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Physics

What is the unit of force in the SI system?

The unit of force in the SI system is the newton. The symbol for the newton is N. The newton is what is referred to as a derived unit. Derived units are composed of a combination of the SI fundamental units. There are seven fundamental SI units: $\hspace{2em}$ metre (m) used for measuring length\ $\hspace{2em}$ second (s) used for measuring time\ $\hspace{2em}$ kilogram (kg) used for measuring mass\ $\hspace{2em}$ kelvin (K) used for measuring temperature\ $\hspace{2em}$ ampere (A) used for measuring electric current\ $\hspace{2em}$ mol (mol) used for measuring quantities\ $\hspace{2em}$ candela (cd) used for measuring luminous intensity Using our known physics equations, we can determine how to express the newton in fundamental SI units. Starting with Newton’s First Law, $\hspace{2em}\ F = ma$ We can substitute the units for the variables to get $\hspace{2em}$ 1 N = (1 kg)(1 m s$^{-2}$) Therefore the unit kg m s$^{-2}$ is equivalent to N

Physics

What is the formula to calculate absolute uncertainty?

Most numerical values in science are measurements, and every measurement has an uncertainty. Scientists communicate the uncertainty of their measurements to show their level of confidence in their measurements. If the length of a ramp had been measured to be 80 cm, the absolute uncertainty of the measurement would likely be ± 1 cm. For a time measurement taken by a person with a stopwatch, the absolute uncertainty is normally ± 0.2 s. These are examples of absolute uncertainty. If a series of measurements is being added together or subtracted, the absolute uncertainty of the calculated value is found from the sum of the individual uncertainties on the measurements. For example, consider the addition of the following lengths and their uncertainties. $\hspace{2em}$(2.0 ± 0.1) cm + (8.2 ± 0.2) cm + (2.5 ± 0.5) cm The result will be the sum of the lengths, and the absolute uncertainty will be the sum of the individual uncertainties: $\hspace{2em}$Length = 2.0 + 8.2 + 2.5 = 12.7 cm $\hspace{2em}$Absolute uncertainty = 0.1 + 0.2 + 0.5 = 0.8 cm $\hspace{2em}$The final answer is expressed as 12.7 cm ± 0.8 cm. If two measurements are being multiplied or divided, then the calculation of the absolute uncertainty on the result is more complex. During multiplication or division operations, the relative or fractional uncertainties need to be added together. To do this, the absolute uncertainties on the original measurements must first be converted into fractional uncertainties, and then the fractional uncertainty of the result must be converted back into an absolute uncertainty. Consider the following calculation of the density of an object. The mass m and volume V of the object are measured to be $\hspace{2em}m$ = 4.5 g ± 0.2 g $\hspace{2em}V$ = 2.5 cm$^3$ ± 0.5 cm$^3$ The density can be found from the formula $\hspace{2em}\rho=\dfrac{m}{V}$ Giving $\hspace{2em}\rho=\dfrac{4.5\ \ce{g}}{2.5\ \ce{cm}^{3} }=1.8 $ g cm$^{-3}$ Now we need to find the absolute uncertainty on the result. First we find the fractional uncertainties on the mass and the volume: $\hspace{2em}$ $\dfrac{\Delta{m}}{m}=\dfrac{0.2}{4.5} \hspace{3em} \ce{and} \hspace{3em} \dfrac{\Delta V}{V} =\dfrac{0.5}{2.5}$ Because we are dividing the terms, we add the fractional uncertainties $\hspace{2em}$ $\dfrac{\Delta{m}}{m} + \dfrac{\Delta V}{V} = \dfrac{0.2}{4.5} + \dfrac{0.5}{2.5} = 0.24$ Knowing that $\hspace{2em}$ $\dfrac{\Delta \rho}{\rho}=0.24$ (this can also be expressed as 24%) We can now solve for ∆p: $\hspace{2em}$ $\Delta \rho = (0.24)\rho =(0.24)(1.8)=0.4\ \ce{g cm}^{-3}$ Our final value for density, with its absolute uncertainty, becomes $\hspace{2em}$ $\rho =1.8\ \ce{g cm}^{-3} ± 0.4\ \ce{g cm}^{-3}$

Physics

How to find total mechanical energy?

