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Inessa05 [86]
3 years ago
15

The potential energy of a particle as a function of position will be given as U(x) = A x2 + B x + C, where U will be in joules w

hen x is in meters. A, B, and C are constants. show answer No Attempt 50% Part (a) Enter an expression for the x-component of the force as function of position F(x), in terms of the constants A, B, and C.
Physics
1 answer:
satela [25.4K]3 years ago
3 0

Answer:

F = - 2 A x - B

Explanation:

The force and potential energy are related by the expression

      F = - dU / dx i ^ -dU / dy j ^ - dU / dz k ^

Where i ^, j ^, k ^ are the unit vectors on the x and z axis

The potential they give us is

     U (x) = A x² + B x + C

Let's calculate the derivatives

    dU / dx = A 2x + B + 0

The other derivatives are zero because the potential does not depend on these variables.

Let's calculate the strength

      F = - 2 A x - B

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They both Move energy  through a Field  without Moving A Substance
7 0
4 years ago
A CO molecule has an intrinsic dipole moment whose magnitude is 4 X 10^-31 C*m. If the separation between the atoms is 0.11 nm,
aleksandrvk [35]
4x10^-31/(0.11x10^-9)= 3,63 x 10^-21
5 0
3 years ago
The amplitude of the electric field for a certain type of electromagnetic wave is 570 N/C. What is the amplitude of the magnetic
vesna_86 [32]

Given Information:  

Amplitude of the electric field = E =  570 N/C

Speed of light = c = 3.0x10⁸ m/s

Required Information:  

Amplitude of the magnetic field = ?

Answer:

Amplitude of the magnetic field =  1.90x10⁻⁶ T

Explanation:

The relation between the amplitude of the electric field and magnetic field is given by

B = E/c

Where c is the speed of light

B = 570/3.0x10⁸

B = 1.90x10⁻⁶ T

Therefore, the amplitude of magnetic field is 1.90x10⁻⁶ T

8 0
3 years ago
The nucleus of an atom can be modeled as several protons and neutrons closely packed together. Each particle has a mass of 1.67
Lady bird [3.3K]

Explanation:

The nucleus of an atom can be modeled as several protons and neutrons closely packed together.

Mass of the particle, m=1.67\times 10^{-27}\ kg

Radius of the particle, R=10^{-15}\ m

(a) The density of the nucleus of an atom is given by mass per unit area of the particle. Mathematically, it is given by :

d=\dfrac{m}{V}, V is the volume of the particle

d=\dfrac{m}{(4/3)\pi r^3}

d=\dfrac{1.67\times 10^{-27}}{(4/3)\pi (10^{-15})^3}

d=3.98\times 10^{17}\ kg/m^3

So, the density of the nucleus of an atom is 3.98\times 10^{17}\ kg/m^3.

(b) Density of iron, d'=7874\ kg/m^3

Taking ratio of the density of nucleus of an atom and the density of iron as :

\dfrac{d}{d'}=\dfrac{3.98\times 10^{17}}{7874}

\dfrac{d}{d'}=5.05\times 10^{13}

d=5.05\times 10^{13}\ d'

So, the density of the nucleus of an atom is 5.05\times 10^{13} times greater than the density of iron. Hence, this is the required solution.

7 0
4 years ago
Two rockets are fired at each other with initial velocities of 150m/s150m/s and are 6000m6000m apart. The first rocket is accele
Nataliya [291]

Answer:

3469.788 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration

First rocket

s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 5\times t^2\\\Rightarrow s=150t+2.5t^2\ m

Second rocket

s=ut+\frac{1}{2}at^2\\\Rightarrow s=150\times t+\frac{1}{2}\times 15\times t^2\\\Rightarrow s=150t+7.5t^2\ m

When this will collide the total distance they would have covered would be 6000 m.

6000=150t+2.5t^2+150t+7.5t^2\\\Rightarrow 6000=300t+10t^2\\\Rightarrow 10t^2+300t-6000=0

t=5\left(\sqrt{33}-3\right),\:t=-5\left(3+\sqrt{33}\right)\\\Rightarrow t=13.72, -43.72

Hence at 13.72 seconds they will collide assuming they are launched at the same time.

s=150t+7.5t^2\\\Rightarrow s=150\times 13.72+7.5\times 13.72^2\\\Rightarrow s=3469.788\ m

The second rocket would have gone 3469.788 m when they collide

7 0
4 years ago
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