physics questions
-
Hi, i will give 2 scenarios that i dont understand, if there are any mistakes pls correct them and tell me which forces acts on the box/hand and why. Thanks for the help
1st Example: when bob holds a box flat on his hand and it is without motion. that means the FBD of the box is weight down and contact force of the hand up on the box right?
Qns 1) that means the FBD of the box is weight down and contact force of the hand up on the box right?
Qns 2) the FBD of the hand will be contact force on hand by box but what else will there be since the hand is also motionless, there is no net force. so when i lift an object up the FBD of the hand will be contact force of box onto hand (Down) and upwards force by hand?
Conclusion question: the contact force on the box is NOT the force acting on the box? while whenever i hold an object i am NOT exerting a force onto the object, rather i am exerting a force on my hand to counter the contact force on my hand by the box. Is these conclusions correct? thanks
example 2: bob then moves his hand upwards with the box still flat on his hand.
Qns 3) the FBD of the object changes to gain a net upwards force but what are all the forces acting on it? is it that the contact force increases? i dont think that that is possible so im completely unsure in this case.
Qns 4) the FBD of the hand will be what? if the contact force increases that means that the contact force acting on the hand by the box will also increase so does that mean that the upwards force by the hand be a lot more? im not sure at all about this situation.
Please explain them to me n thank you for helping me with this.
-
1st eg
1)yes you are right.
2) in addition to contact force downwards of box on hand, there is weight of hand. and the hand is supported by the body where there is ultimately a normal contact force of ground on body.
conclusion:???why not? i mean newton's third law states that forces occur in action reaction pairs.
2nd eg
1) yeah the normal contact force of the hand on box increases, its possible.
2) the normal contact force of the ground on the feet will increase. think of the situation when you raise your hands and push against the ceiling.
-
Ok this is what i think of the FBD all of the sitution
i think that the Contact force can change because of SgStudentStressed post
http://postimage.org/image/41evs2mpt/
in eg1 part 1) the contact force and weight cancel out
part 2) the contact force which is equal to the contact force in part 1+weight=upwards force
in eg2 part 1) the contact force is greater than weight so net upwards
part 2) the contact force and weight is less than the upwards applied force. the contact force is also equal to part 2)
so the applied force onto hand in both case is the greatest force?
-
Hi I have a question regarding longitudinal waves I hope u all will help me clarify my doubts. 1) when plotting a displacement vs distance graph is the disrance the resting position of the particle? 2)when plotting a pressure distance graph is the distance the resting position if the particle? And is the point where the particles change from a moving left ribs right direction the position that have average pressure? Meaning when one rarefraction ends when a compression occur the boundary there where it switches vibration direction? Thanks
-
1) No. A particle rests at its amplitude, and has highest speed at its equilibrium position (displacement = 0)
2) No also.
For displacement - distance or pressure - distance graphs, the distance refers to the distance of the particle from a fixed zero position, usually the distance from the first particle.
-
Originally posted by eagle:
1) No. A particle rests at its amplitude, and has highest speed at its equilibrium position (displacement = 0)
2) No also.
For displacement - distance or pressure - distance graphs, the distance refers to the distance of the particle from a fixed zero position, usually the distance from the first particle.
Um ill use an example. If a particle vibrate to the right by 1cm. Is the displacement 1cm while distance be 0? Thanks -
let ne rephrase the question when plotting displacement distance graphs for longitudinal waves is the distance the position of equilibrium while displace is how much it deviate from the equilibrium position. like a resting point is 0 distance if a particle move 1cm the distance is 0 and displacement is 1? Cos that's how my textbook dots it
-
Originally posted by StudentQns:
Um ill use an example. If a particle vibrate to the right by 1cm. Is the displacement 1cm while distance be 0? ThanksNo. Two errors in this statement:
1) Displacement is a vector, so there is both magnitude and direction. Is it positive to the right or to the left? You have to define it.
