Dear friends, Please read our latest blog post for an important announcement about the website. ❤, The Socratic Team

0
Active contributors today

## Question #ba2ab

MetaPhysik
Featured 5 months ago

Use magnetic field.

#### Explanation:

If you shoot charge particles perpendicular to magnetic field that is pointing upward (from the floor to ceiling), positive charges get deflected to the right, and negative charges to the left. Those that carries no net charge (neutral), go straight through.

We just happened to call right deflecting particles positive, those in the opposite way negative. We could have name positive negative or vice versa. You can call positive yang and negative yin if you like. Their behaviors in magnetic field will be exactly the same.

Another way to tell positive of negative is to have a standard material that you already know its polarity. So those attract to it must have opposite charge, repel by it the same charge. See this example.

## Question #1fd99

Mark C.
Featured 5 months ago

They are similar, in that they both have an inverse square relationship with distance, but Newton’s ULG concerns the effect mass has on other mass, and Coulomb’s law does the same for charges.

#### Explanation:

Newton’s Universal Law of Gravitation shows that mass is (very weakly) attracted to other masses, but because the value of the constant in this relationship (G) is very small, the effect is negligible until we consider masses in the billions of kilograms (moons, planets, stars etc.)

Newton didn’t explain everything in gravity - he would not even attempt to explain why mass should do this (hypothesis non fingo) nor did he find a value for G (that was Cavendish in 1798.) There also remains (unsolved) the reason why there is only ever an attractive force between masses (expressed as a negative sign in the equation, so that the force, $\vec{F}$ is always in the opposite direction to the distance, $\vec{r}$.)

The equation that bears Coulomb’s name is very similar in form, but allows for both positive (repulsive) and negative (attractive) force vectors between charges, as charges can be both of the same sign in charge (repulsive forces) or dissimilar (attractive forces between opposing charges.) In effect, Coulomb’s law is a version of Gauss’s law.

There is generally a factor of ${10}^{18}$ or so in the difference between the two forces (assuming unit masses and charges) with electrostatic forces greater in size than gravitational ones, so the size of the constants dominates these effects.

For reference, the two equations (in scalar form) are, $F = - G \frac{{m}_{1} \times {m}_{2}}{r} ^ 2$ with $G \approx 6.67 \times {10}^{-} 11 N {m}^{2} k {g}^{-} 2$ and $F = k \frac{{q}_{1} \times {q}_{2}}{r} ^ 2$ with $k \approx 8.99 \times {10}^{8} N {m}^{2} {C}^{-} 2$

## What are the evidences in favor of the Nuclear Shell Model ?

Aritra G.
Featured 4 months ago

There are multiple experimentally determined facts that indicate the shell-like structure of the atomic nucleus. Some of them are listed below :

1) Nuclei with Z and N (or both) equal to 2, 8, 28, 20, 50, 82 and 126 are extra stable. These numbers are called the magic numbers and this stability variation is analogous to stability of 2, 8, 18, 32 etc electrons in electronic shells.

2) Nuclei with N = a magic number have smaller neutron absorption cross section as compared to their immediate neighbours indicating a more stable nucleus for those values of N.

3) The electric quadrapole moment is zero for a nucleus with N or Z equal to a magic number indicating the spherical symmetry analogous to that in closed shells of atoms.

4) Nuclear species with one neutron excess of a magic number readily emit a single neutron indicating that the extra neutron is loosely bound to the rest of the nucleus.

5) Nuclei with magic number of nucleons of a particular type have more number of stable isotopes and isotones.

## Question #ea004

Douglas K.
Featured 4 months ago

No. AC is much more efficient for transmission over long distances than DC, because AC allows you to use transformers step up the voltage for transmission and step down the voltage at the load.

#### Explanation:

Let's consider a simple case that I will deliberately make a "best case" scenario.

An industrial customer was a facility 10 km away from your power plant that requires $1 \times {10}^{6} \text{ W}$ at $440 \text{ V}$ and does not care whether you supply it do him using AC or DC.

Your power lines have the incredibly low resistance of $0.1 \Omega \text{/km}$

If you choose DC, the current in the power lines must be the same as the current in the load, we can use $P = V I$ to compute the current in the load:

#1 xx 10^6" W" = (440" V")I#

$I \approx 2200 \text{ A}$

Compute the resistance of the power lines

$R = 2 \left(10 \text{ km")(0.1Omega"/km}\right)$

$R = 2 \Omega$

NOTE: The factor 2 is required, because you need a 2 wires to make a complete circuit, one out and one back.

We can use $P = {I}^{2} R$ to compute the power lost in the power lines:

#P_"lost" = (2200" A")^2(2Omega)#

${P}_{\text{lost"~~ 9.6xx10^6" W}}$

Using DC, you loose 9.6 times the power that you deliver.

Lets consider the AC solution where you step of the voltage to $250000 \text{ VAC}$ for transmission.

