Recent content by bww

1. A garden hose has an internal diameter of 1.5cm. It's connected to a sprinkler that consists merely of an enclosure with 14 holes, each .20 cm in diameter. The water in the hose moves with a speed of 3.0m/s. At what speed does the water leave the sprinkler? OK, so some relevant equations...

...yeah you're right, you guys are a lot of help here. thanks. i figured it out.

oh, no, I'm terribly sorry, my mistake. I meant that all i knew was that the weight of the wrench was 3N lighter in water. I do not have the density of the wrench. I only know that it weighs 3N less in water and that water's density is 1,000kg/m^3 and I need to find the mass of the wrench. sorry...

how do i know the density of steel?

...i guess I'm still confused.

A steel wrench feels 3N lighter in water. a. what is the mass of the wrench? b. how would your answer change if it took a more dense fluid to make the wrench feel 3N lighter? a. ok so i know that density x volume=mass. The density of water is 1,000kg/m^3. That's about all I have. I don't...

A solid cylinder has a mass of 10kg and a radius of .2 meters. Starting from rest it achieves a rotational velocity of 2 revolutions per second after 4 seconds. a. What torque is required to cause this? I found the Inertia, (.5)(10(.2) =.2 and found alpha= 2*2pi/4 and got 3.142. Then I took...

There is no friction, it is being ignored.

yeah that was part of the question, i already did that part. i think I've got it right, but where to now?

A 1kg ball rolls along a circular track that is banked at 20 degrees. The normal force on the ball is 10.43N, tilted inward. The radius of the track is 10m. a. What is the net force on the ball (include direction)? b. How fast is the ball going for it to complete the turn? possible...

well the potential energy will be 196J right? but the kinetic energy will be 0 when the object is at rest. so i use the difference between the PE and work (196-100=96J). That is my W, so 96J=40N(force of friction)d. 96/40 d=2.4m. Ok, so that works. Thanks a ton. Want to help me out on another...

so if work equals the change in kinetic energy, the work done on the rough patch is 40Nx2.5=100J like before. That is the change in kinetic energy after the block travels 2.5m on the rough patch? so i solve for v in the KE formula again, I get 96=5v^2, divide 96 by 5, take the square root of...

ohh, the relationship between work and energy is W=KE+PE right?

so when the block hits the rough patch it has a velocity of 6.3m/s. The frictional force of 40N acts against it, so it will slow down. Work is Fxd right? so a force of 40Nx2.5m=100. But what do I do with that number? Forces working on the block...gravity? so 9.8N pulling down...

ok, well the problem comes from a sheet of problems where i was solving potential and kinetic energy problems. So, this problem being different than those has thrown me off. I attempted to find the acceleration of the block before it hits the rough patch by using the PE formula of MxGxH (10kg x...