Mechanical energy is defined as the sum of an object’s energies due to its motion and position. The amount of mechanical energy an object possesses determines the amount of work it can do on other objects, or in other words, the amount of energy it is capable of transferring to other objects. Given the above definition, we can be more specific about the types of energy that contribute to the total mechanical energy of an object: - Kinetic energy $E_k$ is the energy associated with an object’s motion and is given by the formula $E_k = \dfrac{1}{2}mv^2$ where $m$ is the mass of the object and $v$ is its speed. Note that this is the macroscopic energy of motion of the particle, and does not include the microscopic energy of the movement of its particles which contributes to the object's internal energy. The energy associated with an object’s position is called its potential energy. Two main types of potential energy will contribute to mechanical energy: - Gravitational potential energy $E_p$ is the energy of an object due to its position in a gravitational field. For an object at a height $h$ above a reference point, its gravitational potential energy is calculated with the formula $E_p = mgh$, where $m$ is the mass of the object and $g$ is the gravitational field strength. At the Earth’s surface, the value for $g$ is 9.8 N kg$^{-1}$. - Elastic potential energy $E_H$ is is the energy stored due to the deformation of an elastic object. Work can be done in changing the shape of an object, for example stretching or compressing a string, and energy is stored in the object as a result. This stored energy is released when the object returns to its original shape. The standard formula for elastic potential energy is derived from Hooke's Law: $E_H = \dfrac{1}{2}kx^2$. Where $k$ is the spring constant and $x$ is the displacement from the equilibrium position. Another potential energy that can be considered to contribute to total mechanical energy is electric potential energy. Like gravitational potential energy, a charged object will have stored energy due to its position in an electric field. Because the idea of mechanical energy is normally applied to larger objects and not small charged particles, we will ignore it here. Having discussed the different types of energy that contribute to the mechanical energy of an object, we can create a formula for total mechanical energy. In words, the formula is $\hspace{3em} $ Mechanical Energy = Kinetic Energy + Potential Energy or more specifically $\hspace{3em} $ Mechanical Energy = Kinetic Energy + Gravitational Potential energy + Elastic Potential Energy In the form of an equation, the total mechanical energy can be expressed as $\hspace{3em} E_{tot}=E_k + E_p + E_H$ or $\hspace{3em} E_{tot}=\dfrac{1}{2}mv^2+ mgh+ \dfrac{1}{2}kx^2$

Physics

What is the relationship between frequency and period?

