physics-help

[physics-help]

hi,

is it okay to ask a question about everyday life?

My father was making apple jam. He took lots of sliced apples, some sugar and put it all in a pot without heating it. After one night (still room temperature), apple slices are all sugared-through while the sugar itself is a thick liquid substance. How does it work? Sugar doesn't exactly melt in room temperature, does it? The slices aren't desiccated either, where did the water come from?

Hi!

The effect you were seeing was osmosis - that's the movement of moleculed through a semi-permeable membrane. The driving force behind this is the imbalance of the concentration between two solutions on each side of that membrane.

Basically, if you stand in one corner of the room and spray a bit of deodorant, its contents/molecules will, at some point in the future, be equally distributed in the room. It's not a perfect example because of e.g. different masses and disturbances to the air in the room, but everytime you add [something] to a fluid it will want to be equally distributed.

A semi-permeable membrane in that context is a barrier that can't be penetrated by some of the contents of the solution on one side. To give you another example, imagine a group of people standing on one side of a meshed wire fence - the air can flow through, but the fence holds the people back.

Okay back to the apples + sugar. Semi-permeable membranes can be found in a lot of organisms since it's a great way to e.g. regulate cell pressures, so basically a lot of living cell walls are .. semi-permeable. When you add a lot of sugar to the apple slices (= a higher concentration than what is inside the apple), the water inside the apple will try to "dilute" the solution on the outside and create an equilibrium (=sugar concentration within the apple and outside as equal). The sugar did not melt overnight, its presence just caused some of the water to dry from the apples and dissolve the sugar into some thick syrup. Sugar is also hydrophilic, meaning it attacts water, so this is what really gets this process started.

You mentioned that the slices don't seem to be desiccated - I guess that's because the water content of apples isn't actually that high, but if you put e.g. sugar on some fresh strawberries or salt on some cucumber and wait for a while, you'll see that those now def drier.

Here's an article that focuses a bit more on osmosis in the kitchen!

[physics-help]

Hey! I need help! If jake jumped 36 m (horizontally) and was in the air for 2.4 seconds what was his takeoff speed and angle? For the velocity I got 15 m/sec, but my q is how do I find the angle?

Hey Anon!

What you did was calculating the mean velocity in the horizontal direction (Jake jumps 36m wide which take 2.4sec means 15m/sec in horizontal direction). These 15m/s would be the right answer if Jake jumped /only/ horizontally and moved at a constant speed; yet in reality, you jump to a certain height, and your speed in both directions depend on time … 

This isn’t a case of linear motion, this is projectile motion with a takeoff speed and angle. For illustrations and examples I’d rec the wikipedia article: https://en.wikipedia.org/wiki/Projectile_motion

In Jake’s case, we’ve got the total duration of T=2.4s and a total width of that jump of s_x=36m.

First, you can seperate the velocity in horizontal (v_x) and vertical (v_y) direction:

v_x= v * cos(a), wherein v is the takeoff speed and a the takeoff angle

v_y= v * sin(a) - g * t, wherein the latter term includes the influence of gravity

Now you can integrate those equations re: time so you get a relationship between the distance and the time:

s_x= v * cos(a) * t + c_x

s_y=v * sin(a) * t - g/2 * t^2 + c_y

c_x and c_y are the constants that occur with every integration - in our case, we can just see that Jake starts at the point (0,0) and lands on (36,0), so the initial distance in x- and y-direction is c_x=c_y=0.

We know that when Jake lands, it took 2.4s to cover 36m. So if we set t=2.4s, we know that the distance in x-direction is s_x=36m and that he’s on the ground again (where he started), so s_y=0m:

36 = v * cos(a) *2.4

0   = v * sin(a) * 2.4 - g/2 * 2.4^2

To solve these equations, I rearranged to first one to v=… and substituted v in the second equation. My results are a takeoff speed of 19m/s and an angle of 38°.

I hope that helped!

-sorrel

[physics-help]

I need so much help, I just don't understand physics. How do you draw an velocity time graph from an acceleration time graph. I repeat there is no distance time graph. Just an acceleration graph and I have to draw the velocity time graph. Does the formula V= xf-xi/tf-ti help with anything?

Hey!

The acceleration is the 1st deviation of the velocity over time, as in a(t) = dv(t) / dt. In your case you have give the acceleration over time a(t) and you’re looking for the velocity - you need to integrate!

Speaking on graphs, the graph for the velocity over time is a function, where - when you derive the function - you get the acceleration graph. In case your acceleration graph is a polynomial function (like, linear, quadratic, …), you need to upp the degree of your polynomial (so linear acceleration means quadratic function for velocity … and so on).

