That statement is wrong for mass as U has lots of neutrons as well as protons. Compare the U and H atomic weights of ~238 and ~1.
It is correct for charge when you remove approximately as the U nucleus has 92 protons whereas the H nucleus has only 1 proton.
Mass does not come into your...
Incidentally, you can see how impractical the question actually is.
The path difference is ~5.3 wavelengths of light and the answer is the ~0.3 wavelengths of light.
So, to be accurate, the screen has to be perpendicular to a phenomenal accuracy. If it is just the tiniest bit...
You have had that Aha! moment when you have understood something.
You can now solve any problem like this no matter how much the question is dressed up to make it look more difficult.
Remember what Rutherford said: " All of physics is either impossible or trivial. It is impossible until you...
The point about the laser is that laser light is coherent - all the light is in step or in phase. So, the phase of the light arriving at the top slit is the same as the phase of the light arriving at the bottom slit. Hence, the phase of the light leaving the top slit and the phase of the light...
I didn't check your calculation because I cannot remember if the formula you give is right or wrong.
You don't need to apply a formula blindly to solve (ii) if you think about the physics of what is actually happening.
The slits are very wide apart compared with a normal double slit experiment...
Me too. But don't forget the increased cosmic ray density at that altitude; nor the increased UV intensity - UV rots rubber; nor quantum mechanical effects as all those tyre gas molecules may move in the same direction at the same time and force the tyre off the rim; nor altitude sickness as it...
There is an identical problem when a diver ascends from depth, from a place of high external pressure to a place of lower external pressure. She must exhale to remove air from her lungs or she will damage them.
I think you must have missed my " ... (to a first order). Other things dominate." qualification. I did not say irrelevant.
I was also assuming, as it wasn't stated, it was an early high school leve question and not an undergraduate level question where I would expect a more complete explanation.
The barometer question is an example of an incorrectly designed examination question demonstrating functional fixedness that causes a moral dilemma for the examiner.
Question: Show how it is possible to determine the height of a tall building with the aid of a barometer
Expected answer...
You have all the information you require to answer this question in the thread above.
Your assumption in the Subject that it has something per se to do with the gas laws is leading you astray as it is incorrect: neither the pressure, volume nor temperature of the tyre gas vary (to a first...
The stress in the tyres is due to the difference between the internal pressure trying to burst the tyre and the external atmospheric air pressure supporting the tyres.
Now, the difference in atmospheric air pressure between sea level and the top of any mountain you could ride up will only be a...
I was required to do this experiment at university and I was flabbergasted by the results - my predictions were completely wrong so I know how confusing it appears!
I am assuming the magnet is symmetric about its vertical axis and that all rotation is about that vertical axis.
Rather than...
I think you are misunderstanding what the OP is saying.
He calculates the energy required to be ##mv^2## which is correct. Note it is twice the extra energy acquired by the mass alone, which is ##\frac{1}{2}mv^2##.
So, one half of the power goes to accelerate the mass and one half the power...
If you use the energy equation and calculate the extra kinetic energy acquired by the added mass, you are only calculating the energy given to the mass m to accelerate the mass m from zero speed to v. But this in not the only thing which happens so that calculation is completely irrelevant. **...
You should ignore the vertical direction - it is irrelevant. Any energy gained by dropping will be dissipated into heat and sound or whatever. As is any jumbling around on the belt or any consideration of elastic or inelastic - ignore it all. You could think of the sand being viscous treacle...
Which is why I don't like Archimedes Principle as taught in school - it is a special case.
I responded with pressure because that is what is really happening - the upthrust is caused by the fluid pressure.
Yes the total "horizontal area" is equal to the arae of the circle at the waterline ...
... but the bit of "horizontal area"close to the surface has only a very small pressure applied to it whereas the bit of "horizontal area" close to the tip has a large pressure applied to it as it is deeper...
That is better written as "As long as there is fluid in-contact with surface of the object, where the resultant force from the pressure has a vertical component there will be an upthrust."
Pressure always acts normal to the surface it is pressing on. If a vertical cone is placed point downwards...
The upthrust in Archimedes Principle is caused by the fluid pressure acting on a surface of the body which has a horizontal component.
So, on a floating vertical cylinder, the upthrust is the pressure at the depth to which the cylinder has sunk into the fluid multiplied by the area of the...
The pressure at the surface is atmospheric air pressure which is about 14 lb/in^2.
The absolute pressure in the vessel is the sum of the atmospheric pressure plus the pressure due to the height of the liquid.
The atmospheric pressure also presses against the external surface of the vessel so...
I of course appreciate that this special case circuit can be simplified and solved in my head.
Telling that to the poster helps him or her with this one, special case and is probably quite useless for most other cases.
My solution gave a generic method which can be used in any and every case...
I would tend to solve this problem using "currents in a loop" approach, rather than "currents in a branch" (although it is exactly the same).
See below. Now write the equations for each current loop calculating the voltages as current x resistance.
In the i1 loop, there is a voltage source...
When I was at school I always thought the conservation of momentum was inferior, in some way, to the great conservation of energy law which applies to everything. It was only when I read Principia that I finally understood - that Aha! moment - what momentum is.
Newton defines it beautifully...
I totally agree.