Oscillations are repeating motions that occur at regular time intervals. The time for one complete cycle of the motion before it repeats is called the period. For example, the period of the Earth orbiting the Sun is approximately 365 days, and the period of the Earth’s rotation around its axis is one day. If a person is playing jump rope, the period of the rope’s turns could be approximately two seconds, or for a spinning ceiling fan, it might take just half a second for a complete cycle. The SI symbol for period is $\textit{T}$. The examples here show that period is a time measurement per cycle. The concept of a period can also be written as an equation: $\hspace{3em}$ $\ce{period}=\dfrac{\ce{time}}{\ce{cycle}}$ Frequency is a value that tells us the number of cycles of a repeating motion that occur in a given amount of time. The symbol for frequency is $f$. If we say that a person has a heart rate of 55 beats per minute, we are stating the frequency of their heartbeat. Vinyl records used for playing music typically rotate at 33 rpm or revolutions per minute, which is also a measure of frequency. We can say that the frequency of Earth’s rotation is one rotation per day. Notice that all of the examples above involve a number of cycles per unit time. The definition of frequency can also be written as an equation: $\hspace{3em}$ $\ce{frequency}=\dfrac{\ce{cycles}}{\ce{time}}$ The relationship between frequency and period is that they are inverses of one another. This relationship can be expressed through the equations: $\hspace{3em}$ $f=\dfrac{1}{T} \hspace{3em}$and $ \hspace{3em}T=\dfrac{1}{f}$ We can show how this equation connects the definitions of period and frequency from the first two paragraphs. $\hspace{3em}$ $\ce{period}=\dfrac{\ce{time}}{\ce{cycle}}$ $\hspace{3em}$ $\ce{frequency}=\dfrac{1}{\ce{period}}=\dfrac{1}{\frac{\ce{time}}{\ce{cycle}}}=\dfrac{\ce{cycle}}{\ce{time}}$ Note that in IB Standard Level Physics and IB Higher Level Physics, period is normally given in the units of seconds (s), and frequency has an SI unit of hertz (Hz). Through the relationship between frequency and period, we can explore the relationship between these two units. $\hspace{3em}$ $f=\dfrac{1}{T} \hspace{3em}$therefore$\hspace{3em}\ce{Hz}=\dfrac{1}{\ce{s}} =\ce{s}^{-1}$ As an example, consider an object that oscillates with a frequency of 5 Hz. If we want to find the period, we can apply the equation $\hspace{3em}$ $T=\dfrac{1}{f}=\dfrac{1}{5\ \ce{Hz}}=$ 0.2 s Therefore, it takes 0.2 s for the object to complete one oscillation. Waves are generated by an oscillating source. The source causes a disturbance in the medium, resulting in the particles in that medium undergoing oscillations. The frequency of a wave is equal to the frequency of these individual oscillations of its particles. The period of the wave can be found from the same inverse relationship as before, $\hspace{3em}$ $T=\dfrac{1}{f}$

Physics

Why is specific heat important?

Specific heat is defined as the amount of thermal energy required to raise the temperature of a unit mass of a substance by one degree of temperature. It is important because it determines the amount of energy that needs to be added or removed to heat up or cool down a substance or an object. The IB Physics data booklet formula involving specific heat is $\hspace{2em}$ $Q = mc\Delta T$ Where $\hspace{2em}$ $Q$ is the heat required in J\ $\hspace{2em}$ $m$ is the mass of the sample in kg\ $\hspace{2em}$ $c$ is the specific heat capacity of the substance \ $\hspace{2em}$ $\Delta T$ is the change in temperature in k. This formula can be rearranged to solve for specific heat capacity to get $\hspace{2em}$ $c = \dfrac{Q}{m\Delta T}$ We can see from the formula that the units on $c$ will be J kg$^{-1}\ ^o$C$^{-1}$. Substances with high values of specific heat require more energy for a given change in temperature than substances with a lower value for specific heat. Water is an example of a substance with a high specific heat, with $c$ equal to 4186 J kg$^{-1}\ ^o$C$^{-1}$. This relatively high value means that a significant amount of energy is required to raise water to its boiling point. It also means that a lot of energy is released when water cools down again. Some older heating systems in houses circulate hot water to deliver heat to the rooms. Water’s high carrying capacity for thermal energy makes it ideal for this use. Another example of where water’s high specific heat plays an important role is in weather. Although the land may heat up and cool down relatively quickly as the air temperature changes, bodies of water will take much more time for their temperature to change. In summer, the water will stay cooler than the air and help moderate extreme heat. Similarly, in winter bodies of water can help moderate very cold air temperatures. Because of our understanding of specific heat, we can calculate the heat capacity of objects. This is the amount of energy required to raise an object's temperature by one degree kelvin. Once this is known, we can predict the rate at which objects will change temperature and the amount of heat energy involved. Understanding specific heat is extremely valuable for designing systems where heat transfer is fundamental to their operation.

Physics

What is the gravitational field strength formula?

