Answers by visitors named: Lindsay

So for part A, I got 6.38 hours because I added 3.18, 2.8 (210 km/75) and .40 together. Is this correct?

Ok good. Then, for part B...I got 76.8 for the average velocity. ...I sure hope this is correct...!

Thanks. :)

In order to get the time, would I have to do the vf-vi divided by 2, times the distance?? I guess I really do need a brush-up on algebra!

**Oops, I meant vf+vi. Is that the right equation?

Ok so I plugged in the numbers and got 42.1 for a) and -11.4 for b). However, my book says that I got them switched...that 11.4 s is a) and -42.1 m/s is b). How did this happen?

Ok cool, I got that. Thanks a lot. However, this next one gives me the length of one leg (359) and the adjacent theta of 83. How exactly can I solve this?

I try this and I get 2.62 knots/s^2, but it says this answer is wrong. Is it?

N/m, I needed to convert to knots to meters per second. I got the right answer. Thanks for the help, Quidditch. :)

43 knots, right?

I'm sorry. There was a problem before it where I had to figure out the average acceleration, and that was 1.35 m/s^2. Which fomula can I use to get the length of the runway?

To get how many seconds, would I have to take the 43 knots and divide it by 1.5?

43 knots is 22.097 m/s. So could I now do 22.097 divided by 1.35?

Ok, so I came up with 16.37 s as my answer for the time. So I do I simply plug that number into the equation 'distance=1/2 x a x t^2' and that is my runway length?

THANKS! :)

**Oops sorry, my sister was using this site last night for Chemistry.

Ok great, I get it now. Thanks!

What information must I use to get the velocities?

THANK YOU!

So, am I supposed to figure out v2? And what is g, in this case?

Now, when I'm solving for time, do I not yet know what v/2 is? Like, it would just be "t = v/2 x -9.8?" I'm so sorry...this is just not making any sense to me yet.

What exactly do you mean by "set them equal?"

Ok, the time I got was 2.624 s. Now what?

Which equation am I using for distance?

Right. I plug in "1/2 x 3.05 x (2.58^2") and come out with 10.15 m. But that's obviously not correct. I know I'm doing something wrong, but I'm just not sure what. I'm sorry...this just isn't making much sense yet.

Ohhh ok I see my mistake! Thanks for your patience. :)

Never mind! The mass has nothing do with the problem. I got it. :)

To find the time, would I have to use the formula "x = vt + 1/2at^2?"

I found the time. Now I just have to double it, and that will be the time he spent in the top 15.4 cm?

Ok n/m, I got the answer. However, now it asks for the time in the bottom 15.4 cm. Do I have to use my time from the first problem to figure it out?

When finding the times, do I have to switch the cm to m?

How can I solve for V if I don't yet have the time?

But don't I need time in order to find the initial velocity?

See, I use your method without rounding and now come up with 0.0852 s. So...now I'm a little stuck on which answer is the correct one.

Great I got it, thanks. :)

Thanks, I got it now. :)

THANK YOU!

15.6 m/s. I understand now, thanks a bunch.

Once I then have the initial vel., how can I go about getting the horizintal distance? Don't I need time?

Ahh ok I got it now. Thanks. :)

Ok makes sense. The only part I'm a little confused about is finding the horizontal speed...which equation is that?

But my number was correct, right? 17.714 m/s?

And drwls...where is 0.2041 coming from?

All right. So this is what my equation looks like so far. I need to solve for Vx: 25Vx^2 = 313.786 + Vx^2 Correct?

All right n/m, I finally got the answer. :)

Thanks!

How did you come up with 39.96 Vh^2?

This is probably really obvious, but I'm still not undertanding why you have to subtract one.

Now it asks for the max. height. above the court reached by the ball. Do I use the equation Vf^2 = vi^2 + 2(g)(delta y) ?

Ok so I have the initial speed and the max height. Now it asks "At what initial speed must the ball be hit so that it lands directly on the opponent's back line?" Do I use a process similar to that of the first question?

Would my equation for the x-dir. be Vcos 55*T = 18, since it's the full length of the court? What would my height for the y-dir. be? Vsin 55*T - (g/2)T^2 = ???

V is 41 m/s, right?

I keep getting 64.28 m/s as my answer, but it's not right. What am I doing wrong?

Would 59.5 be the final answer?? B/c that is not right, either...

I keep getting the same answer as you. This is what I did to get my original answer: V^2 = sqrt[Vx^2 + Vy^2] where... Vx = 41 m/s * cos 38 Vy at impact can be calculated using Vy^2 = (41 sin 38)^2 + 2 g H I got 55.56 as my Vy. I have a feeling this is where I got messed up...