I hope that helped!

-sorrel

[physics-help]

deveney2306 : Hi I was wondering if you can give me some help. I’m following along with the solution but i can’t figure out where the arm of W and T values come from. ;;;

Hey! I read this like 5 times and tbh ... I’d say according to the sketch given, the point C in right in the middle of the lath. Thus, AC=CB=40cm, half the length. You know AP from the task, so that’s all you need to know for the arms of W and T.

-sorrel

[physics-help]

Could you please explain these formulas? I'm really confused on what they are and how to use them: d=ut+(1/2)at^2 and v^2=u^2+2ad

Hi! Sorry for the late reply, tumblr is very unreliable when it comes to telling me about asks.

These are constant acceleration formulae, or suvats. Basically, as long as your acceleration is constant (clue is in the name), you can substitute in whatever values to find out what you need to know, for example about a projectile.

d = distance/displacement (sometimes written as ‘s’)

u = initial velocity

v = velocity at time ‘t’

a = acceleration

t = time

d=ut+(1/2)at^2 is a good way of finding displacement or time if you don’t have the final velocity, and v^2=u^2+2ad is useful if you don’t have time but need to know final velocity.

We answered questions using suvat here, here and here, so you can see them in action.

[physics-help]

Hi, I'm so confused and I need help on my physics homework. the question is "What is the speed of the roller coaster at point B and C" I know the equation for speed is distance over time but I don't have the info for either, all I know is that point B is 0 meters above ground level and point C is 160 meters above ground level the coaster weighs 80kg, the coaster is going 20m/s at point A (which is 200 meters above ground level) and I'm supposed to Ignore friction.. Please help I'm so confused!!

The trick to this is the conversion of energy from gravitational potential into kinetic. 

The equation for gravitational potential energy is mass x g x height, so when the rollercoaster is at point A it has 80 x 9.8 x 200 = 156800J of energy.

By point B this will all have been converted to KE (as the GPE will be m x g x 0 = 0) so it has gained 156800J of kinetic energy:

KE = 1/2 x m x v^2

156800 = 1/2 x 80 x v^2

v = 62.6

But this is its gain in speed! So at B v = 20 + 62.6 = 82.6 m/s

I hope this helps, see if you can use to same method for the speed at C.

PS, I’m really sorry I’ve been absent for so long - I’ve only been using tumblr on my mobile and it doesn’t inform me when I get a message, I didn’t realise how many messages were building up in the inbox! I’ll try and get to some of the more recent ones soon, but I’ll assume that the older ones no longer need solving - let me know if otherwise!

[physics-help]

Hi I need help for physics can you help me?

Sure, just shoot us a question!

[physics-help]

I am really confused with apparent weight? What is it exactly? Is it always the normal force?

Solution comes from Freddie:

As far as I understand apparent weight, yes, it is the normal reaction force. Scales measure apparent weight/the amount of upward force they are applying on you. Hence why apparent weight increases if you're in a lift accelerating upwards - the lift floor (and the imaginary scales on it) is not only counteracting the downward force of your weight, it's also applying an upward force to accelerate you upwards as well.

Hmm technically apparent weight is the force you exert downwards on any body with respect to which you are at rest (Aaaargh, grammar) - not sure if there would be any situation in which the counteracting force wouldn't be the normal reaction force but it would probably be quite unusual :).

[physics-help]

Hello. My teacher gave us a worksheet with the instructions to 1) draw the coordinates and recurrent given. 2) draw the components using tip to tail method. 3) find the magnitude of each component and give answer with correct units. For the problems all that's given is R=37.0m, 30*north of east. Is this actually possible to complete? Or is my teacher a nuthead?

Well, from my understanding of the question (we haven't done this using the same terms but I'm assuming it's a vector problem):

You need to draw a line, of length 37 'metres' at an angle of 30* to the horizontal. Then make a right angled triangle (what you've just drawn is the hypotenuse) by drawing in the horizontal and vertical components. You can measure these (use the same scale as your original line!) to find their magnitude, or use trig.

Let me know if you need a more detailed explanation :)

[physics-help]

Hi! I'm a high-school student (sorry for my English, I'm not a native speaker). When my teacher started explaining the electricity, I didn't really understand how can the electric potential be independent from the electric charge, as he said, if its formula is V=U/q ...? I hope I made myself understood

The electric potential is the amount of energy transferred by each unit of electric charge. Of course, this can vary, and it does so no matter how much charge there is. Think of 'v' as a constant of the component (depending on its resistance), and therefore if more charge flows through the component (if 'q' increases) then the amount of energy transferred will increase ('u' will increase) so 'v' stays the same.