A gravitational field is a region in space where a mass will experience a gravitational force. The gravitational field strength at a location is defined as the force per unit mass experienced by a small test mass placed at that location. The field strength can be expressed as an equation as follows: $\hspace{3em} g = \dfrac{F}{m}$ Where:\ $\hspace{3em}$ $g$ is the gravitational field strength in units of N kg$^{-1}$\ $\hspace{3em}$ $F$ is the gravitational force experienced by the mass in units of N\ $\hspace{3em}$ $m$ is the mass of the object experiencing the gravitational force in units of kg. Any object with mass will create a gravitational field in the region around it. This field extends an infinite distance away; however, it decreases proportionally with the square of the distance. We can use Newton’s Law of Gravitation to develop the formula for the gravitational field strength at a given distance from an object. The law states that the force of gravity between two masses is directly proportional to the product of the masses and inversely proportional to the square of the distance of the separation of their centres of mass: $\hspace{3em} F_g = G\dfrac{m_1m_2}{r^2}$ Where:\ $\hspace{3em}$ $F_g$ is the force of gravity in units of N \ $\hspace{3em}$ $G$ is the gravitational constant 6.67 x 10$^{-11}$ N m$^2$ kg$^{-2}$\ $\hspace{3em}$ $m_1$ and $m_2$ are the masses of the two objects in units of kg\ $\hspace{3em}$ $r$ is the separation of their centers of mass in m. We can substitute this equation for gravitational force into the first question for gravitational field to derive a new equation $\hspace{3em} g = \dfrac{F}{m}$ $\hspace{3em} g = \dfrac{G\dfrac{m_1m_2}{r^2}}{m}$ $\hspace{3em} g = \dfrac{Gm_1}{r^2}$ This formula gives the gravitational field strength at a distance $r$ from a the centre of a mass $m_1$ . This mass is normally a planet, star or moon.

Physics

Why does v = fλ?

The universal wave equation v = fλ is one of the fundamental equations in Physics. It relates the speed of a wave, v, the frequency, f, and the wavelength λ. Waves are periodic disturbances that propagate energy through a medium. They are caused by an oscillating source that creates the original disturbance. The frequency of the resulting wave is equal to the frequency of the source. The period $T$ of the oscillation is the inverse of the frequency $f$. One period is the time that it takes for the source to complete a full cycle and return to its original position. The particles in the medium propagate the wave through individual oscillations with this same period. The wavelength $\lambda$ of a wave is defined as the distance between two successive points on a wave that are oscillating in phase. An example of points on a transverse wave that are in phase with each other are crests - at a crest the particles are at the top of their oscillatory motion. Similarly, all points on troughs are in phase with each other. So the distance between adjacent crests or troughs is one wavelength. For a longitudinal wave, a wavelength is the distance between the centers of two adjacent compressions or rarefactions. Now we can think about how far a wave travels in the time that it takes the source disturbance to repeat itself, or in other words, complete one cycle. The time for one cycle is the period $T$. Waves will travel away from the source at a speed $v$. This speed depends on the nature of the medium and is independent of the frequency of the source. If we consider a wave that travels away from the oscillating source, we can use the speed equation $v = \dfrac{s}{t}$ to determine that the distance $s$ that the wave travels in one period $T$ is given by $\hspace{3em} s = vT $ This distance is the separation between two repeating points on the waveform, and so is equal to the wavelength $\lambda$ $\hspace{3em} λ = vT$ Solving for the speed gives $\hspace{3em} v = \dfrac{\lambda}{T}$ And we know that the period is the inverse of the frequency $\hspace{3em}T = \dfrac{1}{f}$ Substituting in, we arrive at the wave equation $\hspace{3em} v = f \lambda$

