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A 20.0 g bullet strikes a 0.600 kg block attached to a fixed horizontal spring whose spring constant is 6.74 103 N/m and sets it into vibration with an amplitude of 21.2 cm. What was the speed of the bullet before impact if the two objects move together after impact? I've been stuck on this...

thanks chi meson...i got part A as -94485 J, but I'm having problems with part B. I'm still confused about how to find the average force exerted on him by the snow to stop him..help please!

A paratrooper pilot fell 370 m after jumping without his parachute opening. He landed in a snowbank, creating a crater 1.0 m deep, but survived with only minor injuries. Assume the pilot's mass was 75 kg and his terminal velocity was 50 m/s. (a) Estimate the work done by the snow in bringing...

thanks guys! i understand b now..but I'm still stuck on c. is this the right equation for part c? Wfr = Ffr (dcos30) = .40(300)(-9.8) (4.6cos30) = 4684.85 J?

A 300 kg piano slides 4.6 m down a 30° incline and is kept from accelerating by a man who is pushing back on it parallel to the incline. The effective coefficient of kinetic friction is 0.40. (a) Calculate the force exerted by the man. i got F = mu(k)mgcos(angle) + mgsin(angle) = 0 so F =...

so are my two unknowns Ffr and a?

A 1300 kg car rounds a curve of radius 70 m banked at an angle of 12°. If the car is traveling at 88 km/h, how much friction force is required? ok so this is what i did so far but i got stuck. the sum of Fy = mg - Fncos(angle) + Ffrsin(angle) = ma = 0 the sum of Fx = Fnsin(angle) +...

thanks tom and cscott! i understand it much better now

so i know I'm trying to find the acceleration..so would i plug in the answer i got for the mass? Fnet = 749.7 = ma Fnet = 749.7a?? but how can i solve from here if i don't know the Fnet and I'm looking for a?

can someone tell me if I'm doing this right for part A? Fnet = 102(9.8) - (.25)(102)(9.8) Fnet = 999.6 - 249.9 Fnet = 749.7 = ma so if I'm doing this right..where do i go from here?

(a) What is the acceleration of two falling sky divers (mass 102.0 kg including parachute) when the upward force of air resistance is equal to one-fourth of their weight? ________ m/s2 (downward) (b) After popping open the parachute, the divers descend leisurely to the ground at constant...

A person stands on a bathroom scale in a motionless elevator. When the elevator begins to move, the scale briefly reads only 0.67 of the person's regular weight. Calculate the acceleration of the elevator. how do i start this problem? help please!

thanks cscott! its correct!

webassign is a homework program on the internet...it tells you right away if you're wrong or right...but doesn't give you the right answer until after its due or if you get the answer right.

sorry...i don't have the answer either..it's due tomorrow night..hmmm...i wander why it won't work...

it doesn't say...its a webassign problem

i'm still confused with this problem..i thought i could just multiply 1800kg by .60 m/s^2? but i don't get the right answer... help please!

How much tension must a rope withstand if it is used to accelerate a 1800 kg car vertically upward at 0.60 m/s2? could someone tell me how to start this problem? thanks so much!

so V1y = -40 and V1x = 80.6 and V2y = 0, right? so Vy = -40 +0 = -40? so how do i find V2x?

what are my original components and my resultant components?

so i understand part a, but I don't understand how to do part b. please help! You are given a vector in the xy plane that has a magnitude of 90.0 units and a y component of -40.0 units. (a) What are the two possibilities for its x component? I got 80.6 and -80.6. (b) Assuming...

ok so i understand the first question now. but I'm still confused about if a vector of magnitude zero have a nonzero component?

During baseball practice, a batter hits a very high fly ball, and then runs in a straight line and catches it. Which had the greater displacement, the player or the ball or neither? Explain. I think it's neither but I'm not sure and i don't understand why. could someone please explain...

still having trouble. so here's what i got..but i got them wrong! V1x= -8.94 V1y = 0 V2x= 4.13cos(35) = 3.38 V2y= 4.13sin(35) = 2.37 Vx= -8.94 +3.38 = -5.56 Vy= 0 + 2.37 = 2.37 V = [the square root of (-5.56)^2 +(2.37)^2] = 6.04 tan (angle)= 2.37/-5.56 = -.426 inverse tan...

ohh ok thanks! but what about the V2x and V2y components. what did i do wrong? because i got those components wrong too so i can't do part B until i finish part A...help please!

Vector V1 is 8.94 units long and points along the -x axis. Vector V2 is 4.13 units long and points at +35.0° to the +x axis. (a) What are the x and y components of each vector? V1x = V1y = V2x = V2y = (b) Determine the sum V1 + V2. Magnitude ________ Direction ______°...

someone help please! i don't know where to go from here! (see above message)

i know that V1x = -150 and V1y = 0. but I'm confused how to find V2x and V2y. do i use the Vx=V1x + V2x? but i don't know what Vx is. and i know that to find Vx you use Vx=Vcos(angle) but i don't have V either. where do i go from here?

vector problem..PLEASE HELP! A car is driven 150 km west and then 30 km southwest. What is the displacement of the car from the point of origin (magnitude and direction)? i have no clue how to start..could someone help me?

Vector V1 is 8.94 units long and points along the -x axis. Vector V2 is 4.13 units long and points at +35.0° to the +x axis. (a) What are the x and y components of each vector? so i got V1x=-8.94 and V1y=0 which is right but i don't know how to get the V2x and V2y i got V2x=4.99 or 5.00...

ohh! thank you so much! it didn't sound right that a cliff was 3.6m high and he hit the water at 44.1m! lol

need help please! what am i doing wrong?? A diver running 1.2 m/s dives out horizontally from the edge of a vertical cliff and reaches the water below 3.0 s later. How high was the cliff and how far from its base did the diver hit the water? so does v = 1.2 m/s and t=3 seconds i used the...

hey could someone tell me how to start this problem? i don't know which formula I'm suppose to use.. A 90 m long train begins uniform acceleration from rest. The front of the train has a speed of 19 m/s when it passes a railway worker who is standing 190 m from where the front of the train...