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Scrat [10]
3 years ago
9

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20

.0 kg. Burt has mass 30.0 kg; Ernie has mass 40.0 kg. Initially, the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Ignore any horizontal force that the water exerts on the log, and assume that neither friend falls off the log.
Physics
1 answer:
Taya2010 [7]3 years ago
5 0

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

   = (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

   = 150/90

   = 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

  = 30/90

  = 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

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Explanation:

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v = \sqrt{2×(-9.8)×(4 - 10)}

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4 0
3 years ago
Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20
Taya2010 [7]

Answer:

The distance the log has moved by the time Ernie reaches Bur is 1.33 m.

Explanation:

give information:

The log is 3.0 m long and has mass 20.0 kg.

Burt has mass 30.0 kg; Ernie has mass 40.0 kg

Ernie has mass 40.0 kg.

to find the distance, first, we have to calculate the center of mass

X = ∑ m x /∑m

   = (20 x (3/2)) + (30 x 0) + (40 x 3)/ (20+30+40)

   = 150/90

   = 5/3

when Ernie walk, the center of the mass is

X = (70 x 0) + (20 x (3/2))/(70 + 20)

  = 30/90

  = 1/3

the distance of log moved = 5/3 - 4/3 = 1.33 m

5 0
3 years ago
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