Ask question

Ask question

All categories

2 years ago

6

In nuclear fission, a nucleus of uranium-238, containing 92 protons, can divide into two smaller spheres, each having 46 protons

and a radius of 5.90 ✕ 10-15 m. What is the magnitude of the repulsive electric force pushing the two spheres apart?

1 answer:

IRINA_888 [86]2 years ago

3 0

Answer:

$F = 3501.34 N$

Explanation:

As we know that electrostatic repulsion force between two protons is given as

$F = \frac{kq_1q_2}{r^2}$

now we know that

$q_1 = q_2 = 46 \times (1.6 \times 10^{-19})$

$q_1 = q_2 = 7.36 \times 10^{-18} C$

now the distance between the two atoms is same as the distance between two center

$r = 5.90 \times 10^{-15} + 5.90 \times 10^{-15}$

$r = 1.18 \times 10^{-14} m$

now from above formula we have

$F = \frac{(9 \times 10^9)(7.36 \times 10^{-18})^2}{(1.18 \times 10^{-14})^2}$

$F = 3501.34 N$

You might be interested in

An electron passes through a point 2.83 cm 2.83 cm from a long straight wire as it moves at 35.5 % 35.5% of the speed of light p

igor_vitrenko [27]

Answer:

The magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Explanation:

Given:

Distance from the wire to the field point $r = 2.83 \times 10^{-2}$ m

Speed of electron $v = 35.5 \%c$

Current $I = 17.7$ A

For finding the acceleration,

First find the magnetic field due to wire,

$B = \frac{\mu _{o}I }{2\pi r }$

Where $\mu_{o} = 4\pi \times 10^{-7}$

$B = \frac{4\pi \times 10^{-7} \times 17.7 }{2\pi (2.83 \times 10^{-2} ) }$

$B = 12.50 \times 10^{-5}$ T

The magnetic force exerted on the electron passing through straight wire,

$F = qvB$

$F = 1.6 \times 10^{-19} \times 0.355 \times 3 \times 10^{8} \times 12.50 \times 10^{-5}$

$F = 21.3 \times 10^{-16}$ N

From the newton's second law

$F = ma$

Where $m =$ mass of electron $= 9.1 \times 10^{-31}$ kg

So acceleration is given by,

$a = \frac{F}{m}$

$a = \frac{21.3 \times 10^{-16} }{9.1 \times 10^{-31} }$

$a = 2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

Therefore, the magnitude of electron acceleration is $2.34 \times 10^{15}$ $\frac{m}{s^{2} }$

7 0

2 years ago

How does the sun's energy most directly influence precipitation in an area?

topjm [15]

The sun's energy influences climate in various ways. For example the latitudes at the equator receive more energy from the sun and therefore have warmer temperatures, On the other hand the sun's energy influences precipitation in a climate by driving the water cycle which determines precipitation.The sun is what makes the water cycle take place. That is the sun provides energy or heat to the earth; the heat causes liquid and frozen water to evaporate into water vapor gas, which rises high in the sky to form clouds ( precipitation), that in turn give us rain

5 0

2 years ago

A car stopped at a red light, not moving?

timofeeve [1]

I think it's a) 1st Newton's law... so sorry if it's wrong...

5 0

3 years ago

You are given a vector in the xy plane that has a magnitude of 84.0 units and a y-component of -67.0 units.

melomori [17]

Answer:

Explanation:

a)Magnitude = $\sqrt{(x1-y2)^{2} + (x1-x2)^{2} }$

84= $\sqrt{(0- (-67))^{2} + (x-0)^{2} }$

x= +50.67 or -50.67 units

b) We are given that the resultant is entirely in the -ve x direction which means that the y-component of the resultant is 0; It means that the y-component of the next vector = -ve of the y component of the initial vector i.e 67.

To make the magnitude 80 units in the negative x direction where the y component is 0, the x component must be -130.67(-50.67 - 80) as the x component is + 50.67units.

Magnitude = $\sqrt{(0- (67))^{2} + (-130.67)^{2} }$ = 146.85 units

c) The direction vector = 67/146.85 i - 130.67/146.85 j where i corresponds to the vector in y direction and j corresponds to the vector in x direction. Or this vector is at an angle of 180 - $Tan^{-1}(67/130.67)degrees$ i.e 152.85 degrees from the +ve x-axis.

5 0

3 years ago

When a voltage difference is applied to a piece of metal wire, a current flows through it. if this metal wire is now replaced wi

Nikitich [7]

The resistance of the cylindrical wire is $R=\frac{\rho l}{A}$ .

Here $R$ is the resistance, $l$ is the length of the wire and $A$ is the area of cross section. Since the wire is cylindrical $A=\frac{\pi d^2}{4}$ .

Comparing two wires,

$R_1=\frac{\rho_1 l}{A_1} \\ R_2=\frac{\rho_2 l}{A_2}$

Dividing the above 2 equations,

$\frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{A_2 }{A_1} \\ \frac{R_1}{R_2}=\frac{\rho_1 }{\rho_2} \frac{d_2^2 }{d_1^2} \\$

Since $d_2=2d_1$

The above ratio is

$\frac{R_1}{R_2}=\frac{1.68(10^{-8}) }{1.59(10^{-8}) } (4)\\ \frac{R_1}{R_2}=4.2264$

We also have,

$\frac{E/R_1}{E/R_2} =\frac{I_1}{I_2} \\ I_2=\frac{R_1}{R_2}I_1 \\ I_2=4.23I_1$

The current through the Silver wire will be 4.23 times the current through the original wire.

8 0

2 years ago