She needed the all the caffeine she could get because, if she stopped on the rim she would have needed to decelerate in, say, 0.3m.
She would have pulled about 6g by applying a force of about 150kg.
Not bad for a 7 year old (the 50 percentile for 25kg girls is 7 1/2 years)...
May I suggest that you think a little more about the angular momentum the child has when she lands on the rim.
Travelling at 6 metres/sec she possesses a huge amount of angular momentum about her feet and will therefore either
a) rotate about her feet and fall towards the centre, in which...
I suggest that the word heading invites the interpretation of continued motion as in "The man is heading for the train station at 6 m/s".
It's a badly worded question.
I am confused.
If the jump is radial the velocity is not necessary - the child lands on the rim without adding angular momentum due to her velocity.. So why give the radial velocity of 6.0 m/s unless the child is intended to continue?
Do the problem in small steps.
Step 1. What happens after the child steps on the roundabout but before she has left the rim?
Step 2. What happens as she moves from the rim towards the centre?
Step 3. What happens when the child reaches the centre? This bit is more tricky as you need to...
That does rather assume that trains run in only one direction.
Whereas a train running north from the equator to the North Pole needs to be decelerated to lose its eastward equatorial velocity of 1.035 miles/hour, a train traveling south from North Pole to the equator needs to be accelerated...
Your speaker diaphragm is 1mm in diameter and you are measuring 10 metres away.
The waves will be launched as planar waves, and be planar very, very close to the diaphragm (a millimetre?) but will become approximately spherical very, very quickly.
Don't expect any base frequencies :smile...
And that assumption opens up a whole can of worms we had better not get into here - it deserves a thread of its own.
And a word of advice for KittyCat1534. If in doubt, give both answers and explain the assumptions leading to them. " If I assume he is a point then ...", and "If I assume he is...
Correct
Not quite correct. It will be rotating slightly slower than originally.
He will be rotating when he is at the centre so he will have some angular momentum. Hence the roundabout must have given him some, so the roundabout must be rotating slightly slower than it was originally...
1. Once the boy is on the roundabout there are no external forces being applied to the roundabout/boy system. You can consider it a closed system (ignoring friction, air resistance etc).
2. If there are no external forces being applied the angular momentum of the system cannot change.
So...
He was not rotating before he jumped on so he had no angular momentum relative to the axis of rotation..
He is rotating when he is standing at the centre so he now has angular momentum.
Where did he get it from? The roundabout which must, therefore, slow down to give him his angular motion...
1. Imagine a train starts traveling north from the equator on a line which runs to the North Pole.
2. Viewed from outside earth, the train has an eastward velocity due to the Earth's rotation. At the equator, this velocity is 24,860 miles/24 hours, or 1,035 miles/hour.
3. Let the train...
Your diagram is not very good for doing an analysis. Where is the eye you were asked to put in? It's all about angles - how can you measure angles in your diagram?
Perhaps some was my error as, when I said draw a picture, I should have said draw a diagram.
You should have drawn a diagram...
Draw a picture with an eye below the surface of the water.
Draw a ray of light from air to the eye, passing through the water. Draw another ray at a different angle of incidence. And another. And another.
What happens as you change the angle of incidence of the ray?
Look up "total...
A disk and a stick are both cylinders.
A uniform disk floats with its axis vertical.
A uniform stick floats with its axis horizontal.
So, as a cylinder's ratio of length to diameter changes, it will transition from one mode to the other.
Googling metacentre / metacentric height will probably...
The effect is quite significant.
Take a nominal one second pendulum with a length of about 40 inches and a spherical bob of radius 3 inches. It has a period some 100 seconds/day longer than a 40 inch simple pendulum which is equivalent to lowering the bob by 0.1 inch.
Not so.
The sphere is in neutral equilibrium. The pencil is in unstable equilibrium.
See wiki for an explanation of the difference between neutral and unstable equilibrium.
" Think of a body which is at rest but not in equilibrium. Give explanation as well as figure/diagram. Relevant Equations: Σ F = x (x is not equal to 0)"
A sphere on a horizontal surface is at rest and in neutral equilibrium. Without a force it will remain at rest indefinitely. The...
The more I read about Newton and what he did the more impressed I am. He had the critical insight on this problem as Definition III in Principia. III in Principia.
We are electronic and electrical engineers respectively so mechanics was not a large part of the syllabus.
The only methodology we have been taught is (it was 50 years ago!)
1. work out what is happening and how it moves by examination
2. once you know what happens calculate the values.
We...
Thank you for your responses as the fog is beginning to lift.
I am still surprised how difficult this problem is for me which is because it uses techniques I haven't been taught so it's back to school for me. Can anyone recommend something I should read or will searching on dynamical equations...
I have been struggling with this as a recreational, not homework problem.
Q1. Does the disk rotate?
Q2. If so, about which axis does it disk rotate?
Q3. And why that axis?
I can make several arguments but cannot see which is correct.
1. The disk rotates about its centre
2. The disk...
I have been struggling with as a recreational, not homework problem.
Does the disk rotate? If so, about which axis does it disk rotate? And why?
I can make three arguments but cannot see which is correct.
1. The disk rotates about its centre
2. The disk rotates about the point opposite...