Solution:
The approximate distance from Earth to the moon is 384,400 kilometers. The time it takes for the light to reach the moon is
, taking into account significant figures. [Solution]
cm? [Solution]
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ANSWERS:
Part 1:
Treating two-dimensional forces as 2 separate motions helps break down the force into two vectors for the x and y values respectively. This helps when adding vectors together so you can determine how two forces might act on each other. In a real-world application, the first thing I can think of is a sailboat and its relation to the wind. When sailing you don’t typically go in a “straight” direction (being with or against the wind I mean) as this could put you in “irons” and you could effectively stall the boat. Typically you set sail perpendicular to the wind and you would “tack” (or go back and forth) upwind. I find this to be an effective representation of vectors for multiple reasons. It shows that with multiple forces acting on your boat, the direction is still in one seemingly straight line and when broken down you could see how multiple vectors are added to produce the line of which the boat follows. Also when considering tracking from an aerial view It helps emulate how vectors could be added. If I wanted to go upwind I have to move my boat along the “x-coordinate” and then along the “y-coordinate” and these two motions add up to the total displacement of my boat.
Part 2:
The first law states that a moving object will continue in its course of motion unless a force is acted on it, This directly relates to the second law as the second law explains exactly what a “force” is. a force requires both mass and acceleration. A=F/m is the better way and more often way to use it because acceleration isn’t really multiplied by a mass to get force (well it is, but you cant apply an acceleration to something and I’ll explain that a bit more) you get acceleration by applying force to a given object. This force as stated in the first law is what changes the acceleration either positively or negatively. The reason I say you can’t apply an acceleration to something is that you apply force to change its acceleration. just as F=ma, and a=F/m m=f/a and you cant apply the mass either. If you know the force and acceleration you can calculate for mass, but you can’t make an object by a force being divided by acceleration, what does that even physically mean? These numbers are all related directly but when figuring for real-world problems I believe it is important to understand that it is a force acting upon an object, and the output of this scenario is acceleration on the object. and that is why a=f/m is more applicable and makes more sense when applying to a real-world problem.
ANSWER 2:
I would like to present my example before I explain how different motions work together. There is a missile defense system called the Iron Dome. Basically, this system locates incoming missiles and then works as a Brainiac Physics freak by determining the trajectory using calculations dealing with velocity, gravity, angles, speeds, magnitudes, ect., the whole nine yards. The system then shots something like a back as myth, or a mirror like calculation and fires its own rockets heading to intercept the incoming rounds. It does this to multiple targets simultaneously, and in a split second. Different values must be measured and combine in order for this to work such as time, displacement, angle, gravity, acceleration. They must then be combined to find the path in which the rocket travels. All of these forces, motions, and magnitudes are connected. If one changes, it effects the other. Calculations must be accurate in order to intercept the potentially deadly rounds.
In newtons first law, he addresses that an object will stay in motion at the same direction and speed unless it is acted upon by an unbalanced force. In his second law he dives deeper and explains that the size, or mass, of an object effects its acceleration. When I think about this, I look at football players. I see a big guy and a little guys running at each other head on because they don’t see each other for what ever reason. When these objects meet head on, the guy with the bigger mass will stop the little guy if they are headed about the same speed and its gonna be ugly in my estimation.
When I was in high school playing football, I remember a single play when I locked up with a guy and saw the quarter back peeling off to the outside. I managed to spin off of the player in front of me and had a free sprint towards the quarterback. My tunnel vision did not all me to see the guy who was about 9 inches taller and 60lbs heavier then me in my path. I found out when I was hit so hard that I ended up on my back with a bloody nose. My motion stopped when I met a force greater then myself and man oh man did it hurt!
Answer 3:
The one big way they are tied together is by time. We measure both of the axis with respect to time. The applications can include calculating a bullets travel distance, dropping supplies from above, sports applications, etc.
Both laws are similar that they talk about objects in motion and force behind them. His first law states how with no force object will stay in its current state. The second law talks and discusses what happens when a force is applied to the object. Mass and acceleration also effect the objects state. It is the most fundamental relationship in physics. It is the basis and is used it building applications, vehicle, pulley systems, etc.
Answer 4:
Well Fellow travelers I might have something interesting for both part 1 and Part 2 of this discussion. So we shall start with Part 1.
