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3 years ago

The study of the human mind and how it behaves in certain situations is called

Physics

2 answers:

Bingel [31]3 years ago

8 0

Answer:

Psychology

Explanation:

The study of the human mind and how it behaves in certain situations is called Psychology.

Ilia_Sergeevich [38]3 years ago

3 0

Psychology because it’s the study of mind and behavior

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A charge of -8.00 nC is spread uniformly over the surface of one face of a nonconducting disk of radius 1.05 cm.

gavmur [86]

Answer:

(a) $E = -1.02 \times 10^5~N/C$

(b) $E = -9.7 \times 10^4~N/C$

Explanation:

(a) The electric field for a point charge is given by the following formula:

$\vec{E} = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}\^r$

Since this formula is valid for point charges, we have to choose an infinitesimal area, da, from the disk. Then we will calculate the E-field (dE) created by this small area using the above formula, then we will integrate over the entire disk to find the E-field created by the disk.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{(\sqrt{z^2 + r^2})^2}$

Here, z = 0.025 m. And r is the distance of the infinitesimal area from the axis. dQ is the charge of the small area, and should be written in terms of the given variables.

In cylindrical coordinates, da = r dr dθ. So,

$\frac{Q}{\pi R^2} = \frac{dQ}{da}\\\frac{Q}{\pi R^2} = \frac{dQ}{rdrd\theta}\\dQ = \frac{Qrdrd\theta}{\pi R^2}$

Hence, dE is now:

$dE = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}$

The surface integral over the disk can now be taken, but there is one more thing to be considered. This dE is a vector quantity, and it needs to be separated its components.

It has two components, one in the vertical direction and another in the horizontal direction. By symmetry, the horizontal components cancel out each other in the end (since it is a disk, each horizontal vector has an equal but opposite counterpart), so only the vertical component should be considered.

Let us denote the angle between dE and the horizontal axis as α. This angle can be found by the geometry of the triangle formed by dE, vertical axis of the disk, and horizontal plane. So,

$\sin(\alpha) = \frac{z}{\sqrt{z^2 + r^2}}$

Therefore, vertical component of dE now becomes

$dE_z = \frac{1}{4\pi\epsilon_0}\frac{Q}{\pi R^2}\frac{rdrd\theta}{z^2 + r^2}\frac{z}{\sqrt{z^2+r^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 \int\limits^R_0 {\frac{rdrd\theta}{(z^2+r^2)^{3/2}}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2} 2\pi(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}})$

Substituting the parameters, z = 0.025 m, Q = - 8 x 10^(-9) C, and R = 0.0105 m, yields the final result:

$E_z = \frac{1}{2\epsilon_0}\frac{Qz}{\pi R^2}(\frac{1}{z} - \frac{1}{\sqrt{z^2+R^2}}) = -1.02 \times 10^5~N/C$

(b) We will have a similar approach, but a simpler integral.

$dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{z^2 + R^2}\\\frac{Q}{2\pi R} = \frac{dQ}{Rd\theta}\\dQ = \frac{Qd\theta}{2\pi}\\dE = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qd\theta}{2\pi(z^2 + R^2)}\frac{z}{\sqrt{z^2+R^2}} = \frac{1}{4\pi\epsilon_0}\frac{Qzd\theta}{2\pi(z^2 + R^2)^{3/2}}\\E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{Qz}{2\pi(z^2 + R^2)^{3/2}}2\pi$

$E_z = \frac{1}{4\pi\epsilon_0}\frac{Qz}{(z^2 + R^2)^{3/2}} = -9.07\times 10^4~N/C$

Note that, in this case the source object is a one dimensional hoop rather than a two dimensional disk.

3 0

3 years ago

A bike race against the clock takes place on a straight road. Yan drives at 37 km / h and he starts the course 30s before Christ

joja [24]

Given data:

Yan speed;

$u_1=37\text{ km/h}$

Christopher speed;

$u_2=38.9\text{ km/h}$

Christophe starts 30 s later than Yan. Therefore, Christophe takes 30 s less than Yan to reach the same distance.

Part (A)

The distance is given as,

$d=ut$

Let both Yan and Christophe meet at d distance from the start position. Therefore,

$u_1t=u_2(t-30)$

Substituting all known values,

$\begin{gathered} (37\text{ km/h})t=(38.9\text{ km/h})\times(t-30) \\ \frac{(37\text{ km/h})}{(38.9\text{ km/h})}=\frac{(t-30)}{t} \\ 0.95=1-\frac{30}{t} \\ \frac{30}{t}=1-0.95 \\ \frac{30}{t}=0.05 \\ t=\frac{30}{0.05} \\ t=600\text{ s} \end{gathered}$

Therefore, 600 s after Yan's departure Christophe will join him.

Part (B)

The distance is given as,

$d=u_1t$

Substituting all known values,

$\begin{gathered} d=(37\text{ km/h})\times(600\text{ s}) \\ =(37\text{ km/h})\times(600\text{ s})\times(\frac{1\text{ hr}}{3600\text{ s}}) \\ \approx6.17\text{ km} \end{gathered}$

Therefore, Christophe joins Yan after 6.17 km from the start.

3 0

1 year ago

Radiation from the Sun reaching Earth (just outside the atmosphere) has an intensity of 1.4 kW/m2. (a) Assuming that Earth (and

baherus [9]

To solve the problem we will require the concept of Force as a definition of pressure and Area, and the concept of light pressure itself determined by the relationship between intensity and the speed at which light travels. We will match the terms and find the desired force value,

$F = PA$