You might have to correct me if I'm wrong, but I think... In order to calculate overtime pay, you must take your amount earned per hour and multiply it by 1.5. In your case, Eric's overtime rate would be $13.86/hour ($9.24 x 1.5 = $13.86). For double time, take the amount earned per hour and multiply by 2. So Eric's double time rate would be $18.48/hour ($9.24 x 2 = $18.48). Hope that helps. :)

There really is no formula for this. All you must do is add 8 to both sides. What you do to one side, you must do to the other. w-8 = 9 w-8(+8) = 9(+8) The eights cancel out on this side. Now you are simply left with w = 17. Make sense?

Hmm yeah I figured Vfy was wrong. But I'm still a little confused on which I equation I DO need to use.

I finally got the right answer. And I was using the right equation for Vfy, just not using the right numbers.

16.73 is the final?? Because again, it's not right. Hmm...

Ahh ok I see. There isn't anything I should add to them? I just want to make sure they are as complete as possible.

Great, thanks. That helps a lot. But now I have to do a graph of velocity and the x/y directions. My velocity in the x-direction is a staight line, and the vel. of the y continually decreases. They ask me to explain why each line has the shape that it does in terms of slope and acceleration. For the y, is what you said...that the slope is a constant? What about for the x-direction?

I come up with 13.6 N as my net force. Is this correct?

17? I come up with 19, but you might want to double check that.

C2H6 = 14 valence electrons There are 4 valence electrons in a C, so in 2 C there is 8. H has only one. 8 + (1 x 6) = 14 v.e.

Now it asks for the normal force exerted on the mass by the floor. Do I need to use the acceleration for that?

m is 68 kg?

I come up with 0.96 m/s^2. Is this correct?

N/m I finally got it. Thanks for all your help. :)

For lift I got 1320, and for deltamass I got 304.76. Is this right? If so, do I have to add those to the netforce?

How do I find the mass?

A railroad flatcar is loaded with crates. The coefficient of static friction between the crates and the floor is 0.27. If the train is moving at 59.7 km/hr, in how short a distance can the train be stopped with a constant acceleration without causing the crates to slide? Ok so I understand that mu*mass*g= mass*v^2/2d, and I need to solve for d. But how do I get the mass?

See, I thought that was right, too. But I plug that into my computer and it says it's not right. So I'm not sure what I'm doing wrong...

I come up with 0.233 m as my answer, but that's not right. What am I doing wrong?

So wait...do I use cos 15 to find the acceleration?

Ohh ok great I understand. Thanks a lot. :)

I came up with 5079 N as my answer to a. Is that correct?

Yeppp n/m it is correct. Thanks a lot. :)

Ahh ok, I didn't realize the masses cancel out. But of course you're right. That really was easy. :)

What is V, in this case?

A youngster shoots a bottle cap up a 15.0° inclined board at 1.92 m/s. The cap slides in a straight line, slowing to 0.95 m/s after traveling some distance. If the coefficient of kinetic friction is 0.35, find that distance. So I get that (1/2)MV^2 = MgX*sin 15 + Mg*muk*cos 15*X, but what is V, in this case?

I keep getting 0.315 m, but that's not right. What am I doing wrong?

I've entered this into the computer, but it says it's not right. Hmmmm...

Ok I understand that part. Now it asks for the minimum magnitude of F that will start the block moving up the plane, and the magnitude of F that is required to move the block up the plane at constant velocity. Is it a similar process for both?

49% is 0.49 as a decimal, and 80% is 0.80 as a decimal. 0.49 x 0.80 = 0.392, or 39.2%

Do you come up with an answer of 33.91? Because that's what I get.

Why is it sin and cos of 23.5? I thought the angle was 21.6...

Using 21.6 as the angle for both cos and sin, I come up with 33.91, which is the incorrect answer. Am I still doing something wrong?

That's right. Now it asks for the minimum value of M2 for which the system will remain at rest. Do I still use mgSinTheta-mg*mu*CosTheta?

Ok I got it, thanks. The last part is this: Suppose the coefficient of kinetic friction between the block and the inclined plane is muk = 0.36. If M2 = 217.1 kg, what is the magnitude of the acceleration of M1? Which equation am I using for this?

Ahhh ok that was what was wrong. Thanks a ton. :)

I tried both of the ways and got the same answer. However, the answer I got using bobpursley's equation got me -5.11E-8, when it was really positive. Was there an error in that equation, or did I just mess up with the math somewhere along the way?

Great I got it, thanks for all your help.

Oh wait...do I need to use F = GM1M2/d^2?

Ok I understand that. But what is the distance from craft to earth?

Oh ok I see. Well, my online homework site would only accept it as postive, but it has been known to make errors before.

Actually, that just made me even more confused. I still don't know what I'm supposed to use as the distance from the spaceship to the earth. Thanks, tho.

Right. But w/o knowing the actual mass of the earth and moon, only the ratio, how can I do this?

Ohhh I see now! Sorry, I just wasn't understanding before. Thanks. :)

(6.67E-11)(2.16)(0.0088) / (0.5068)^2 = 4.94E-12 N Is that correct?