Part1:
Imagine you have a deck gun from a WW2 battle ship that is attached to another gun section. The gun first section is fully attached to the second gun section with a precision made burst petal assembly that is placed between them sealing the two barrel sections. In the first section you have an explosive charge in the form of a gun powder in powder bags like the WW2 battle ship used that is placed into the powder breach that is part of the gun. On the other side (the gun barrel side) is a a projectile that is ready to go down the guns barrel. That projectile is in the shape of a piston with a slightly concaved face and that portion of the barrel is filled with nitrogen gas between it and the burst petal as a trapped atmosphere.
Now on the other side of the Burst petal you have another object called a Sabot that is sitting in a vacuum facing a large chamber and in that chamber is a target of a metallic type material that is held in place by a very stout structure with a myriad of measurement devices that do not touch the surface. The Sabot is made out of a Lexan like material and is layered and is shaped like a hammer head with the face of the Sabot made of material that is used to simulate a very hard face to hit the target with and this Sabot is going to be used to strike the target like a hammer strikes an anvil and it has a well know weight or mass.
Now we light the fuse like days of old to ignite the powder charge of the deck gun. The explosion fires the first projectile/piston. The piston accelerates and compresses the nitrogen gas in the barrel with increasing pressure until just before the projectile/piston gets to the other end the burst petal does what its name suggests and bursts apart nearly immediately with a very high exit pressure! This extremely high pressure propels the Sabot with an increasing velocity toward the second barrels exit aperture. There is equipment setup to measure crossing times as the Sabot travels down the second barrel and these measurements are made at known fixed intervals. This crossing time data is used to calculate the Sabots velocity as it travels down the barrel until just before it exits into the vacuum chamber. This provides the needed acceleration and velocity but then you use one more measurement to confirm the velocity.
Now here is Part 2.
You have the known measured mass of the Sabot and you have the acceleration just before exit but you need to confirm confirm the velocity. The Sabot exits the gun with terrific velocity due to the expanding gas but there is some turbulence at exit so you need to verify the Sabot’s position and attitude in space as well as confirm the calculated velocity.
To do this you use and x-ray beam that is soft enough in energy to produce a shadow graph of the Sabot that you capture with instrumentation based on the timing of the calculated speed of the Sabot. Now you have the needed data to predict the Sabot time of impact and therefore its timing of its strike the target. With this data you can calculate the Sabot impact Force that it will impart when it strikes the target. Now with the Force known and the arrival time known you can then measure the shock wave produced by the Sabot and this gives you the measurement of the elasticity of the material under test.
This process is what I would call fun Physics and it produces measurements for material scientists as well as shock diagnostics for Shock Physicist. This data can then be used for say predict what might happen to the International Space Station as to its structural integrity of the outer metallic structure to being hit by Meteorites.This could also be used to study the materials crushing capability for say a car body or many other things that can be deformed, stretched, broken or shattered.
This whole process came from an actual test facility that is used for material science and physics investigation’s and was one of the many things I was involved in at one time in my career. I hope that this was interesting and interesting illumination to projectile physics and what can be done with it.
Answer 5:
PART 1:
Two dimensional motion:
Newton’s second treatment of acceleration and force: ( Force = mass * acceleration)
A body moving at constant velocity can be described by a sum of velocities in two directions, typically x and y co-ordinates.
The path of an airplane may combine constant speed in the horizontal direction (x-coordinate) with acceleration due to gravity in the vertical direction (y-coordinate). This independence of vertical and horizontal motions is constant, and can be seen in the way an airplane flies in the air horizontal to the ground and vertical to the ground at the same time at a certain velocity and acceleration.
Accelerations have the same additive properties. They too are vectors that can be added by constructing a parallel-space, like the airplane. Forces too are vectors and obey the same addition rule. In other words: when several forces act on a body, each produces its own effect on motion. One force does not interfere with the motion produced by another force.
PART 2
Newton’s first law of motion says that an object at rest or constant movement remains that way until acted upon by an external force.
Newton’s second law is an extension of the first law and says that the object will have a change in momentum over time as a result of the external force.
Acceleration is the change in velocity over time. This is key to solving problems with gravity, force, mass, and movement. This plays a part in transportation, especially airlines sharing airspace. A minor error in air-traffic-control could cause a collision of airplanes. That is a real-world situation.
