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EleoNora [17]
2 years ago
11

Jeff is a landscaping contractor and lifts a rock weighing 600 pounds by wedging a board under the rock. Jeff weighs 150 pounds

and puts all of his weight on the lever. How much mechanical advantage did the lever provide to Jeff in lifting the rock? A) 0 B) 2 C) 4 D) 8
Physics
2 answers:
jonny [76]2 years ago
6 0

Answer: 4

The mechanical advantage is the ratio of the force exerted  by the object to the force applied to do work on it.

Here, Jeff tried to lift a rock weighing 600 pounds by wedging board under the rock. Jeff who weighs 150 pounds uses all his weight to exert force on lever and lift rock.

Mechanical advantage, M.A.=\frac{weight\hspace{1mm}of\hspace{1 mm}rock}{weight\hspace{1mm}of\hspace{1 mm}Jeff}=\frac{600 pounds}{150 pounds}=4.

Therefore, the mechanical advantage that lever provided to Jeff in lifting rock is 4.

attashe74 [19]2 years ago
6 0
Answer: 4

The lever enabled Jeff to lift 600 pounds with merely 150 pounds.
By definition, the mechanical advantage is 600/150 = 4.
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Answer:  91.4 J

Explanation:

Kinetic energy is the energy possessed by a body due to virtue of its motion.

K.E. = 0.5 m v²

Mass of the continent is given, m = 1.819 × 10²¹ kg

Side of the block of continent, s = 4150 km = 4150000 m

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m = 2780 kg/m³× (4150 × 10³ m)²× 38 × 10³ m = 1.819 × 10²¹ kg)

The continent is moving at the rate of, v = 1 cm /year = 0.01 m / 31556926 s = 3.17 × 10⁻¹⁰ m/s

⇒ K.E. = 0.5 × 1.819 × 10²¹ kg × (3.17 × 10⁻¹⁰ m/s)²= 91.4 J

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A 2.4 kg block is dropped onto a spring and platform of negligible mass. The block is released
statuscvo [17]

The speed of the block when the compression is 15 cm is 9.85 m/s.

The given parameters;

  • <em>mass of the block, m = 2.4 kg</em>
  • <em>height of the block, h =  5 m</em>
  • <em>compression of the spring, x = 25 cm = 0.25 m</em>

The spring constant is calculated as follows;

F = kx\\\\mg = kx\\\\k = \frac{mg}{x} \\\\k = \frac{2.4 \times 9.8}{0.25} \\\\k = 94.08 \ N/m

The speed of the block when the compression is 15 cm can be determined by applying the principle of conservation of energy;

\Delta K.E = \Delta P.E\\\\\frac{1}{2} m(v^2  - v_{0 }^2 ) = mgh - \frac{1}{2} kx^2\\\\\frac{1}{2} mv^2  = mgh -   \frac{1}{2} kx^2\\\\mv^2   = 2mgh - kx^2\\\\v^2 = \frac{2mgh - kx^2}{m} \\\\v = \sqrt{\frac{2mgh - kx^2}{m}} \\\\v = \sqrt{\frac{(2 \times 2.4 \times 9.8 \times 5) - (94.08 \times 0.15^2)}{2.4}} \\\\v = 9.85 \ m/s

Thus, the speed of the block when the compression is 15 cm is 9.85 m/s.

Learn more here:brainly.com/question/14289286

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2 years ago
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