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mass cancels

(mu)(g)(d)=1/2v^2

sqrt((0.27)(9.8)(19)*2)=v

v=10.03 m/s

v^2=2gh

(10.03)^2=2(9.8)h

h= 5.13 m

Ff = 027 * m * 9.8 = m * 2.646

a = -2.646 m/s^2

vf^2 = vi^2 + 2 * a * d, vi = 0

0 = vi^2 + 2 * -2.646 * 19

vi^2 = 100.548

vi = √100.548

This is approximately 10.027 m/s. Since the hill is frictionless, we can use conservation of potential and kinetic energy to determine its height. Let’s determine her kinetic energy at the bottom of the hill.

KE = ½ * m * 100.548 = m * 50.274

PE = m * 9.8 * h

m * 9.8 * h = m * 50.274

h = 50.274 ÷ 9.8 = 5.13 meters

**a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27. ,**a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27.,a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27.### Answers

Best Answer: (mu)(mass)(gravity)(distance)=1/2m(veloc...mass cancels

(mu)(g)(d)=1/2v^2

sqrt((0.27)(9.8)(19)*2)=v

v=10.03 m/s

### that means her initial velocity as she left the slide was 10.03 m/s

v^2=2gh

(10.03)^2=2(9.8)h

h= 5.13 m

### The friction force causes her velocity to decrease from her velocity to 0 m/s at she slides 19 meters.

Ff = 027 * m * 9.8 = m * 2.646

a = -2.646 m/s^2

### Let’s use the following equation to determine her velocity at the bottom of the hill.

vf^2 = vi^2 + 2 * a * d, vi = 0

0 = vi^2 + 2 * -2.646 * 19

vi^2 = 100.548

vi = √100.548

This is approximately 10.027 m/s. Since the hill is frictionless, we can use conservation of potential and kinetic energy to determine its height. Let’s determine her kinetic energy at the bottom of the hill.

KE = ½ * m * 100.548 = m * 50.274

PE = m * 9.8 * h

m * 9.8 * h = m * 50.274

h = 50.274 ÷ 9.8 = 5.13 meters

a child sleds down a frictionless hill with vertical drop h. at the bottom is a level stretch where the coefficient of friction is 0.27.

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