Tag Info

New answers tagged

0

Incomplete answer: assuming the wall is a perfect reflector, we can remove it and ask at what point the sound reaches a second car driving away from the wall to the northwest, in a path reflected from the original car's through the line of the wall. We know the position $O$ of the first car when the driver pressed the horn, and at any given time $t$ we can ...


0

Sound leaves the vehicle. When it strikes the wall, it reflects at a 45 degree angle and returns to the car. The triangle you will be using has legs that are length X, with a hypotenuse that is length Xsqrt2. That is for evey 1 foot of length in a leg, I will have 1.414 feet of length in the hypotenuse. What you want to solve for is time based on this ...


1

The problem is that you have not solved the question yet. What you have found is not the friction between the boxes. It is something else. As you actually state yourself, you have instead found the maximum [static] friction. This is just the maximum possible value and not at all necessarily equal to the actual friction. Static friction can be anything from ...


1

If it is static friction, then the two blocks are stuck together and both have the same acceleration. In that case, the top block has a net force of 40 N (100 N pull - 60 N friction) and the bottom block has 60 N (just from the friction). Since 60N < 80N max friction, then the ansatz that this is static friction is consistent.


1

I maybe seriously underestimating the OP's question (and not quite sure if I understand it) but I believe he/she is simply asking why we only need two quantities (rotation and translation) to describe 3D space. I'll start with 2 dimensions: Imagine you're playing the game Asteroids and you want to move your spaceship a little to the right to get out of the ...


0

To change the velocity of the ball a force must be applied against it. In your example the force is applied by the wall on the ball and this force does work. And this work which is not same in all frames of reference, according to your example, gets completely converted to heat.To explain it simply, lets say that the ball is deformed and becomes completely ...


0

I think the real reason is because if you change your velocity then everything that used to be at rest should now continue just as they did before you changed your velocity. You can call it indifference, nothing cares how you move unless you interact with it. Or you can call it relativity. What this does is reduce the case 1) of things at constant motion ...


0

Timaeus has given the full technical answer: the kinetic energy of the wall itself changes a tiny bit. Since kinetic energy scales as $v^2$, this is totally negligible in the first case (where the wall starts with $v = 0$) but actually significant in the second case (where the wall starts with $v = 2$), and that's where the missing energy goes. Luckily, in ...


0

Imagine a ball ($mass= 1kg$) moving at a velocity $2 m/s$ towards a wall. When it hits the wall, it suddenly stops, thereby liberating all its KE as heat. Here, $Initial K.E. = (1/2)*m*v^2 = 2J$, and final KE is obviously zero. So heat liberated ($Final KE - Initial KE$) equals $2J$. Now, suppose I hop on to a car moving at $2 m/s$ towards the oncoming ball ...


0

Initially your ball has some energy ($2J$) and some momentum ($2Ns$). And the wall has some energy ($0J$) and some momentum ($0Ns$). And there is some internal energy, $U=U_0,$ the thing that heat increases. Afterwards the ball has some energy ($0J$) and some momentum ($0Ns$). And the wall has some energy ($0J$) and some momentum ($2Ns$). And there is some ...


-1

In the car frame the ball initially has 8J of kinetic energy relative to you, however it has only 2J of kinetic energy relative to the wall which is also moving in the cars frame.


0

The carbon dioxide panic is so widespread that these days, virtually all carmakers make the data available for the CO2 emissions per kilometer directly. For example, this is the table for all Škoda models https://www.fleetnews.co.uk/cars/co2-emissions-calculator/skoda/ Replace "skoda" in the URL by "audi" or another brand to get the numbers you're ...


1

If this is a correct description of what happens, can we conclude that g does same work on P and on P'? Yes. This is correct. If g acts perpendicularly to the velocity, it performs work of magnitude zero. This is also correct. The reason the two statements above are not contradictory is that the work done by the gravity changes the direction of ...


0

Let's look at the kinetic energy in both cases. Subscript $i$ is for initial, $f$ is for final and $\Delta\text{KE}$ is the change in kinetic energy. For P (shot horizontally): $\text{KE}_i=\frac{1}{2}mv_i^2=\frac{1}{2}m(v_{xi}^2)=\frac{1}{2}m (9.8^2)$ $\text{KE}_f=\frac{1}{2}mv_f^2=\frac{1}{2}m(v_{xf}^2+v_{yf}^2)=\frac{1}{2}m (9.8^2+9.8^2)$ ...


0

Our equation for work follows from the conservation of energy. If we consider some object then we expect that if we do work $W$ on it then its kinetic energy must increase by $W$. So the requirement for the equation for work is that it must be equal to the change in kinetic energy. Proving this is usually done using integral calculus, but since you give the ...


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 ...


-1

There are some physical quantities that are usefull (and this is under statement), like energy. It is conserved, it is a function of some other very important quantities that can help you describe the motion of the body etc. If you can justify energy, there should be no problem in justifying work, which is energy transfered to a body by some force. Quantity ...


0

Will the rotational speed of the spaceship be the same as the angular velocity of the circle it is traveling in? I would put it another way. The rotating force creates an acceleration that causes it to move in a circle with the same angular velocity. It's not just that they're the same, but the rotating force causes the circular motion. (I'm assuming ...


0

Define the $y$ axis as the vertical axis and the $x$ axis as the horizontal axis. A projectile flying through the air has a velocity vector $v$, let's say at an angle $\alpha$ to the horizontal: The velocity vector $v$ can be decomposed into two components: The $x$ component: $v_x=v \cos\alpha$. The $y$ component: $v_y=v \sin \alpha$. It's important to ...


