Recent content by marissa12

oo that's right stupid parentheses.. but i still can't get the first part?

the s is still in the equation though

maybe i screwed up the math..because that didnt work 0.5*m*v^2=(-u_k*m*g*D)-(m*g*D*sin(6)) 9187500=-58860d-146344d 9187500=-205204d d=12.4 and that was still wrong, and i can't see where i messed up in the math

Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loose gravel are constructed to stop runaway trucks that have lost their brakes. The combination of a slight upward slope and a large coefficient of friction in the gravel brings the truck safely to a halt. Suppose a...

the 2nd part of the problem is w= intergral of { mv*dv} between v_final and v_initial and we are supposed to express that in m_vinitial and v_final and i got the answer ((m*v_final)^2-(m*v_initial)^2)/2 and that didnt work. i used the basic intergral formula fint{t*dt between a and b}...

a particle of mass m moving in the x direction at constant acceleration a . During a certain interval of time, the particle accelerates from v_initial to vfinal , undergoing displacement s given by s=x_final-x_initial. Express the acceleration in terms of v_initial, v_final, and s: that was...

hah yay.. i had it right the whole time.. its just i wrote my work soo sloppy i messed up the numbers. thank you !

You see the boy next door trying to push a crate down the sidewalk. He can barely keep it moving, and his feet occasionally slip. You start to wonder how heavy the crate is. You call to ask the boy his mass, and he replies " 51.0 kg." From your recent physics class you estimate that the static...

is the Ft in the cable attached to the block: Ft-fg(2000*9.81)=ma Ft-(2000*9.81)=(2000)(a) if its in equilib. then the net force is zero.

The cable cars in San Francisco are pulled along their tracks by an underground steel cable that moves along at 9.5 mph. The cable is driven by large motors at a central power station and extends, via an intricate pulley arrangement, for several miles beneath the city streets. The length of a...

a disk is 8.0 cm in diameter. A reference dot on the edge of the disk is initially located at theta =45 . The disk accelerates steadily for 1/2 second, reaching 2000 rpm, then coasts at steady angular velocity for another 1/2 second. What is the speed of the reference dot at t=1 ok so you...