Physics

How to calculate percentage uncertainty

In science, all measurements have an associated uncertainty. This uncertainty communicates the precision to which the measurement was taken. If a scientific result is the product of a series of calculations of measured values, then the result itself will have an associated uncertainty to show the degree of confidence in the result. The percent uncertainty of a value is the same as the fractional or relative uncertainty of the value expressed as a percentage. It can be found from the following formula: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% $ For example, if the time for a cart to roll down a ramp is measured to be 3.6 s ± 0.2 s we can see the absolute uncertainty on the measurement is 0.2 s. We want to express this as a percentage of the measured value: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% = \dfrac{0.2}{3.6}=5.6 \%$ The method used to find the percentage uncertainty of a calculated value depends on the mathematical operation being performed. If the values involved are being added or subtracted, then the absolute uncertainties need to be added to find the absolute uncertainty on the result, and then the percentage uncertainty can be found. For example, if three length measurements are being added together: $\hspace{3em}$ (2.0 ± 0.1) cm + (8.2 ± 0.2) cm + (2.5 ± 0.5) cm The result will be the sum of the lengths and the absolute uncertainty will be the sum of the individual uncertainties: $\hspace{3em}$ Length = 2.0 + 8.2 + 2.5 = 12.7 cm $\hspace{3em}$ Absolute uncertainty = 0.1 + 0.2 + 0.5 = 0.8 cm $\hspace{3em}$ The result is expressed as 12.7 cm ± 0.8 cm. We can now find the percentage uncertainty on the answer as follows: $\hspace{3em}$ % uncertainty = $\dfrac{\textrm{absolute uncertainty}}{\textrm{measured value}} \times 100 \% = \dfrac{0.8}{12.7}=6.3 \%$ If two values are being divided or multiplied together, their individual percentage uncertainties must be found. The final result's percentage uncertainty will be the sum of the individual percentage uncertainties. For example, consider the following calculation of the density of an object. The mass $m$ and volume $V$ of the object are measured to be $\hspace{3em}$ $m$ = 4.5 g ± 0.2 g $\hspace{3em}$ $V$ = 2.5 cm$^3$ ± 0.5 cm$^3$ The density can be found from the formula $\hspace{3em}$ $\rho= \dfrac{m}{V}$ Giving $\hspace{3em}$ $\rho=\dfrac{4.5}{2.5}=1.8$ g cm$^{-3}$ Now we need to find the percentage uncertainty of the result. First, we find the percentage uncertainties on the mass and the volume: $\hspace{3em} \dfrac{\Delta m}{m} \times 100 \% = \dfrac{0.2}{4.5} \times 100 \% =4.4 \% \hspace{2em} \hspace{2em}$ and $\hspace{3em}\dfrac{\Delta V}{V} \times 100 \%= \dfrac{0.5}{2.5} \times 100 \%=20 \%$ Because we are dividing the terms, we add the percentage uncertainties: $\hspace{3em}$ density % uncertainty = mass % uncertainty + volume % uncertainty = 4.4% +20% $\hspace{10em}$ = 24%

Physics

What determines gravitational pull?

Gravitational pull, otherwise known as the force of gravity, acts between all objects due to their mass. Around any individual mass there exists a gravitational field that will cause any other mass placed within it to be attracted to the original mass. The gravitational field strength is represented by the symbol $g$ and has units of N kg$^{-1}$. The gravitational pull on an object depends on its mass m and the strength of the gravitational field that it is in. The formula for the gravitational force $F_g$ is $\hspace{3em} F_g = mg$ The gravitational pull between two masses can also be found using Newton’s Universal Law of Gravitation. This law shows that the gravitational force between two objects is directly proportional to the product of their masses and inversely proportional to their separation distance. $\hspace{3em} F_g = G\dfrac{m_1m_2}{r^2}$ Where: $\hspace{2em}F_g$ is the force of gravity in units of N \ $\hspace{2em}G$ is the gravitational constant 6.67 x 10$^{-11}$ N m$^2$ kg$^{-1}$\ $\hspace{2em}m_1$ and $m_2$ are the masses of the two objects in units of kg\ $\hspace{2em}r$ is the separation of their centres of mass in units of m

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