# Make the internet a better place to learn

2

## What does wave refraction cause?

Mark C.
Featured 5 days ago

Generally, a change in both wavelength and velocity of the wave.

#### Explanation:

If we look at the wave equation we can get an algebraic understanding of it: $v = f \times \lambda$ where $\lambda$ is the wavelength. Clearly if v alters, either f or $\lambda$ must change. As the frequency is determined by the source of the waves, it remains constant. Due to conservation of momentum the direction alters (provided the waves are not at ${90}^{\circ}$)

Another way of understanding this is to consider the crests as being lines of soldiers - an analogy I’ve use during many times. The soldiers are marching at an angle (say ${45}^{\circ}$) from the concrete parade ground onto,the grass lawn. As they move onto the lawn each soldier slows down, this means the distance between those slower soldiers is reduced.

Look at the diagram below (imagining the wave crests as the soldiers) and you’ll see that because velocity falls, the wavelength changes too (and the direction of travel.) That means frequency must remain constant -

1

## Explain thermal expansion of water?

Mark C.
Featured 1 week ago

Like all materials, an increase in temperature (average non-translational kinetic energy of the particles) will cause them to increase the average distance between particles.

#### Explanation:

Particles have temporary forces between them due to mutual coulombic repulsion of the electron ‘clouds’ that surround them. As temperature rises the oscillation of the mass (effectively the nucleus) in the system becomes more violent hence occupy a larger effective volume.

So far, so normal, but water is unusual as a liquid because of the polarity (and relatively small size) of the molecule. This means the forces between particles can also include hydrogen bonding (still weak, temporary but a bond with both attractive and repulsive effects.) This means that water’s expansivity is unusually variable with temperature and reaches a minimum not at the freezing point, but at ${4}^{\circ}$C where it is most dense.

This provides some more detail on the data, but the anomalous effects in water are better explained here.

It truly is the weirdest fluid - but the one essential ingredient for life as far as we know.

1

## You push a crate full of nooks across the floor at a constant speed of 0.5 meter per second. You then remove some of the books and push exactly the same as you did before. How does the crate’s motion differ, if at all?

Gió
Featured 1 week ago

I would say that the crate now is accelerated.

#### Explanation:

During the first part you exert the exact force $\vec{F}$ to maintain the box in uniform motion, i.e., constant velocity (acceleration$= 0$). In doing so you are winning against kinetic friction ${f}_{k}$ but just! So your force and friction balance (almost) so that there is no acceleration but only constant velocity.

When you remove some books the box becomes lighter so that the normal reaction $N$ decreases as well reducing the contribution of kinetic friction. Now if you push with the same force as before you face less friction so that you get a component of force that gives you acceleration!

1

## The rise of a line in a distance-versus-time graph is 900 m aand the run is 3 min. What is the slope of the line?

Nimo N.
Featured 1 week ago

#### Explanation:

Anna found that the distance v.s. time graph has a slope of 900.

As you are likely aware, the general form for a linear equation is:
$y = m \cdot x + b$, where
y is the dependent variable (graphed on the vertical axis),
m is the slope of the line,
x is the independent variable, (graphed on the horizontal axis), and
b is the y-intercept, (the y-coordinate where the line crosses the vertical axis).

Then, an equation for your line would be:
$d = 300 \cdot t + b$.

However, the value of b most likely would be zero - if the movement started at rest, so the formula would likely be:
$d = 300 \cdot t$.

As to the choice of which of the two quantities, d or t, is the dependent variable, the formula for the slope tells the tale:
The rise for the slope is the numerator, representing a change in the vertical direction. So d is the dependent variable in the equation.

1

## Derive a relation between Torque and Moment of Inertia?

MetaPhysik
Featured 1 week ago

$I = {\tau}_{\text{net}} / \alpha$

#### Explanation:

Momentum of inertia is similar to mass in Newton's 2nd Law.

Mass is the inertia of an object in response to an unbalanced (net) force that causes the object to accelerate translationally.

$m = {F}_{\text{net}} / a$

Momentum of the inertia is the rotational inertia in response to an unbalanced (net) torque that causes the object to accelerate its rotation on an axis (i.e., angular acceleration).

$I = {\tau}_{\text{net}} / \alpha$

So it fact, it's like Newton's 2nd law for angular motion.

${F}_{\text{net}} = m a - -$ Newton's Second Law for translational motions.

${\tau}_{\text{net}} = I \alpha - -$Newton's Second Law for angular motions.

However, $I$ is more complex, it depends on mass of the object and the geometric shape rotating on its axis of rotation.

2

## The energy found in food that you eat is known as?

MetaPhysik
Featured 1 week ago

It's known as a Calorie, which is then stored as potential energy in the form of chemical bonds or chemical energy

#### Explanation:

The food that you eat is eventually broken down into three main typical of chemicals: carbohydrates, proteins(amino acids), and fats. The energy is stored in the chemical bonds of the these molecules. These molecules are eventually processed to make the energy molecules called ATP that provides energy that the body needs.

So food energy is also called chemical energy. Fundamentally, chemical bonds are electrical in origin, as molecules are made of nuclei with positive charge and electrons with negative charges. When chemical reactions happened, nuclei and electrons are reshuffled, giving up the electric potential energy stored in these molecules.

The pathways food into energy is best illustrated here.

2

## What are Gs? And why do they increase with speed?

Mark C.
Featured 1 week ago

Firstly, the previous answer (Steve’s) is correct, I just wanted to add something on the end in relation to speed.

#### Explanation:

For this to make sense you (probably) need to alter your understanding of how forces work a little.