Or you have to say it is displacement 1cm to the right.2) Distance, as previously mentioned, refers to the distance of the particle from a fixed zero position, usually the distance from the first particle. So your distance is only zero if you are referring to the very first particle where you arbituary define as zero distance.
-
Originally posted by StudentQns:
let ne rephrase the question when plotting displacement distance graphs for longitudinal waves is the distance the position of equilibrium while displace is how much it deviate from the equilibrium position. like a resting point is 0 distance if a particle move 1cm the distance is 0 and displacement is 1? Cos that’s how my textbook dots it
The textbook is correct.
Distance is the position of the equilibrium position of the particle from a fixed zero location.
Displacement is the deviation from equilibrium position.
And final sentence yes.
-
oh ok thanks for the help! i have another physics question to ask. its under electromagnetism which the school haven't taught so this is all self taught. When a magnet attracts a bunch of pins directly it can attract 10 of them. when a iron block is attracted first, only 8 pins can be attracted. 1) so does this mean that the induced magnet (iron block) is weaker than the magnet itself? 2) if 1) is right, (i think so) then wont there still be the same attraction in total? as taking the attracting side of the magnet to be north, the induced south pole will be weaker. so some attraction will still be left on the magnet? This is what i think for 2): (taking the attracting side to be north pole)The forces of attraction on the magnet directly (CASE 1) is stronger than the forces of attraction on the iron block and on the pins(CASE 2). this is because the induced magnet is weaker than the magnet itself. however, as the induced magnet is weaker, there will still be a net north pole attraction force at the magnet itself as the south pole on the iron block cannot completely cancel itself out with the north pole on the magnet as the induced magnet's south pole is weaker than the north pole on the magnet itself. however, the net attraction in case 1 is still stronger as the pins are directly attracted to the magnet, while in the case 2, besides the weaker induced magnet's attraction, the left over attractivce force by the magnet will be weakened as the pins are further away from the magnet than in case 1. thus due to the increased distance the attraction of the magnet itself attracting the pins will weaken. this causes the net attraction in case 1 to be stronger than the net attraction in case 2. key point: so the key idea of my own understanding 1) induced magnet is weaker 2) case 2 weaker due to the extra distance to attract the iron pins which causes the attractive forces to weaken too. hence the forces of attraction is weaker unlike in case 1 where they are directly attracted to each other. thanks for all the help guys! i really appreciate it. currently i do not have a physics tuition teacher so i really need to rely on the forum to answer my questions :) thanks!
-
I need to see the graphic of how the iron block is attracted.
Typically, a magnet, be it induced or not, has the strongest magnetism at its poles. If the pins are attracted at the centre of the iron block, it follows that the magnetic field will be at its weakest, and hence attract lesser pins.
-
Originally posted by eagle:
I need to see the graphic of how the iron block is attracted.
Typically, a magnet, be it induced or not, has the strongest magnetism at its poles. If the pins are attracted at the centre of the iron block, it follows that the magnetic field will be at its weakest, and hence attract lesser pins.
no, I mean that a magnet can attract 10 pins but when a magnet attracts a iron block which then attracts the pins, the iron block end attracts only 8 pins. since the induced magnet is weaker, then won't there be through same net attractive force? As for eg if the magnets end is north then the end of the iron block touching it will be south. But since the south pole is weaker than the north pole of the magnet itself, won't there be the same attraction as the north and south will not completely neutralize itself? -
I'm not sure how it works, but I guess it might be weaker due to the individual magnetic domains not perfectly aligned in the induced magnet.
More info on magnetic domains:
-
oh ok thanks for the help. Also, i have a physics questions about sound Sound waves are used to detect faults in girders. a cro is used to detect these faults, a pulse producer and a detector is placed on opposite sides of the detector.
The safe detector has a upwards spike at 0microseconds and at 10 microseconds.
