It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..

8 0

2 years ago

The de broglie wavelength of an electron with a velocity of 6.00 × 106 m/s is ________ m. The mass of the electron is 9.11 × 10-

WINSTONCH [101]

Answer: $1.212(10)^{-10} m$

Explanation:

The de Broglie wavelength $\lambda$ is given by the following formula:

$\lambda=\frac{h}{p}$ (1)

Where:

$h=6.626(10)^{-34}\frac{m^{2}kg}{s}$ is the Planck constant

$p$ is the momentum of the atom, which is given by:

$p=m_{e}v$ (2)

Where:

$m_{e}=9.11(10)^{-28}g=9.11(10)^{-31}kg$ is the mass of the electron

$v=6(10)^{6}m/s$ is the velocity of the electron

This means equation (2) can be written as:

$p=(9.11(10)^{-31}kg)(6(10)^{6}m/s)$ (3)

Substituting (3) in (1):

$\lambda=\frac{6.626(10)^{-34}\frac{m^{2}kg}{s}}{(9.11(10)^{-31}kg)(6(10)^{6}m/s)}$ (4)

Now, we only have to find $\lambda$ :

$\lambda=1.2122(10)^{-10} m$ >>> This is the de Broglie wavelength of the electron

8 0

2 years ago

The maximum allowable resistance for an underwater cable is one hundredth of an ohm per

Studentka2010 [4]

Answer:

A = 1.54 x 10⁻⁵ m² = 15.4 mm²

Explanation:

The resistance of a wire can be given by the following formula:

$R = \frac{\rho L}{A}\\\\A = \frac{\rho L}{R} = \frac{\rho}{\frac{R}{L} }$

where,

A = smallest cross-sectional area = ?

ρ = resistivity of copper = 1.54 x 10⁻⁸ Ωm

$\frac{R}{L}$ = resistance per unit length of wire = 0.001 Ω/m

Therefore,

$A = \frac{1.54\ x\ 10^{-8}\ \Omega m}{0.001\ \Omega/m}$

6 0

2 years ago