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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...

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...

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 at...

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...

Don't worry about it, I figured it out :D

So for a I know that M (molarity) = moles of solute/liters of sol'n

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.

alright so i got: dx = 1.5t^2 + vcos0t + 20 dy = 2t^2 + vsin0t + 40 where the 0 is theta

so would i integrate a to find position and then velocity?

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...

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...

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 I am 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 I am 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?

wait.. wouldn't gravity be -9.8 m/s^2 so that the answer would be 19.6 ? and i don't know what units it'd be

so: - initial position (y0) would be 0 m - initial velocity (v0) would be 0 m/s - time would be 2.0 sec - force of gravity 9.8 m/s^2 so my answer would be -19.6 m/s??

y = y0 + v0y*t - 1/2*g*t^2 ?

now how do i find vertical component?

alright, i know that v = d/t so i'd get v = 40m/2sec = 20 m/sec but do i use the horizontal distance or vertical distance ? and what does velocity give me? is that the horizontal component?

or is that only when an object is thrown directly upwards and falls directly downwards.

can i do this: y = 1/2 * g* t^2 y = 1/2 (9.8) (2)^2 y = 19.6

how do i find the horizontal velocity component?

so would saying the following make any sense: the horizontal velocity is constant so the projectile is moving 40m/s right the vertical velocity changes 9.80 m/s downward because of gravity

2 seconds after being projected from ground level, a projectile is displaced 40 m horizontally and 53 m vertically above its launch point. What are the horizontal and vertical components of the initial velocity of the projectile? At the instant the projective ahieves maximum height above...

Alright so based on the website you provided and my notes i calculated the average speed to be 644 km/h and the average velocity to be 214.7 km/h in the south east direction. v = (483 + 966)/2.25 = 644 km/h vave = (966-483)/2.25 = 214.7 km/h south east What do you think??

Basic physics help Alright, so to find the magnitude and direction of the plane's average velocity i used the vave = delta r/delta t formula to find average velocity for vx and vy and then i used pythagorean theorem to find v and i got 910.8 km/h... does that make any sense? Now I am assuming...

based on your last reply, what's the definition for magnitude? b/c now I've got to find magnitude and direction for average velocity and then average speed

Basic physics help ok.. so i used pythagorean theorem and got this: r = 1080.020833 which i'll round to 1080 for 4 sig figs. Now finding r gave the the magnitude would i use the tan formula to find direction? now i found the angle to be = tan-1 (966/483) and got 63.4 * but how do i...

ok.. so i used pythagorean theorem and got this: r = 1080.020833 which i'll round to 1080 for 4 sig figs. Now finding r gave the the magnitude would i use the tan formula to find direction?

Basic physics help A plane flies 483 km east from city A to city B in 45.0 min and then 966 kim south from city B to city C in 1.50 h. For the total trip, what are the magnitude and direction of the plane's displacement, the magnitude and direction of its average velocity, and its average...

Would this be correct: I calculated r the distance between points A and C by r = (483 km) + (- 966 km) to get - 483 km So the planes magnitude was 483 km in the negative direction, or south east. Is this correct?

A plane flies 483 km east from city A to city B in 45.0 min and then 966 kim south from city B to city C in 1.50 h. For the total trip, what are the magnitude and direction of the plane's displacement, the magnitude and direction of its average velocity, and its average speed?

a = (3t-15.0)/t is what i got for a Now i need to sub that into above to solve for t

1/2at2 - 1/2(1.5)t2 + 15t -35.0 = 0 what do i do from there

i've got another one too.. A motorcycle is stationary at traffic light. At t=0s, light turns green and it accelerates at 3m/s2 when light turns gree, a car is 40 m behind the motorcycle and traveling at 15m/s. At that instant, car slows down at a constant rate so it is never closer than 5 m to...

HELP ! Assignment due in an hours A gold ball is hit from ground @ 25* to the horizontal. When it lands, it has a velocity of 20 m/s @ 30* below horizontal. A) find initial speed B) time that ball is in air C) height of landing pt. relative to start pt, which may be positive or negative

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i've figured out an answer, not sure if its right but if anyone's willing to check here is it: The explorer must travel 12. 61 km. I got 67* but that doesn't entirely make sense now that i look at it. Anyhoo let me know what you think.