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Natali5045456 [20]
3 years ago
5

A sled with a mass of 20 kg slides along frictionless ice at 4.5 m/s. It then crosses a rough patch of snow that exerts a fricti

on force of 12 N. How far does it slide on the snow before coming to rest?
Physics
1 answer:
uysha [10]3 years ago
3 0

Answer:The sled slides 16.875m before rest.

Explanation:

a=\frac{F}{m} =\frac{12N}{20kg}

a=0.6 m/s²

Vf=0=Vi-a.t

t=\frac{Vi}{a} =t=\frac{4.5m/s}{0.6 m/s2} =t=7.5seg

d= Vi.t - \frac{a.t^{2}}{2}

d= 4.5 * 7.5 - \frac{0.6*7.5^{2} }{2} \\\\d=16.875m

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A small mailbag is released from a helicopter that is descending steadily at 3 m/s.
mario62 [17]

<u>Answer:</u>

a) Speed of mailbag after 3 seconds = 32.4 m/s

b) Package is 44.1 meter below helicopter

c) If the helicopter was rising steadily at 3.00 m/s

       Speed of mailbag after 3 seconds = 26.4 m/s

       Package is 44.1 meter below helicopter

<u>Explanation:</u>

a)  We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

   Initial velocity = 3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

   v = 3+9.8*3 = 32.4 m/s

  Speed of mailbag after 3 seconds = 32.4 m/s

b) We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

 Velocity of helicopter = 3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= 3*3+\frac{1}{2} *0*3^2\\ \\ s=9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = 3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= 3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 53.1m

  Distance traveled by package  = 53.1 meter.

So package is (53.1-9)meter below helicopter = 44.1 m

c) Initial velocity = -3 m/s, acceleration = 9.8 m/s^2 and time = 3 seconds.

  v = -3+9.8*3 = 26.4 m/s

  Speed of mailbag after 3 seconds = 26.4 m/s

 Velocity of helicopter = -3 m/s, time taken = 3 seconds, acceleration = 0 m/s^2.

    s= -3*3+\frac{1}{2} *0*3^2\\ \\ s=-9m

    Distance traveled by helicopter = 9 meter.

 Velocity of package = -3 m/s, time taken = 3 seconds, acceleration = 9.8 m/s^2.

  s= -3*3+\frac{1}{2} *9.8*3^2\\ \\ s= 35.1m

  Distance traveled by package  = 35.1 meter.

So package is (35.1+9)meter below helicopter = 44.1 m

4 0
3 years ago
A hiker determines the length of a lake by listening for the echo of her shout reflected by a cliff at the far end of the lake.
ArbitrLikvidat [17]

Answer:

L = 499 m

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        l_{lake} = \frac{\Delta x}{2} = \frac{998m}{2} = 499 m

  • The length of the lake is 499 m.
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