To get off a frozen lake, a 70 kg person removes his shoe of mass 0.175 kg and throws it horizontally away from the shore at a velocity of 3.2 m/s. if the person is 5.15 m from the shore, how long do they take to reach the shore?

the person will be in the shore at 10.73 minutes after launch the shoe.

Explanation:

For this we will use the law of the lineal momentum.

Also,

L = MV

where M is de mass and V the velocity.

replacing,

wher Mi y Vi are the initial mass and velocity, Mfp y Vfp are the final mass and velocity of the person and Mfz y Vfz are the final mass and velocity of the shoe.

so, we will take the direction where be launched the shoe as negative. then:

(70)(0) = (70-0.175)() + (0.175)(-3.2m/s)

solving for ,

=

= 0.008m/s

for know when the person will be in the shore we will use the rule of three as:

1 second -------------- 0.008m

t seconds-------------- 5.15m

solving for t,

t = 5.15m/0.008m

t = 643.75 seconds = 10.73 minutes

the old price per ounce is c/4.

the new price per ounce is c/3.6.

the ratio of the old price to the new one is

(c/4)/(c/3.6) = (c/c)×(3.6/4) = 0.9 . . = 9/10 . . "nine tenths" in either form