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30

45 degree angle for a projectile gives you the maximum distance in a vacuum, but air resistance, as pointed out, changes that a little. With air resistance slowing the ball, you need to to throw a tick under 45 degrees for maximum distance. Also Also, since you throw from above the shoulder, not from the ground, the ball is usually thrown a foot or ...


28

X-rays do warm you up. It's just that the X-rays are more dangerous per photon (they can do major damage to cells and DNA, and are known to cause tumors and cancer), so they limit the amount of time you're exposed to the bare minimum needed for a clear image. The total energy from standing in the sunlight for several seconds is much higher than the energy ...


23

I am a baseball fan (and a physicst), and your coach is misleading you a little. First, in the absence of air resistance, a 45-degree launch will get the ball there with miminum energy expenditure. But not, as the other answers suggest, minimum time. And time matters. A lot. :-) Your coach should also be telling you to plan your longer throws such ...


20

In addition to the answer from @MichaelS, you need to consider where the energy from each source is deposited: Sunlight energy is deposited on/in the skin where there are numerous nerve endings. An increase in skin temperature is "measured" and your brain is aware of it. X-ray energy which is absorbed by the body is mainly absorbed by bones and some ...


10

Why does my sports coach tell me that when I'm fielding I should throw the baseball 'flat' to get the maximum distance? I thought from physics that you get the most distance from throwing at a 45 degree angle? The second question first, this is true if you are a robot throwing a ball on the Moon (no atmosphere) that releases the ball at the same speed ...


4

Let's analyze with some simplifications. Ignoring friction for a moment... if you throw a ball at angle $\alpha$ with velocity $v$ such that it will fly for a distance $d$, and it is caught at the same height as it is released, then we can quickly calculate the velocity and time of flight. Time of flight: $$t = 2\frac{v_y}{g} = \frac{2v\sin\alpha}{g}$$ ...


4

Think of it like this: if anything about the velocity is changing in time then the acceleration is non-zero. That thing can be magnitude or direction or both. Therefore, if the change of $|\vec{v}|$ with respect to time is not zero, then the acceleration is, by definition, not zero. However, we can have a case where the direction of the velocity is changing ...


4

How did you integrate acceleration to get velocity? Note that $\Delta v = \int_{t_i}^{t_f} a(t) dt$ But you have an acceleration that is a function of position, not time. So you can't naively integrate this and get velocity. There is a trick. Notice that you can rewrite acceleration as $a = \frac{dv}{dt} = \frac{dv}{dx} \frac{dx}{dt} = v ...


4

I am not a baseball expert, but the time, seconds, will maybe matter here. You are correct, the maximum distance, theoretically, is when you throw at a 45 degree angle. However, in baseball you might want the maximum distance compared to how long time it will take. You maybe want the ball to get as far as possible, as fast as possible? This might make your ...


3

Deceleration is a "special case" of acceleration. More precisely, acceleration is given by the vector $\vec a$ which has both a magnitude and a direction. Sometimes the same vector $\vec a$ increases the velocity $\vec v$ – when they are oriented in "mostly the same direction" – and we speak about "real acceleration" in the sense of an increasing speed. And ...


3

To get velocity from acceleration, you need to integrate with respect to time. But your expression of acceleration is given with respect to position. Thus, your current calculation is not correct. You need to figure out how to convert the position-dependent information to time-dependent information. Since they give you the solution and you just have to ...


3

No, all your reasoning is totally right. The conclusion isn't that the graphs are wrong, it's that the time of impact is less than 0.1 second. In this video, for example, the time of impact is just about 0.01 seconds.


3

There's two main things to consider - energy and absorption charasteristics of different photon wavelengths. The Sun emits a lot of energy, obviously. Even at Earth's distance from the Sun, the energy concentration is still far from negligible - when this energy impacts your body and is absorbed, it mostly causes heating (a bit complicated by wavelength, ...


2

The notion of work in physics was first formulated by the French mathematician Gustave Coriolis in Calculation of the Effect of Machines, or Considerations on the Use of Engines and their Evaluation published in 1829. Coriolis defined work as "weight lifted through a height". He was concerned with developing a term that could measure the units of work ...


2

It turns out that people have been crazy enough to build bicycles with smaller "spare wheels" alongside their normal wheels that can be "ramped up" to speeds that have the same angular momentum in the opposite direction. When you do this with a unicycle it becomes difficult to balance even at speed, which means that the primary mechanism is gyroscopic. But ...


2

Your image shows what is going on at the microscopic level between two surfaces. To understand why friction works, you have to look smaller, at the atomic level: and when you get to that point you're no longer taking about "friction" as we know it, but about physiochemical interactions between atoms and molecules. Those interactions are mediated by ...


2

With regards to intuition, it might help to think about situations of mechanical advantage. For example, consider a simple pulley system. modified from "Pulley1a". Licensed under Public Domain via Commons - https://commons.wikimedia.org/wiki/File:Pulley1a.svg#/media/File:Pulley1a.svg You can work out using force that the weight $W/2$ balances the weight ...


