A solid metal sphere of radius a = 2.5 cm has a net charge Qin = - 3 nC (1 nC = 10-9C). The sphere is surrounded by a concentric

Furkat [3]

Answer: 630 V

Explanation: In order to solve this problem we have to consider the potential given by:

In the region 0<a<b

V(r)= -∫E*dr where E can be calculated by Gaussian law then E= k*qa/r^2

where qa=-3 nC

then

The V(r)=k*qa/r+C C is zero since V(r=∞)=0

Finally we have

V(r)= k*qa (1/r)-(1/b)

thus V(a)= k*qa (1/a)-(1/b)

Replacing the values V(a) = 630 V, as the solid metal sphere is a equipotential object thus at the center have the same value of V that in r=a ( 630 V).

4 0

1 year ago

A spring attached to the ceiling is stretched one foot by a four pound weight. The mass is set in motion by pulling down 2 feet

Marina CMI [18]

Given that,

Weight = 4 pound

$W=4\ lb$

Stretch = 2 feet

Let the force be F.

The elongation of the spring after the mass attached is

$x=2-1=1\ feet$

(a). We need to calculate the value of spring constant

Using Hooke's law

$F=kx$

$k=\dfrac{F}{x}$

Where, F = force

k = spring constant

x = elongation

Put the value into the formula

$k=\dfrac{4}{1}$

$k=4$

(b). We need to calculate the mass

Using the formula

$F=mg$

$m=\dfrac{F}{g}$

Where, F = force

g = acceleration due to gravity

Put the value into the formula

$m=\dfrac{4}{32}$

$m=\dfrac{1}{8}\ lb$

We need to calculate the natural frequency

Using formula of natural frequency

$\omega=\sqrt{\dfrac{k}{m}}$

Where, k = spring constant

m = mass

Put the value into the formula

$\omega=\sqrt{\dfrac{4}{\dfrac{1}{8}}}$

$\omega=\sqrt{32}$

$\omega=4\sqrt{2}$

(c). We need to write the differential equation

Using differential equation

$m\dfrac{d^2x}{dt^2}+kx=0$

Put the value in the equation

$\dfrac{1}{8}\dfrac{d^2x}{dt^2}+4x=0$

$\dfrac{d^2x}{dt^2}+32x=0$

(d). We need to find the solution for the position

Using auxiliary equation

$m^2+32=0$

$m=\pm i\sqrt{32}$

We know that,

The general equation is

$x(t)=A\cos(\sqrt{32t})+B\sin(\sqrt{32t})$

Using initial conditions

(I). $x(0)=2$

Then, $x(0)=A\cos(\sqrt{32\times0})+B\sin(\sqrt{32\times0})$

Put the value in equation

$2=A+0$

$A=2$ .....(I)

Now, on differentiating of general equation

$x'(t)=-\sqrt{32}A\sin(\sqrt{32t})+\sqrt{32}B\cos(\sqrt{32t})$

Using condition

(II). $x'(0)=0$

Then, $x'(0)=-\sqrt{32}A\sin(\sqrt{32\times0})+\sqrt{32}B\cos(\sqrt{32\times0})$

Put the value in the equation

$0=0+\sqrt{32}B$

So, B = 0

Now, put the value in general equation from equation (I) and (II)

So, The general solution is

$x(t)=2\cos\sqrt{32t}$

(e). We need to calculate the time

Using formula of time

$T=\dfrac{2\pi}{\omega}$

Put the value into the formula

$T=\dfrac{2\pi}{4\sqrt{2}}$

$T=1.11\ sec$

Hence, (a). The value of spring constant is 4.

(b). The natural frequency is 4√2.

(c). The differential equation is $\dfrac{d^2x}{dt^2}+32x=0$

(d). The solution for the position is $x(t)=2\cos\sqrt{32t}$

(e). The time period is 1.11 sec.

5 0

1 year ago

PROBLEMA DI FISICA:il volume di un attizzatoio da camino in ferro, a temperatura ambiente (20°C), è 12,2 cm cubi. A contatto con

lesantik [10]

Responder:

20.3 ° C

Explicación:

<u>Según la ley de Charles</u>: <em>cuando la presión sobre una muestra de gas seco se mantiene constante, la temperatura y el volumen estarán en proporción directa. </em>

Paso uno:

datos dados

Temperatura T1 = 20 ° C

Temperatura T2 =?

Volumen V1 = 12.2 cm ^ 3

Volumen V2 = 12.4 cm ^ 3

Aplicar la relación temperatura y volumen

$\frac {V_{1}}{T_{1}}=\frac {V_{2}}{T_{2}}$