gina decided to order some clothes from a catalog. she ordered 3 pairs of jeans at $39 dollars each, 4 t-shirts at 15 dollars ea
1 answer:
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Answer:
$216
Step-by-step explanation:
Do = 0.2. Now you do 270 x 0.2, which equals 54. Finnaly you do 270-54, which equals 216. So now you have your answer.
Answer:
α ≥ 48.2°
Step-by-step explanation:
The complete question is given as follows:
" A skier starts at the top of a very large frictionless snowball, with a very small initial speed, and skis straight down the side. At what point does she lose contact with the snowball and fly off at a tangent? That is, at the instant she loses contact with the snowball, what angle α does a radial line from the center of the snowball to the skier make with the vertical?"
- The figure is also attached.
Solution:
- The skier has a mass (m) and the snowball’s radius (r).
- Choose the center of the snowball to be the zero of gravitational potential. - We can look at the velocity (v) as a function of the angle (α) and find the specific α at which the skier lifts off and departs from the snowball.
- If we ignore snow-ski friction along with air resistance, then the one work producing force in this problem, gravity, is conservative. Therefore the skier’s total mechanical energy at any angle α is the same as her total mechanical energy at the top of the snowball.
- Hence, From conservation of energy we have:
KE (α) + PE(α) = KE(α = 0) + PE(α = 0)
0.2*m*v(α)^2 + m*g*r*cos(α) = 0.5*m*[ v(α = 0)]^2 + m*g*r
0.2*m*v(α)^2 + m*g*r*cos(α) ≈ m*g*r
m*v(α)^2 / r = 2*m*g( 1 - cos(α) )
- The centripetal force (due to gravity) will be mgcosα, so the skier will remain on the snowball as long as gravity can hold her to that path, i.e. as long as:
m*g*cos(α) ≥ 2*m*g( 1 - cos(α) )
- Any radial gravitational force beyond what is necessary for the circular motion will be balanced by the normal force—or else the skier will sink into the snowball.
- The expression for α_lift becomes:
3*cos(α) ≥ 2
α ≥ arc cos ( 2/3) ≥ 48.2°