[physics-help]

Hey, I'm really sorry no one's asks are getting answered at the moment - I was away all summer and am on temporary hiatus while I get my shit together for A2s. Hopefully we'll be up and running soon!

[physics-help]

I have my physics final coming up and I'm terrified it's over ten different topics and I'm finding it difficult to rme e reverythjng

[sirens in the background]

Hey Nonnie, I’ve got the first aid kit ready, what do you want me to do?

[physics-help]

My second question is , "When two identical charged spheres are placed 0.36 m apart, there is a repulsive force between them of 0.024 N. find the charge on each sphere." I know the formula is F = K(q1)(q2)/r^2 and my teacher included the answer on the bottom of the worksheet, which is 5.9 x 10 ^ -7, but I can't figure out how to get the answer since the charges are identical. I've tried substituting 2q, q/2, and just q for q1 and q2 in the equation, but none of them yield the right answer!

Well, Anon, I have no idea what you’ve done wrong because I got the same result…

F = q² / 4*PI*EPSILON*r²

0.024 N = q² / 4*PI*EPSILON* (0.36m)²

q= 5.88*10^-7 C

Have you found your mistake? Not sure if that helped …

[physics-help]

Hi! I have two questions on my physics homework that I can't figure out! The first question is "If you stick a piece of transparent tape on your desk then quickly pull it off, you'll find that the tape is then attracted by an uncharged piece of paper. What is happening in the paper that causes this attraction? What is this phenomenon called?" The second question won't fit, so I'll ask it in a separate message. Thank you!!!

Hi, so let’s take this apart step by step:

When you pull that tape quickly off the desk, there’s something happening called triboelectric effect - that’s when two surfaces of different materials touch and match up their electrochemical potential by exchanging electrons. So now we have a slightly charged tape.

Now, as it is charged, there is an eletrical field. This field has an effect on the paper, which is called electric polarisation. In any material the center of charge is right in the middle, but the elctrons can move freely around the core. Now with the field this center shifts slightly so in the end you have a slightly charged (opposite charge of the tape!) surface of the paper. You can try this out yourself here (at the very bottom of this page there is some sort of interactive graphic - I can’t copy it because the guy wrote it in Java; but it’s really good! Don’t get distracted though just because the site is in German …)

All in all, if you need more visualisation, there is this wonderful video with a comb, a towel and some pieces of paper.

[physics-help]

For that person at the amusement park, I'm pretty sure you could measure the period and then find omega and use SHM equations to find the length of the pendulum (aka the radius)

Thanks, Anon! (:

[physics-help]

So I'm in high school in America. On May 20, 2014 my physics class is going to an amusement to analyze a ride. I have a ride called the "pirate ship." This ride is most like a pendulum. (Search kennywood pirate ship for more information) I'm comparing the forces felt for different snapshots during the ride. I need a way to find the radius for this ride. Now obviously it is a big ride so I can't just measure it and I don't know anything else about the circle.I don't know circumference or diameter

This is one of the more interesting questions I’ve recieved.

This could be done with SHM equations, I hear, but you would need to measure the angular or linear velocity of the ride fist.

Alternatively, you could use the good old trigonometry-and-shadows method, which this page explains:

http://www.wikihow.com/Measure-the-Height-of-a-Tree

This would assume that the ride touches the ground, however. If you can measure the shadow and the angle (you’d need one of those things which you look into and tilt which I can’t remember the name of - anyone?), you can work out the height of the ride (the ‘opposite’ side of the triangle) using tan.

I hope this helps - it’s certainly a tricky problem! If you need and trigonometry help, just tell us.

[physics-help]

Helpp A walker leaves a car park and walks at a steady speed of 1.2m/s for 1 hour. How far did the walker travel? I done that, I got 4320m, the next part of the question; a runner leaves the same car park 40 minutes after the walker and catches up with the walker at a distance of 4km from the car park, what's the runners speed? Ok, so I firstly converted the 4km into m, getting 4000m, then as it's 40minutes to seconds, I did 40x60 and got 2400, speed =d/v so I got 1.7 which is wrong. help?

The first part of the question is right.

For the second part of the question, you need to work out how long the walker takes to travel 4000m:

time = distance / speed = 4000 / 1.2 = 3333 seconds

The runner set off when the walker had done 40 minutes of this walking, so he had to go as far in 40 minutes less time. The amount of time the runner had is:

3333 - (60 x 40) = 933 seconds

We now know he travel 4000m in 933 seconds, therefore:

speed = distance / time = 4000 / 933 = 4.3 m/s

Hope this helps!