When an object turns a corner, or moves in circular motion there only needs to be one force acting (there may be more, but only 1 is required) and that force is centripetal i.e. acting towards the centre of the circle. A good example is the orbit of the moon (for this argument, a circle) around the earth. The only force acting is gravity and that acts towards the earth (centre of mass, close to the earth, but again we’ll ignore that for now) causing it to accelerate.

Now, to develop this idea we need to clarify what we mean by an acceleration, and Newton’s 1st law. An object can have a speed in a straight line, but this tells you nothing about the direction of travel. For that, you need a concept called velocity which describes both the speed and direction (in Physics we call these vectors.)

Acceleration is defined as a change in velocity per unit time, and that allows either the speed or the direction to alter as time goes by. In other words, turning a corner at constant speed means you are accelerating (because the direction changes) and thus requires a net force. Newton’s 1st law states that an object will continue in uniform motion (a straight line at the same speed) unless acted upon by an external force. So if there is no net force, an object keeps going in the same direction at the same speed.

Secondly, why does this not tally with your experience of being in a car on a roundabout? You seem to feel a force outwards, that people call centrifugal, but is in fact just the effect of mass not wanting to change where it is (or where it is going) called inertia. Masses, in effect, resist acceleration, including turning their direction, and the greater the mass, the more resistance there is.

Now we can finally relate the question (say an F1 car taking a high speed turn) to the so called “g force” experienced by the driver. As the mass requires a force to turn it, given by $F = \frac{m {v}^{2}}{r}$. This shows that as speed increases the force required to turn the mass increases exponentially (as a squared term) and as the radius reduces (the turn on the track becomes ‘tighter’) so the force also rises. This required force (on a level track) comes from the friction between tyre and track and points inwards (centripetally) towards the inside of the turn.

The inertia of the mass (driver) creates an apparent force pointing outwards, but we now know this is just mass being “stubborn” and not willing to accelerate. This “force” is described in units of “g” or “G” to compare it to the force we experience as gravity, in other words in units of 9.81 Newtons for every kilogram of mass.

Sorry for the long answer, but it isn’t simple! Any clearer?

2

## A 6 kg shell is fired with a 30° angle,v= 40 m/s. At the highest point it devides into 2 parts,one 2 kg and the other 4 that move in the horizontal line and 2 kg hits the ground at the start point.where the 4 kg piece hits the ground?

ali ergin
Featured 5 days ago

$\text{please check the math operations.}$

#### Explanation:

• We will use the principle of momentum conservation to solve the problem.
• In the solution of the problem, we must assume that the conditions are ideal and that there is no energy loss.
• Suppose that C is the maximum height. At this point, the object instantaneously has only a horizontal velocity component.
• When the object is at point C, it only has instant horizontal momentum.

• momentum vector and direction are seen below.

$\vec{P} = m . {\vec{v}}_{x}$

$P = m v \cdot \cos \theta = 6 \cdot 40 \cdot 0.866 = 207.85 \text{ } k g \cdot m {s}^{-} 1$

• we have to rethink the momentum of each piece when the bullet is divided into two.
• the vectorial sum of the vertical momentum of the parts is zero.Because the bullet only had horizontal momentum.
• Both parts of the bullet must have horizontal momentum.
• The momentum vectors for the part are given below.

$\vec{P} = {\vec{P}}_{1} + {\vec{P}}_{2}$

• Because the first part is returned to its original position, the velocity of this part is equal to and opposite to the horizontal velocity component.

${P}_{1} = {m}_{1} v \cos \theta = 2 \cdot 40 \cdot 0.866 = 69.28 \text{ } k g . m {s}^{-} 1$

• We can find the momentum of the second part.

$207.85 = - 69.28 + {P}_{2}$

${P}_{2} = 207.85 + 69.28 = 277.13 \text{ } k g \cdot m {s}^{-} 1$

• Now we can find the velocity of the second part.

${P}_{2} = {m}_{2} {v}_{2}$

$277.13 = 4 {v}_{2}$

${v}_{2} = \frac{277.13}{4} = 69.28 \text{ } m {s}^{-} 1$

• Objects thrown horizontally from the same height reach equally long.

$t = \frac{{v}_{i} \cdot \sin \theta}{g} = \frac{40 \cdot 0.5}{9.81} = 2.04 \text{ } \sec$

$x = {v}_{i} \cdot t \cdot \cos \theta = 40 \cdot 2.04 \cdot 0.866 = 76.67 \text{ } m$

x_1=v_2*t=69.28*2.04=141 .33" "m

$A M = x + {x}_{1} = 76.67 + 141.33 = 218 \text{ } m$

• falls 218 meters from the point where it is thrown.

1

## What is difference between Newton law of gravitation and columbs law?

Mark C.
Featured 2 days 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 $k \approx 8.99 \times {10}^{8} N {m}^{2} {C}^{-} 2$

1

## What is the bleeder current??

A08
Featured 2 days ago

For propose of safety, a bleeder resistor is connected in parallel with the output of a high-voltage power supply circuit. This resistor discharges the electric charge stored in the filter capacitor of the power supply when the load is switched off. Refer to the figure below.
Bleeder resistor provides the following safety features in the electronic power supplies

1. Acts as a fuse for excess voltage
2. Discharges the filter capacitors
3. Removes shock hazards to operating personal from induction coils
4. Eliminates ground-loop current

The $R C$ discharge current flowing through the bleeder resistor, when the equipment is in off state, is called bleeder current. Value of bleeder resistor will depend upon $R C$ time constant and how quickly one wants the filter capacitor to be discharged.