Answer 6:
With two dimensional motion, we are dealing with the vertical and horizontal planes of motion. The vertical plane is affected by gravity, in most circumstances, and the horizontal is unaffected. The best illustration of this is the throwing of a ball, it has its acceleration from the originators hands as they release it at whatever speed they let go of it at, and gravity immediately begins to take hold as it leaves the individuals hands.
The two laws are very similar because, well for starters, they deal with the basic idea of how fast and how long an object moves, or why it comes to a stop. If an object is stimulated to move then it will, given the force applied is strong enough to overcome the forces holding the object in place. Acceleration is important in solving real world problems because it gives us the basis for calculating when and where an object will be at any given time based off of its mass and acceleration.
Answer 1:
If an equation is dimensionally correct, it does not mean that the equation must be true. On the other hand, when the equation is dimensionally correct, the equation cannot be true.
Dimensional analysis is a technique used to check whether a relationship is correct. So, it can only tell you if a relationship is correct or not, but it can not tell you if it is completely right because of the numerics that may be involved in the calculations.
Answer 2:
An equation being dimensionally correct doesn’t mean that the equation is true. For instance, when calculating the area of a circle you can replace pi with another number and it would still be wrong while being dimensionally correct. However, if an equation is not dimensionally correct, the equation cannot be true. This is because if it is not dimensionally correct it would equate into something looking like 4 grapes = 4 bananas which could not happen.
Answer 3:
In order for an equation to be valid, the dimensions on the left side must match the dimensions on the right side, in which case it is dimensionally correct. An equation can be dimensionally correct but still can be wrong. However, if an equation is dimensionally incorrect, it must be wrong.
Answer 4:
No, a true equation must be dimensionally correct but some dimensionally correct equations are not true. Yes, unless the results of the equation produces the correct units, the equation cannot be correct.
Answer 5:
For an equation to be valid, the dimensions on the left side must match the dimensions on the right side (just like our oranges example.) It is then dimensionally correct.However an equation can be dimensionally correct but still wrong.
For example if I say the area of a circle = 2 x radius^2:
– this is dimensionally correct (both sides have dimensions L^2)
– but it is wrong, as ‘2’ should be ‘pi’.
On the other hand, if an equation is dimensionally incorrect, it must be wrong.
Solution:
The known values are a=-9.80\:\text{m/s}^2; v_o=15.0\:\text{m/s}; y=7.00\:\text{m}
The applicable formula is.
y=v_ot+\frac{1}{2}at^2Using this formula, we can solve it in terms of time, t.
t=\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a}Substituting the known values, we have
\begin{align*}
t & =\frac{-v_0\pm \sqrt{v_0^2+2ay}}{a} \\
t & =\frac{-15.0\:\text{m/s}\pm \sqrt{\left(15.0\:\text{m/s}\right)^2+2\left(-9.80\:\text{m/s}^2\right)\left(7.00\:\text{m}\right)}}{-9.80\:\text{m/s}^2} \\
t&=\frac{-15.0\:\text{m/s}\pm 9.37\:\text{m/s}}{-9.80\:\text{m/s}^2}
\end{align*}We have two values for time, t. These two values represent the times when the ball passes the tree branch.
t_1=\frac{-15.0\:m/s+9.37\:m/s}{-9.80\:m/s^2}=0.57\:sec \\
t_2=\frac{-15.0\:m/s-9.37\:m/s}{-9.80\:m/s^2}=2.49\:secTherefore, the total time between passing the branch is the difference between 2.49 seconds and 0.57 seconds.
t_2-t_1=2.49 \ \text{s} - 0.57 \ \text{s}=1.92 \ \text{s} \ \qquad \ \color{DarkOrange} \left( \text{Answer} \right)
Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 2-1
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 2-3
SOLUTION:
The parallelogram law of the force system is shown.
Consider the triangle AOB.
