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8090 [49]
2 years ago
15

A carpenter lifts a 10-kg piece of wood to his shoulder 1.5 m above the ground. He then sets the wood on his truck at 1.0 m abov

e the ground and makes his delivery going 10 m/s. The kinetic energy of the wood during delivery is _____ J
Physics
1 answer:
FinnZ [79.3K]2 years ago
8 0

Answer:

<em>500Joules</em>

Explanation:

Kinetic energy = 1/2mv²

m is the mass of the wood

v is the velocity

Given

Mass = 10kg

Velocity v = 10m/s

Substitute into the formula and get KE

KE = 1/2 * 10 * 10²

KE  = 1/2 * 1000

KE = 500Joules

<em>Hence the kinetic energy of the wood during delivery is 500Joules</em>

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A 5.0 kg block hangs from the ceiling by a mass-less rope. A Second block with a mass of 10.0 kg is attached to the first block
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The tension in the first and second rope are; 147 Newton and 98 Newton respectively.

Given the data in the question

  • Mass of first block; m_1 = 5.0kg
  • Mass of second block, m_2 =10kg
  • Tension on first rope; T_1 =\ ?
  • Tension on second rope; T_2 =\ ?

To find the Tension in each of the ropes, we make use of the equation from Newton's Second Laws of Motion:

F = m\ *\ a

Where F is the force, m is the mass of the object and a is the acceleration ( In this case the block is under gravity. Hence ''a" becomes acceleration due to gravity  g = 9.8m/s^2 )

For the First Rope

Total mass hanging on it; m_T = m_1 + m_2 = 5.0kg + 10.0kg = 15.0kg

So Tension of the rope;

F = m\ * \ g\\\\F = 15.0kg \ * 9.8m/s^2\\\\F = 147 kg.m/s^2\\\\F = 147N

Therefore, the tension in the first rope is 147 Newton

For the Second Rope

Since only the block of mass 10kg is hang from the second, the tension in the second rope will be;

F = m\ * \ g\\\\F = 10.0kg \ * 9.8m/s^2\\\\F = 98 kg.m/s^2\\\\F = 98N

Therefore, the tension in the second rope is 98 Newton

Learn More, brainly.com/question/18288215

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If only 10 pounds is required to lift a 500-lb block, how much chain must be played out to lift the engine 3.0 inches?
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Answer:

150 inches (12.5 ft)

Explanation:

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The following is the equation one needs to solve:

10 \,lb\,* \,x\,=\,500\,lb\,*\,3\,in\\10 \,lb\,* \,x\,=\,1500\,lb\,in\\

therefore solving for the distance "x" gives as the answer (in inches):

10 \,lb\,* \,x\,=\,1500\,lb\,in\\x\,=\,\frac{1500\,lb\,in}{10\,lb} \\\\x\,=150\,in

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