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3 years ago

5

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 v

elocity of 3.2 m/s. If the person is 5.15 m from the shore, how long do they take to reach the shore?

1 answer:

MakcuM [25]3 years ago

3 0

Answer:

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.

$L_i = L_f$

Also,

L = MV

where M is de mass and V the velocity.

replacing,

$M_i V_i = M_{fp}V_{fp} + M_{fz}V_{fz}$

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)( $V_fp$ ) + (0.175)(-3.2m/s)

solving for $V_fp$ ,

$V_fp$ = $\frac{(3.2)(0.175)}{69.825}$

$V_fp$ = 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

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3 years ago