2

Sorry, but the currently accepted answer is not precise enough. In the situation you discussed, there is no applied force acting on the object, but there is still the normal force $N$, which prevents the object from "falling through" the surface due to gravity. The most common model of kinetic friction is $F_k = \mu N$, where $\mu$ is the coefficient of ...


2

A 45 degree angle gives you the maximum distance to the point where the ball first hits the ground. That's not where a baseball will stop. If you hit the ball at a lower angle, more of the total speed will be in the horizontal direction, and the ball will still continue with a fairly high horizontal speed after bouncing. Thus, the point where friction ...


1

When you write: Derivations (or at least, convincing arguments) of the kinetic energy formula that didn't require the work formula required relativity to make sense, which is unbelievable considering that Newtonian mechanics were established well before relativity. I assume you are referring to arguments like Ron's argument. Although such ...


1

$$ m \ddot{\vec{r}} = \vec{F} $$ multiply by $\dot{\vec{r}}$ and integrat over t: $$ m \int_{t_0}^t \ddot{\vec{r}} \cdot \dot{\vec{r}}~ dt = \int_{t_0}^t \vec{F} \cdot \dot{\vec{r}}~ dt$$ With $\frac{1}{2} \frac{d}{dt} (\dot{\vec{r}}^2) = \ddot{\vec{r}} \cdot \dot{\vec{r}}$ it follows: $$ \frac{1}{2} m v^2 + \left( - \int_{t_0}^t \vec{F} \cdot d\vec{r} ...


1

Using polar coordinates it holds $ |d{\bf r}| = \sqrt{(d|{\bf r}|)^2 + |{\bf r}|^2(d{\bf \phi})^2}$. From this equation you can see, the two expressions you are asking about are actually only equal (in absolute value) for a straight line through the origin, thus otherwise different. For them to be exactly equal, the $d|{\bf r}|$ should be moreover pointing ...


1

If $$ \overrightarrow{r}=r_{x}\widehat{i}+r_{y}\widehat{j} $$ then $$ \left | \overrightarrow{r} \right |=\sqrt{r_{x}^{2}+r_{y}^{2}} $$ and $$ d\left | \overrightarrow{r} \right |=\frac{r_{x}dr_{x}+r_{y}dr_{y}}{\sqrt{r_{x}^{2}+r_{y}^{2}}} $$ on the other hand $$ d\overrightarrow{r}=dr_{x}\widehat{i}+dr_{y}\widehat{j} $$ and $$ \left | d\overrightarrow{r} ...


1

When neglecting air drag, the vertical and horizontal motions decouple and we have the equations \begin{align} \ddot{x} &= 0,\\ \ddot{y} &=-g. \end{align} The initial conditions (at $t=0$) are $x=x_0$, $y=y_0$, $\dot{x}=v_0\cos\alpha$, and $\dot{y}=v_0\sin\alpha$ with $v_0$ the initial speed and $\alpha$ the angle of projection. This gives the ...


1

Friction comes into play whenever there is relative motion between the surfaces in contact or a tendency of motion between the same.There need not necessarily be an externally applied force on either of the bodies,that is,there need not necessarily be a relative acceleration initially,merely relative motion or a tendency for the same. It is the frictional ...


1

What is missing: this must be recognized as a relativistic process and be treated accordingly. Why is it a relativistic process: energy transforms into mass or conversely. This is something that cannot be accounted for using only classical mechanics. In older times I would've added "because it needs to account also for relativistic mass", but since using ...


1

Since this is a specific question, I can't answer it directly for you but I may give you a hint: you already have the displacement formula $\stackrel{\to }{r}\left(t\right)$. From there you may find the relation y(x) using t as a parameter and use the formula for the arc length from calculus: $L=\underset{a}{\overset{b}{\int }}\sqrt{1+{\left({y}^{\prime ...


1

Length: $\int_{t_0}^{t_1}| \vec{r} '(t)|dt$. In your case: $ \int_{0}^{t}\sqrt{(20+4t')^2+(15)^2}dt'$ Use Wolframalpha if you are now sure how to deal with it: https://www.wolframalpha.com/input/?i=int+sqrt(%2820%2B4x%29^2%2B%2815%29^2)dx


1

Imagine a space ship ran put of fuel and a second is going to help. They will do a docking maneuver to hand out some fuel. Now, the second ship has to adjust its speed when it's near the first (because the first can't change its speed). A passenger on the second ship feels the acceleration. But does the ship accelerate or decelerate? If the first ship ...


1

The ball was in flight for four seconds: we can safely say that the ball reached maximum height at $t = 2$. (The gravitational pull is constant and there are no other forces acting, so the flight path is symmetrical). The ball was stationary at $t=2$ so its speed is $=0$ So now use the formula $v= u + at$, where $a$= acceleration, $t$= time, $u$= initial ...



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