# Search results

1. ### The Forces of Rolling

The question is: A wheel is rolling smoothly on horizontal surface. There is a constant horizontal force of magnitude 10 N. Wheel has mass of 10 kg, and radius of 0.3 m. The accel. of its centre of mass has mag. of 0.6 m/s^2. I am asked to find frictional force in unit-vector notation. This...
2. ### Net Ionic Equation clarification

I obtained a net ionic equn' as follows Ba +2 (aq) + SO4 2- (aq) ----> BaSO4 (s) Now I need clarification for the following question and my answer to it. What is the max amount of solid product that could be obtained by reacting one mole of each reactant? I believe that b/c of the...
3. ### Car Question

Car Question !!! Two Jeeps P and B race along straight lines, across flat terrain, and past stationary border guard A. Relative to the guard, B travels at a constant speed of 20.0 m/s at the angle theta=30.0*. Relative to the guard, P has accelerated from rest at a constant rate of 0.400 m/s^2...
4. ### Help with Stoichiometry and Volumetric Analysis

I need some help with this question. A piece of chalk (mainly calcium carbonate) is placed in 250 mL of 0.293 M HCl. All the CaCO3 reacts, realising CO2 gas, and leaving a clear solution. 50.00 mL of the solution is pipetted into another flask. 87.1 mL of 0.0567 M NaOH is required to titrate...
5. ### Standardization of An Acid

Don't worry about it, I figured it out :D
6. ### Standardization of An Acid

So for a I know that M (molarity) = moles of solute/liters of sol'n
7. ### Standardization of An Acid

It requires 46.5 mL of 0.215M NaOH to neutralize 15.0 mL of aqueous H2SO4. What was the molarity (M) of the acid sample? I'm not quite sure how to approach this quesiton, if anyone has and starter tips please let me know.
8. ### The acceleration of a particle on a horizontal xy plane

alright so i got: dx = 1.5t^2 + vcos0t + 20 dy = 2t^2 + vsin0t + 40 where the 0 is theta
9. ### The acceleration of a particle on a horizontal xy plane

so would i integrate a to find position and then velocity?
10. ### The acceleration of a particle on a horizontal xy plane

The acceleration of a particle on a horizontal xy plane is given by a=3ti + 4tj, where a is in meters per second-squared and t is in sec. At t=0, the particle has the position vector r=(20.0m)i + (40.0m)j and the velocity vector v=(5.00m/s)i +(2.00m/s)j. At t=4.00 s, what are its position...
11. ### Basic Physics Volcano Question

During volcanic eruptions, chunks of solid rock can be blasted out of the volcano; these projectiles are called volcanic bombs. At what initial speed would a bomb have to be ejected, at angle 35* to the horizontal, from the vent at A in order to fall at the foot of the volcano at B, at vertical...
12. ### Basics Physics VECTOR Question

The acceleration of a particle on a horizontal xy plane is given by a=3ti + 4tj, where a is in meters per second-squared and t is in sec. At t=0, the particle has the position vector r=(20.0m)i + (40.0m)j and the velocity vector v=(5.00m/s)i +(2.00m/s)j. At t=4.00 s, what are its position...

if i multiply both sides by 2 to get rid of the 1/2 i get 106 = -(9.8) t^2

i plugged in all my values but get stuck at one point with a negative square root: 53 = -1/2(9.8)t^2

well the max height is 53 m, and it reaches that height when v = 0.0 m/s

so i can solve for t in the y=y0 + v0-1/2gt^2 equation and sub t into x=x0 +v0t equation and solve for x

and b/c im looking for how far its been displaced horizontally from the launch pt. i am looking for x?

Now i have to calculate how far is its displacement horizontally from launch pt, at the instant it achieves max height. So i know that its v will be 0 m/s at max heigh.

So the horizontal component of my initial velocity is 20 m/s, and the vertical component of initial velocity is 36.6 m/s

Alright, so i got 36.3 m/s for v0

:O I just figured something out... Don't know if im on the right track but let me know: If i solve for v0 in equation x=x0 + v0t, then i can sub that into the other equation?

alright so solving that equation would give me v0. Now what do i do with that? would i also need to solve for v0 in equation x=x0 + v0t?

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

you said i left out the vertical displacement at time, so that would be y in the case of this equation y = y0 + v0y*t - 1/2*g*t^2 , and y is 53 m ?

so all the - demonstrates is the negative direction?

its on the ground?