It is easy to obtain a transformer with an efficiency of #95%# so we shall use that number; this makes the power in the transmission lines at the primary of the step down transformer $1.1 \times {10}^{6} \text{ W}$

At $250000 \text{ VAC}$, the current in the transmission lines is:

$I = \left(1.1 \times {10}^{6} \text{ W")/(250000" VAC}\right)$

$I \approx 4.5 \text{ A}$

Compute the power lost in the same two transmission lines:

#P_"lost"= (4.5" A")^2(2Omega)#

${P}_{\text{lost" = 40.5" W}} \leftarrow$ this is insignificant

Compute the power lost in the step up transformer:

#P_"generated" = (1.1xx10^6" W")/0.95#

${P}_{\text{generated" ~~ 1.2xx10^6" W}}$

Using the step up / step down transformer, you loose approximately #20%# of the power generated; this is far better than #960%# for DC.

## A projectile launcher is set on a table so that the ball becomes a projectile at a height of 1.2 m above the floor. The mass of the ball is 0.01 kg. A plunger is used to push the ball into the barrel of the launcher compressing a spring a distance of...?

A08
Featured 4 months ago

Let the initial velocity of the projectile be $= u$. Let the origin of coordinates be located at the point of projection.

Horizontal motion.
Distance traveled $= 4 \setminus m$ during time of flight $t$, with initial horizontal velocity $= u \cos {30}^{\circ}$

$\therefore u \cos {30}^{\circ} t = 4$
$\implies \frac{\sqrt{3}}{2} u t = 4$ .......(1)

Vertical motion.
To calculate the time of flight we use the kinematic expression

$s = u t + \frac{1}{2} a {t}^{2}$

Taking $g = 9.8 \setminus m {s}^{-} 2$, noting that gravity acts in the downwards direction and inserting given values we get

$- 1.2 = u \sin {30}^{\circ} t + \frac{1}{2} \left(- 9.8\right) {t}^{2}$
$\implies 4.9 {t}^{2} - 0.5 u t - 1.2 = 0$ ........(2)

Eliminating $u t$ from (2) with the help of (1) we get
$\implies 4.9 {t}^{2} - 0.5 \times \frac{8}{\sqrt{3}} - 1.2 = 0$
$\implies t = \pm 0.846 \setminus s$ ........(3)

Ignoring the $- v e$ root as time can not be negative. From (1) we get the value of $u$ as

$u = \frac{8}{\sqrt{3} \times 0.846} = 5.46 \setminus m {s}^{-} 1$ ......(4)

Now the kinetic energy of the projectile at the time of projection is provided by the potential energy of the compressed spring. We know that

$K {E}_{\text{projectile}} = \frac{1}{2} m {u}^{2}$ and
$P {E}_{\text{spring}} = \frac{1}{2} k {x}^{2}$
where $m$ is mass of the projectile, $k$ is the spring constant and $x$ is the compression of the spring.

Equating both we get

$\frac{1}{2} k {x}^{2} = \frac{1}{2} m {u}^{2}$
$\implies k = m {u}^{2} / {x}^{2}$

Inserting various values we get

$\implies k = 0.01 {u}^{2} / {\left(0.1\right)}^{2}$
$\implies k = {u}^{2}$

Using (4)

$k = 29.8 \setminus N {m}^{-} 1$

## Find an expression for the displacement?

A08
Featured 4 months ago

Given expression for velocity of a particle by

#v=(t−2)\ ms^-1#

Comparing with kinematic expression

$v = u + a t$ ...............(1)

we get

$u = - 2 \setminus m {s}^{-} 1 \mathmr{and} a = 1 \setminus m {s}^{-} 2$

(1). Kinematic expression for displacement is

$s = {s}_{0} + u t + \frac{1}{2} a {t}^{2}$

Inserting given conditions we get the expression for displacement $r$ as

$r = {r}_{0} + \left(- 2\right) t + \frac{1}{2} \left(1\right) {t}^{2}$
$r = {r}_{0} - 2 t + \frac{1}{2} {t}^{2}$

To find ${r}_{0}$, use the given condition at $t = 6 \setminus s$

$r \left(6\right) = 11 = {r}_{0} - 2 \left(6\right) + \frac{1}{2} {\left(6\right)}^{2}$
$\implies 11 = {r}_{0} - 12 + 18$
$\implies {r}_{0} = 5 \setminus m$

$\therefore$ Expression for $r$ becomes

$r = 5 - 2 t + \frac{1}{2} {t}^{2}$ ........(2)

(2). From (2)

$r \left(8\right) = 5 - 2 \times 8 + \frac{1}{2} {\left(8\right)}^{2}$
$\implies r \left(8\right) = 5 - 16 + 32$
$\implies r \left(8\right) = 21 \setminus m$

(3). From (2)

$53 = 5 - 2 t + \frac{1}{2} {t}^{2}$
$\implies {t}^{2} - 4 t - 96 = 0$

Solving the quadratic by split the middle term we get

$t = - 8 \mathmr{and} 12$,

Ignoring $- v e$ root as time can not be negative and inserting in (1) we get

$v \left(12\right) = - 2 + 1 \times 12 = 10 \setminus m {s}^{-} 1$

##### Questions
• · 52 minutes ago
• · 10 hours ago
• 12 hours ago · in Defining Force
• · 14 hours ago
• · 14 hours ago
• · 14 hours ago
• · 14 hours ago
• · 16 hours ago
• · 17 hours ago
• · 17 hours ago
• · Yesterday
• · Yesterday · in Conservation of Energy
• · Yesterday · in Defining Force
• · Yesterday · in Kepler's Laws
• · 2 days ago · in Electric Force