The unsafe detector has a upwards spike at 0 microseconds and 10 mircroseconds and also one additional but smaller upwards spike at 4 microseconds.
why is the the unsafe detector considered unsafe?
my understanding: there is a crack in the girder so when the sound waves is sent in, some of the waves will have to go around it so it causes it to be slower than the sound waves which do pass the crack. because it is slower, so it is shown in the 4 microseconds. meaning that if another pulse is sent out one of the recorded one will be at 10microseconds from the previous one and another which is 4 microseconds ahead of just detected pulse.
is this correct? if not pls tell me why is it unsafe.
for the mods of this forum- i post here cos i do not have a physics tutor right now. i cant ask my school teacher because shes very mean and favours some students only. unfortunately im not one of them. when 3 of my classmates (including me) wanted ask our own respective question, she answered the 2 of them but when i wasnted to ask mine, she said she needed to meet another student -.- she just went back to the staffroom. so i need all the help i can get. thanks for the help!
-
Originally posted by StudentQns:
oh ok thanks for the help. Also, i have a physics questions about sound Sound waves are used to detect faults in girders. a cro is used to detect these faults, a pulse producer and a detector is placed on opposite sides of the detector.
The safe detector has a upwards spike at 0microseconds and at 10 microseconds.
The unsafe detector has a upwards spike at 0 microseconds and 10 mircroseconds and also one additional but smaller upwards spike at 4 microseconds.
why is the the unsafe detector considered unsafe?
my understanding: there is a crack in the girder so when the sound waves is sent in, some of the waves will have to go around it so it causes it to be slower than the sound waves which do pass the crack. because it is slower, so it is shown in the 4 microseconds. meaning that if another pulse is sent out one of the recorded one will be at 10microseconds from the previous one and another which is 4 microseconds ahead of just detected pulse.
is this correct? if not pls tell me why is it unsafe.
for the mods of this forum- i post here cos i do not have a physics tutor right now. i cant ask my school teacher because shes very mean and favours some students only. unfortunately im not one of them. when 3 of my classmates (including me) wanted ask our own respective question, she answered the 2 of them but when i wasnted to ask mine, she said she needed to meet another student -.- she just went back to the staffroom. so i need all the help i can get. thanks for the help!
poor thing.then go and find a physics tutor or go to eagle's tuition centre lah. I am sorry i not that strong in pure physics yet,cant help you .cheers
-
Originally posted by StudentQns:
oh ok thanks for the help. Also, i have a physics questions about sound Sound waves are used to detect faults in girders. a cro is used to detect these faults, a pulse producer and a detector is placed on opposite sides of the detector.
The safe detector has a upwards spike at 0microseconds and at 10 microseconds.
The unsafe detector has a upwards spike at 0 microseconds and 10 mircroseconds and also one additional but smaller upwards spike at 4 microseconds.
why is the the unsafe detector considered unsafe?
my understanding: there is a crack in the girder so when the sound waves is sent in, some of the waves will have to go around it so it causes it to be slower than the sound waves which do pass the crack. because it is slower, so it is shown in the 4 microseconds. meaning that if another pulse is sent out one of the recorded one will be at 10microseconds from the previous one and another which is 4 microseconds ahead of just detected pulse.
is this correct? if not pls tell me why is it unsafe.
for the mods of this forum- i post here cos i do not have a physics tutor right now. i cant ask my school teacher because shes very mean and favours some students only. unfortunately im not one of them. when 3 of my classmates (including me) wanted ask our own respective question, she answered the 2 of them but when i wasnted to ask mine, she said she needed to meet another student -.- she just went back to the staffroom. so i need all the help i can get. thanks for the help!
Sorry I don't understand your question.
Please do not paraphrase the question. Give the question in its original form. Include graphics where possible because certain questions require the graphics for understanding.
Discrepancy in your question:
Your question mentions a detector. However, later, there is a safe detector and an unsafe detector.
Furthermore, I don't think there is such a thing as a safe detector or an unsafe detector.
