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klemol [59]
2 years ago
13

A small box of mass m1 is sitting on a board of mass m2 and length L (Figure 1) . The board rests on a frictionless horizontal s

urface. The coefficient of static friction between the board and the box is ?s. The coefficient of kinetic friction between the board and the box is, as usual, less than ?s.
Throughout the problem, use g for the magnitude of the acceleration due to gravity. In the hints, use Ff for the magnitude of the friction force between the board and the box.

Find Fmin, the constant force with the least magnitude that must be applied to the board in order to pull the board out from under the the box (which will then fall off of the opposite end of the board).

Express your answer in terms of some or all of the variables ?s, m1, m2, g, and L. Do not include Ff in your answer.
Physics
1 answer:
chubhunter [2.5K]2 years ago
3 0

Explanation:

Whole system will accelerate under the action of applied force. The box will experience the force against the friction and when this force exceeds then the box will move. so

Ff = μs×m1×g

m1×a = μs×m1×g

a = μs×g

The applied force is given by

F = (m1 + m2)×a so

F = μs×g×(m1+m2)

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D) Synthesizers have always had a well-established presence in standard ensembles

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An object of mass M is dropped near the surface of Earth such that the gravitational field provides a constant downward force on
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Answer:

The answer is: c. It does not move

Explanation:

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2 years ago
A 56 kg diver runs and dives from the edge of a cliff into the water which is located 4.0 m below. If she is moving at 8.0 m/s t
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Answer:

1) 2197.44 J

2) 0 J

3) 2197.44 J = Constant

4) 2197.44 J

5) Approximately 8.86 m/s

Explanation:

The given parameters are;

The mass of the diver, m = 56 kg

The height of the cliff, h = 4.0 m

The speed with which the diver is moving, vₓ = 8.0 m/s

The gravitational potential energy = Mass, m × Height of the cliff, h × Acceleration due to gravity, g

1) Her gravitational potential energy = 56 × 4.0 × 9.81 = 2197.44 J

2) The kinetic energy = 1/2·m·u²

Where;

u = Her initial velocity = 0 when she just leaves the cliff

Therefore;

Her kinetic energy when she just leaves the cliff = 1/2 × 56 × 0² = 0 J

3) The total mechanical energy = Kinetic energy + Potential energy

The total mechanical energy is constant

Her total mechanical energy relative to the water surface when she leaves the cliff = Her gravitational potential energy = 2197.44 J = Constant

4) Her total mechanical energy relative to the water surface just before she enters the water = 2197.44 J

5) The speed with which she enters the water, v, is given from, v² = u² + 2·g·h

Where;

u = The initial velocity at the top of the cliff before she jumps= 0 m/s

∴ v² = 0² + 2 × 9.81 × 4 = 78.48

v = √78.48 ≈ 8.86 m/s

The speed with which she enters the water, v ≈ 8.86 m/s

7 0
2 years ago
If the top circuit has an oscillation frequency of 1000 Hz, the frequency of the bottom circuit is:_______.
kiruha [24]

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1410 Hz

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Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper
vladimir1956 [14]

Answer:

Total distance does the grasshopper travel before the cars hit is 150 m

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distance=V_{gh} \times t\\distance=15 \times 10\\distance=150 m

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