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satela
10 days ago
12

What is the magnitude of the force, directed parallel to the ramp, that he needs to exert on the crate to get it to start moving

up the ramp
Physics
1 answer:
ValentinkaMS [3.3K]10 days ago
3 0

The full question reads;

Jason is employed at a moving company. A wooden crate weighing 75 kg is positioned on the wooden ramp of his truck, inclined at an angle of 11°.

What is the force magnitude, directed parallel to the ramp, that he needs to apply to initiate the upward movement of the crate?

Answer:

F = 501.5 N

Explanation:

We have the following information;

Mass of the wooden crate; m = 75 kg

Incline angle; θ = 11°

To move the wooden crate up, we must consider that friction is acting in the opposite direction of the movement along the inclined surface. Therefore, the force required can be expressed by;

F = mgsin θ + μmg cos θ

Using online resources, the coefficient of friction between wooden surfaces is μ = 0.5

Thus;

F = (75 × 9.81 × sin 11) + (0.5 × 75 × 9.81 × cos 11)

F = 501.5 N

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Answer:

The separation between Earth and the star is growing.

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7 0
2 days ago
Coulomb's law for the magnitude of the force FFF between two particles with charges QQQ and Q′Q′Q^\prime separated by a distance
kicyunya [3148]

Response:

Clarification:

The force between two charges, q₁ and q₂ at a distance d is represented by the formula

F = k q₁ q₂ / d²

Here, the force between charge q₁ = -15 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C with distance d = (1.66 - 1.24) = 0.42 mm

k = 1 / (4π x 8.85 x 10⁻¹²)

Substituting the values into the equation

F = 1 / (4π x 8.85 x 10⁻¹²) x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -15 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 35969.4 x 10⁻³ N.

For the force between charge q₂ = 34.5 x 10⁻⁹ C and q₃ = 47 x 10⁻⁹ C at a distance d = (1.24 - 0) = 1.24 mm.

Substituting the values into the expression

F = 1 / (4π x 8.85 x 10⁻¹²) x 34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 9 x 10⁹ x -34.5 x 10⁻⁹ x 47 x 10⁻⁹ / (0.42 x 10⁻³)²

= 82729.6 x 10⁻³ N

Both forces direct towards the left (away from the origin, towards the negative x-axis)

Total force = 118699 x 10⁻³

= 118.7 N.

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