-
StudentQns posted this question at physicsforum.com
here's the illustration
-
Second image link:
-
Yea I posted there to, I also have another physics Qns about em waves n penetration ability. I learnt that higher frequency means higher generative ability but then y microwave can cook faster than infrared waves? I thought that infrared waves are more generative so they should be able to cook faster? I have researched quite a lot on this n I have no gotten.a answer such as Wikipedia. Thanks for the help all of u. My physics tutor is injured so he's unable to teach me now. N my school teacher is very bad to me. So im completely replying on this forum for ky ohysics qns. Thanks again for all ur help
-
Hi, the reason why microwaves are used for cooking is out of O level syllabus.
In a nutshell, it is based on the concept of resonance. The frequency of microwaves are very similar to the natural frequency of the water molecules present in food. When microwaves are passed through food, due to resonance, water molecules in food vibrate vigorously, generating heat which will heat up the food.
-
Oh ok thanks. I was revising on practical electricity and I thought of this. they say as resistance increases more heat is produced as P=I^2R but then power is equal to V^2/R too. So as R increases P increases but also as R increases P decreases. so j don't really understand this. thanks for the help and happy Chinese new year
-
P is not directly proportional to R.
As R increases, V and I may change accordingly. It then depends on how much they change before you can decide whether power will change.
-
oh ok i understand. um, i have another question now its related to dynamics. I'm not very sure if its part of my O level syllabus but I'm quite confused about this, here it is: For example, when I punch a wall, what determines the amount of "impact" on the wall? Is it force (or something else, like momentum P=mv?!)? Since F=ma, the force generated would be proportional to how much mass my hand has as well as its acceleration. However, it seems counterintuitive to say that my punch would not produce a force if my hand moves at a constant velocity (a=0). Certainly, a car moving at constant velocity would produce an "impact" when it hits a passer-by. Is the acceleration in F=ma then referring to the decceleration upon my hand hitting the wall? From e.g. 2m/s to 0m/s. If this is so, then whether or not previously my hand was accelerating or not doesn't matter, only the 'final velocity' immediately before impact. However, since a = (final - initial velocity)/time taken, what is the time taken for the decceleration then? Is it instantaneous? How could it be, since that would entail the denominator of the equation being 0? Even if it's a minute fraction of a second, doesn't this mean that the acceleration would be quite huge (e.g. -2/0.001) and so will the resulting force? This doesn't seem right since a punching hand shouldn't cause so much force? Thanks so much for helping!
-
For example, when I punch a wall, what determines the amount of “impact” on the wall? Is it force (or something else, like momentum P=mv?!)? Since F=ma, the force generated would be proportional to how much mass my hand has as well as its acceleration. However, it seems counterintuitive to say that my punch would not produce a force if my hand moves at a constant velocity (a=0). Certainly, a car moving at constant velocity would produce an “impact” when it hits a passer-by.
This is under the concept of impulse, which is about the time interval in which a force is delivered.
It is not in O levels syllabus, but in A levels syllabus.
Is the acceleration in F=ma then referring to the decceleration upon my hand hitting the wall? From e.g. 2m/s to 0m/s. If this is so, then whether or not previously my hand was accelerating or not doesn’t matter, only the ‘final velocity’ immediately before impact. However, since a = (final – initial velocity)/time taken, what is the time taken for the decceleration then? Is it instantaneous? How could it be, since that would entail the denominator of the equation being 0? Even if it’s a minute fraction of a second, doesn’t this mean that the acceleration would be quite huge (e.g. -2/0.001) and so will the resulting force? This doesn’t seem right since a punching hand shouldn’t cause so much force?
Yes, the deceleration can be quite large. The shorter the time, the larger the force. This is the basis for impulse of a force.
-
Oh ok thanks for the help eagle! I have some magnetism questions hope u dont mind 1) how will the magnetic field look like when a bar magnet is placed on a iron wall? The north and south pole are in contact with the iron wall. I know that that bar magnet will be attracted so I know that a north and south pole will form. However, I'm not sure how it will look like. I would greatly appreciate it if u could explain this to me with a diagram. Also, I'm not very sure because the north and south pole are made up and since they have the same properties, the founder of magnets can name the poles which ever way way. So I'm not sure how to draw it.