Login or Sign Up

Ask a New Question

A skier is going down a hill with a coefficient of friction of 0.21 and an angle of 7.5 degrees. if the skier has an initial velocity of 30 m/s, how fast will she be going after 10 s

vf=vi+at

but a= netforce/mass

a= (mg*cosTheta-mg*mu*sinTheta)/m
a= g(cosTheta-mu*sinTheta

solve for vf.

Similar Questions

A 50.0-kg skier starting from rest travels 200 m down a hill that has a 20.0° slope. When the skier reaches the bottom of the

Top answer: Oh, downhill skiing! Let's calculate the amount of work done by friction and the magnitude of the Read more.
A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates.

Top answer: Well, this winter wonderland problem seems like a snowed-in brain teaser! So let me clown around Read more.
A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates.

Top answer: To determine the height h of the first hill so that the skier just loses contact with the snow at Read more.
A skier starts from rest at the top of a hill. The skier coasts down the hill and up a second hill, as the drawing illustrates.

Top answer: The speed V at the crest of the second hill must satisfy V^2/R = g where g = 9.8 m/s^2 and R = 39.6 Read more.
The coefficient of staticfriction between a skier and the snow is 0.16, with a coefficient of kinetic friction of 0.13. The

Top answer: M*g = 550 N. = Normal force(Fn). Fs = u*Fn = 0.16 * 550 = 88 N. = Force of static friction. Fap-Fs = Read more.
A skier starts from rest at the top of a hill that is inclined at 10.3° with the horizontal. The hillside is 210 m long, and

Top answer: energy from gravity: mgh=mg(210sin10.3) energy into snow: 210*mg*cos10.3+mgL set the two energies Read more.
A skier starts from rest at the top of a hill that is inclined at 9.9° with the horizontal. The hillside is 160 m long, and the

Top answer: the potential energy the skier had, mg*160Sin9.9 is all converted to heat in friction. PE= friction Read more.
A skier starts from rest at the top of a hill that is inclined at 9.8° with the horizontal. The hillside is 170 m long, and the

Top answer: during the slide down the hill: use a=gSin9.8-mu*gCos9.8 to find vel. at the base of the incline. Read more.
a 60 kg skier with an initial velocity of 12 m/s coasts up a hill. at the top of the hill, the skier is traveling at 9.75 m/s.

Top answer: Okay so GPE = KE (Gravitational Potential Energy = Kinetic Energy) GPE = mgh (Gravitational Read more.
A 60 kg skier with an initial velocity of 12 m/s coasts up a hill. At the top of the hill, the skier is traveling at 9.75 m/s.

Top answer: B. 0.25 m Read more.