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Question #cb233

Nathan L.
Featured 1 month ago

In idealized projectile motion, a particle is at its maximum height when its instantaneous $y$-velocity is equal to zero.

How come?

Well, for projectile motion, after the particle is launched, the only factor affecting its motion in any way (ideally) is the downward gravitational force exerted by the Earth, which gives the object a constant downward acceleration of magnitude $9.81$ ${\text{m/s}}^{2}$.

With that being said, if the particle is thrown upward, it has an initial $y$-velocity that is positive (taking positive direction to be upward), and this velocity is constantly being changed due to the gravitational acceleration.

Since the acceleration is directed opposite to the initial $y$-velocity, the velocity will decrease at a constant rate, and will intuitively become $0$ at some point (and then negative afterward).

You might recall that the slope on a position vs. time graph at any point is the instantaneous velocity at that point. For a projectile thrown upward, its trajectory will resemble that of an inverse parabola like one shown here:

Notice that the slope of a projectile's path is positive until we reach the maximum height. What has happened is the initial velocity is constantly being decreased (i.e. the slope of the trajectory is decreasing) due to the negative acceleration, and at the maximum on the graph, the slope there (and thus the instantaneous velocity) is $0$, and therefore this is the maximum height, because after this, the motion (and velocity) is directed downward.

Question #5d31a

Aritra G.
Featured 2 months ago

Diffraction refers to the phenomena that occurs when a wave encounters an obstacle or a slit. It is defined as the bending of light around the corners of an obstacle or aperture into the region of geometrical shadow of the obstacle.

Since, light shows wave like properties, it shows diffraction.

If the size of the obstacle (diffracting element) is comparable with the wavelength of light, diffraction effects are significant and have been observed. There is a bending of light which leads to partial illumination in the region of geometrical shadow of the obstacle.
However, accounting to very small wavelengths of visible radiation ${10}^{- 7} m$, diffraction effects are insignificant in daily life.

Diffraction of light is broadly classified as :

1) Fresnel diffraction : This happens where wave-fronts encountered are not plane wave-fronts. The source and the screen are at finite distances from the diffracting element.

2) Fraunhofer diffraction : Both the source of light and the screen are at infinite (sufficiently large) distances from the diffraction slit or obstacle, The wave-fronts encountered are plane ones. The light is parallel to the diffracting slit!

In Fraunhofer diffraction, a single slit or a number of slits are employed with light being parallel to the slit(s). This can be done by Schuster's method for adjusting a spectrometer.

The pattern so obtained on the screen shows successive bright and dark bands with varying width and intensity of bright bands falling off rapidly as shown.

The dark regions are known as the diffraction minima while the bright ones are the maxima.

The central bright region is the principal maxima. The dark regions immediately on both sides are called the first diffraction minima, the successive bright regions constitute secondary maxima.

How could you find the density of an irregular object which floats in water?

A08
Featured 1 month ago

We know that density $\rho = \text{Mass"/"Volume"=m/V_"o}$ ....(1)

One of the methods to find volume of an irregular shaped object is with the help of water displacement method. Method becomes complicated if the objects floats.

For such an object we use the sinker method. A sinker is a weight tied to the floating object to hold it down and submerge it completely in graduated container. Following steps will help to find the density $\rho$ of the object.

1. Find mass $m$ of the object with the help of a Physical balance and record it.
2. Pour water in to the graduated container about one third of its depth and record the volume of water ${V}_{w}$
3. Tie the sinker with a thread and lower it gently in the container without splashing. Record the new volume ${V}_{1}$.
4. Attach the sinker to the object with a thread and lower both in the container gently, without splashing. Record the new volume ${V}_{2}$

Volume of sinker$= {V}_{1} - {V}_{w}$
Volume of sinker and object $= {V}_{2} - {V}_{w}$
$\therefore$ Volume of object ${V}_{\text{o}} = \left({V}_{2} - {V}_{w}\right) - \left({V}_{1} - {V}_{w}\right) = {V}_{2} - {V}_{1}$

Calculate $\rho$ with equation (1).

Explain thermal expansion of water?

Mark C.
Featured 1 month 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.

Question #0a763

MetaPhysik
Featured 1 month 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.

State the factors that influence gravity within the surface of the earth ?

1s2s2p
Featured 3 weeks ago

Your altitude and the position of the centre of gravity of the Earth.

Explanation:

The equation for $g$ on Earth is given by:
${g}_{E} = \frac{G {M}_{E}}{r} ^ 2$, where:

• ${g}_{E}$ = acceleration due to free fall on Earth ($m {s}^{-} 2$)
• $G$ = gravitational constant (#~6.67*10^-11Nm^2kg^-2#)
• ${M}_{E}$ = mass of the object (#~5.972*10^24kg#)
• $r$ = distance between the centre of gravities of the two objects ($m$)

Since $G$ and ${M}_{E}$ are constants $g \propto \frac{1}{r} ^ 2$

$r$ is possible to change even without you moving since many things like magma flow through the Earth which have very tiny changes in the position of the centre of gravity that will slightly change $r$.

Let's say you were 7000km away from the centre of gravity from the Earth:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{7 \cdot {10}^{6}} ^ 2 = 8.129 m {s}^{-} 2$

Now 5000km:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{5 \cdot {10}^{6}} ^ 2 = 15.93 m {s}^{-} 2$

$r$ is usually 6371km

$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{6371 \cdot {10}^{3}} ^ 2 = 9.813646787 m {s}^{-} 2$

But if $r$ was to get 1m less due to movements in the Earth's mantle (for example):
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{\left(6371 \cdot {10}^{3}\right) + 1} ^ 2 = 9.813643707 m {s}^{-} 2$

A 1m change has a slightly small change in the value for $g$

Also, $r$ could change if the ground was to become raised or lowered,movements of liquids like magma can raise and lower the ground, changing the distance between the person and the centre of gravity for the Earth (assuming that hasn't changed). If the ground was to sink by $1 m$ then $g$ will become:
$g = \frac{\left(6.67 \cdot {10}^{-} 11\right) \left(5.972 \cdot {10}^{24}\right)}{\left(6371 \cdot {10}^{3}\right) - 1} ^ 2 = 9.813649868 m {s}^{-} 2$ as you can see, $g$ has increased by a very small amount, but not enough to have any noticeable effect on us.

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