Using cosine law to solve for the resultant force \textbf{F}_{\text{R}}
\begin{align*}
\textbf{F}_\text{R} & =\sqrt{\left(250\right)^2+\left(375\right)^2-2\left(250\right)\left(375\right) \cos\:75^{\circ} }\\
& =393.2 \ \text{lb}\\
& =393\:\text{lb}\\
\end{align*}The value of angle θ can be solved using sine law.
\begin{align*}
\frac{393.2}{\sin\:\left(75^{\circ} \right)} & = \frac{250}{\sin\:\theta } \\
\sin \theta & = \frac{250 \ \sin75 \degree}{393.2}\\
\theta & =\sin^{-1} \left(\frac{250 \ \sin75 \degree}{393.2}\right)\\
\theta & = 37.89^{\circ}\\
\end{align*}Solve for the unknown angle \phi .
\phi =360^{\circ} -45^{\circ} +37.89^{\circ} =353^{\circ} The resultant force has a magnitude of 393 lb and is located 353º measured counterclockwise from the positive x-axis.
Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-21
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-20
Solution:
Part A
From the formula, \text{W}=\text{mg}, we can solve for the mass by dividing the weight by the acceleration due to gravity. That is
\begin{align*}
\text{m} & = \frac{\text{W}}{\textbf{g}}\\
& = \frac{155\ \text{lb}}{32.2 \ \text{ft/s}^2}\\
& = 4.81 \ \text{slug}\\
\end{align*}Part B
Convert the slug to kilograms, knowing that 1 slug = 14.59 kg.
\begin{align*}
\begin{align*}
\text{m} & = \left( \frac{155}{32.2} \text{slug}\right)\left( \frac{14.59 \ \text{kg}}{1 \ \text{kg}} \right)\\
& = 70.2 \ \text{kg}\\
\end{align*}
\end{align*}Part C
Convert the 155 lb to newtons using 1 lb = 4.448 N.
\begin{align*}
\textbf{W} & = 155 \ \text{lb}\times \frac{4.448 \ \text{N}}{1 \ \text{lb}}\\
& = 689 \ \text{N}\\
\end{align*}Part D
Using the same formulas, but now \textbf{g}=5.30 \ \text{ft/s}^2.
\textbf{W}=155\left(\frac{5.30}{32.2}\right)=25.5\:\text{lb}Part E
\textbf{m}=155\left(\frac{14.59\:\text{kg}}{32.2}\right)=70.2\:\text{kg}Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-20
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-21
Solution:
The force of gravity acting between them:
\begin{align*}
\textbf{F} & =\textbf{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\
& =66.73\left(10^{-12}\right) \text{m}^3/ \left( \text{kg} \cdot \text{s}^2 \right) \left[\frac{8 \ \text{kg} \left(12\ \text{kg}\right)}{\left(0.8\ \text{m} \right)^2}\right]\\
&=10\left(10^{-9}\right)\ \text{N}\\
& =10.0 \ \text{nN}\\
\end{align*}The weight of the 8 kg particle
\textbf{W}_1=8\left(9.81\right)=78.5\:\text{N}Weight of the 12 kg particle
\textbf{W}_2=12\left(9.81\right)=118\:\text{N}Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-19
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-17
Solution:
\begin{align*}
\rho _w & =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =\left(\frac{1.94\:\text{slug}}{1\:\text{ft}^3}\right)\left(\frac{14.59\:\text{kg}}{1\:\text{slug}}\right)\left(\frac{1\:\text{ft}^3}{0.3048^3\:\text{m}^3}\right) \\
& =999.6\:\frac{\text{kg}}{\text{m}^3}\\
& =1.00\:\text{Mg/m}^3\\
\end{align*}Engineering Mechanics: Statics 13th Edition by RC Hibbeler, Problem 1-18
Engineering Mechanics: Statics 14th Edition by RC Hibbeler, Problem 1-15
Solution:
To prove that F is in Newtons, we have
\begin{align*}
\text{F} & =\text{G}\cdot \frac{\text{m}_1\text{m}_2}{\text{r}^2}\\
& =\left(\frac{\text{m}^3}{\text{kg}\cdot \text{s}^2}\right)\left(\frac{\text{kg}\cdot \text{kg}}{\text{m}^2}\right)\\
& =\frac{\text{kg}\cdot \text{m}}{\text{s}^2}\\
& =\text{N}
\end{align*}Now, if we substitute the given values into the equation
\begin{align*}
\text{F} & = 66.73\left(10^{-12}\right)\left[\frac{200\left(200\right)}{0.6^2}\right]\\
& = 7.41\left(10^{-6}\right) \text{N}\\
& =7.41\ \mu \text{N}\\
\end{align*}
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