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0

The last second of it's upward motion means that, at the end of that second, the object is at rest. It will start falling back downwards after that. Now since we know that the body experiences constant deceleration $g$ (= 9.8 $m/s^2$), it will have a velocity $u$ of 9.8 $m/s$ in the upward direction at the beginning of that second. Now since we have the ...


1

There are four kinematic equations that apply to this type of problem. One of those equations can be used to solve for acceleration when you don't how long it took for the car to stop. The equation is: $v_f^2 = v_i^2 + 2a \Delta\ x $ The initial velocity is given, the final velocity is zero, and the distance traveled is given. The only unknown, ...


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The total distance traveled during the constant acceleration is 60 feet. Over the course of 60 feet the velocity has to go from 50mph to 0mph. That means the average velocity is 25mph. So, what amount of time going 25mph is equal to 60 feet? Then take that time and divide the change in velocity by it to arrive at the acceleration.


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Well let's pick the $(Ox)$ axis as a reference of frame and take the origin of time the instant the car starts deccelerating with a magnitude $a$: Since we are talking about decceleration it is clear that: $$ \frac{dv}{dt}=-a $$ So: $$ v(t)=v_0-at $$ And: $$ \frac{dx}{dt}=v_0-at\\ x(t)=v_0 t -\frac{a}{2}t^2 $$ Now we know that the car will stop at the ...


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For motion along a circular trajectory the acceleration changes its direction, and is not constant.


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This problem can mean two things: Either it is to be treated using general physics (thus assuming ideal conditions like zero friction in sliding) or it has to be evaluated very specifically (by considering all the energy lost in friction etc.). I believe that this one has to be evaluated using the former scheme. And under that scheme, the work would be ...


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Take the kinematic relationship $$ x_{end} - x_{start} = \frac{ v_{end}^2 - v_{start}^2 }{2 a }$$ which applies to constant acceleration and use it twice. First on the first interval to calculation the acceleration $a$ and then again on the second interval to find the distance traveled.


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Condition for traversing the whole circle: Let the particle of mass $m$ is attached to an inextensible string of length $R$ which provides the necessary centripetal force along with gravity. Let the initial velocity be $u$ at the lowest point of the vertical circle. After time $dt$, it moves to some other point transversing angle $\theta$. The height at ...


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1."The particle will complete the circle when at the highest point if the string doesn't slack at the highest point when θ=π." - Is it because if the rope slacks, then the mg component will pull it inwards and so the particle will not move as a circle? Yes. Gravity (the weight $W=mg$) is then strong enough to pull it back from the "swing". The rope is ...


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Compute the ratio $k$ between the maximum speed and current speed: $$ k = \frac{v_{max}}{v} = \frac{v_{max}}{\sqrt{v_x^2+v_y^2}} $$ Then multiply both horizontal and vertical component with this ratio: $$ v_{x}' = k v_x $$ $$ v_{y}' = k v_y $$ The new speed will be $v_{max}$ and the direction will be preserved, because ratio of $v_x/v_y$ will remain ...


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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 ...


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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 ...


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To solve these, it's enough that you know: $$ v = v_0 + at $$ $$ x = x_0 + v_0 t + \frac{1}{2} a t^2 $$ Fill in the values you know, then solve for the value you want.


1

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 ...


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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 ...


1

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.


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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

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 ...


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.


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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 ...


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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 ...


0

Assume the ball slides down the slope without friction and that it starts from stationary ($v=0$) at a height above the horizontal $h$. During the friction free slide the object's potential energy $mgh$ is converted to kinetic energy $\frac{mv^2}{2}$, so that: $mgh=\frac{mv^2}{2}$ and: $v=\sqrt{2mgh}$. This speed vector is of course oriented parallel to ...


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

Another way to do this is to consider the energy (as you do not seem to be losing energy anywhere, eg with friction). You have an initial velocity (perhaps 0) which corresponds to a kinetic energy, and a potential energy (which is $mgh$). Then the potential energy is converted to kinetic energy, and you can simply figure it out from there. One important ...


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 ...



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