1

"Rest, in physics, refers to an object being stationary relative to a particular frame of reference or another object." - Wikipedia (emphasis mine) While on Earth, the planet is often treated as the default frame of reference. It is not a perfect frame of reference, but for many purposes it is good enough. Since there is no absolute frame of reference, ...


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, ...


0

Gravity, as you know it from your daily experience, is an everywhere parallel force, pointing downwards. In this field, each body is accelerated downwards, regardless where the body is, an where it is moving. But as you know, gravity is not parallel on a larger scale, the force always points to the center of the earth. Imagine your bullet is shot in space ...


1

There are multiple "kinds" of photons - different wavelengths have different effects on you. X-ray works somewhere around the 1nm range of the spectrum. It is ionizing radiation which can mostly go through soft materials but can harm cells when passing. So you usually get only the minimal needed amount of photons to create the image and not much more. The ...


0

What other answer? Yes an object falling accelerates due to gravity and gains KE. An object at height X has potential energy. It took work to raise the object to height X because of gravity.


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 ...


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 ...


0

No matter how many spacial dimensions, there is only one kind of line. Two points describe a line and between them is only one dimension. For rotation, again it doesn't matter which way it's rotating, but the number od possibilities increases with dimensions. The question is whether any combination of rotating (instantaneous) can be combined to a single ...


20

"A state of rest" is a relative term. Relative means - measured in comparison to the things around it. When you sit in a train and sip from a cup of coffee, you can do so because the cup is still relative to you even though both of you might be hurtling through the countryside at 200 km/h. For most experiments, objects can be considered "at rest" if they ...


5

Revolving around the sun is equivalent to free fall around the sun, so the revolution allows you not to 'feel' the sun's gravity. The rotation of the earth is something that can be measured: (i) a centrifugal force which is a small offset on gravity, and (ii) causes the coriolis force. Both these are small effects, so can often be ignored for laboratory ...


0

When you're in a moving vehicle and see trees or buildings, who is moving? Are you moving forward, or are the trees and buildings moving backward? Its counter-intuitive for beginners but both these views are absolutely correct. We can only describe the motion of an object from a reference frame. A reference frame is a specific configuration from where you ...


0

Solving lots of physics problems involved you choosing a frame of reference. As long as all your formulations are from a particular frame of reference, for example the energy of objects then the laws will work. Since you are also travelling with the object, (as the earth rotates) it is easier to say you are at rest. If you are travelling at constant velocity ...


1

Yes your reasoning is completely right but is important to add some information. The diagram you show has no information on the ball size and assumes all the mass is concentrated in the center of mass, which for a spherical and isotropic ball should be in the geometrical center. Now, at the moment of impact $t_1$, the border of the ball will start touching ...


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.


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 ...


0

$\Delta y=78.4m$ is the correct answer. The golf ball was fired horizontally from a height of $78.4m$. With $\Delta y= \frac{1}{2}gt^2$ and $t=4s$ we get $\Delta y=78.4m$. The horizontal velocity component of $30m/s$ was a distraction, not needed for calculating the solution because the vertical component and horizontal component of velocity are ...


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 ...


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 ...


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 ...


1

Work does depend on frame of reference, but so does change in kinetic energy. Work done and changes in kinetic energy should either both bother you or neither bother you. To know how much the kinetic energy changes from one location to another you need to know the force (if constant) and how far apart the locations are: ...


1

As you point out, work done is a function of the frame of reference. More specifically, if you apply a force on an object, that force typically connects two different objects, and it's the relative velocity of these two objects that really concerns us. Example: you are walking in a train, and pull a suitcase behind you. The friction between the suitcase and ...


0

A conventional car work with high energy density fuel. But the drive works efficient in some revolutions only. That's way there is a gear box. Working with electric drives is not new. It was invented at the same time (to be precise a little before) the Benz developed his car. But there are weak points too. Accumulators are heavy and the energy density is ...


0

So why if applied force is increased the normal force can't withstand the increased force? The other answers answer this well. Also is there any possibility that in a situation the frictional force could always cancel the applied force? In that case you would need a material (surface) which is infinitely strong. If you put a heavy stone on a ...


0

To add a little to Asher and Gert's answers: The atoms on surfaces are constantly in motion. If you're used to thinking of solids as being really rigid and difficult to deform, try this viewpoint instead: think of them like a bunch of grains of sand held together by tiny Slinkies. The atoms can move out of the way and slide back into place, or even be ...


0

Why can't friction withstand any force? Because the amount of friction that inter-surface interactions can provide is limited, not unlimited. In the idealised force diagram below, only the object's weight $mg$ provides a vertically downward (aka 'Normal') force, causing a friction force between the object and the plane. This friction force is ...


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 ...


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} ...


0

What about using Galilean free-fall? From $S= \frac 12 g t^2$ and $v = g t$ you get that velocity after a fall $h$ follows $$h= \frac 1{2g} v^2$$ We conclude that if the ball is consuming some essence to get velocity from the line of fall, this essence must be "stored" in space as the square of the velocity. The idea works because if we have already a body ...


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 ...


0

Actually @Ocelo7 has already answered it; I am just showing that in a concise way: Velocity can be found by differentiating the position-vector:$$\dot{\mathbf{r}}= \mathbf{v}(t) = 10\mathtt{\hat{ i}} + (20 - 10t)\mathtt{\hat{j}}$$ . Assuming the initial time is $t_0$ & the required time is $t$, we shall use the property $$\mathbf{v}_{t_0}\cdot ...


0

There's a good way to get an estimate out of this. First, perform dimensional analysis to get an estimate. Start with, finding ${{\Delta a} \over {\Delta v}}$, which is the change in the drag acceleration with respect to velocity. The argument that follows is physical rather than technical, so beware. The above expression, should be proportional to the ...



Top 50 recent answers are included