}^2}\\F_x = \frac{6.67*10^{-11}*4.6*2.6}{5.7^2}\\F_x = 2.46*10^{-11}N$

Applying the previous equation at Y-Axis,

$F_y = \frac{Gm_2m_3}{R_2^2}\\F_y = \frac{6.67*10^{-11}*7.2*2.6}{4.2^2}\\F_y = 7.08*10^{-11} N$

Therefore the angle can be calculated as,

$tan\theta = \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{F_y}{F_x}\\\theta = tan^{-1} \frac{7.08*10^{-11}}{2.46*10^{-11}}\\\theta = 71\°$

Then in the measure contrary to the hands of the clock the Force in the particle 3 is in between the positive direction of the X and the negative direction of the Y at 71 ° from the positive x-axis.

5 0

4 years ago

Why do windows of distant houses appear to reflect the sun only when rising or setting

alisha [4.7K]

most importantly, the sun is reflected back at the same angle as the angle of incidence, so with vertical panes of glass and the sun overhead, the reflection will just hit the ground before traveling any distance.

3 0

3 years ago

Read 2 more answers

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

Dafna1 [17]

Answer:

- pendulum use to determine acceleration