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Gemiola [76]
2 years ago
15

Which component of weight is cause for S.H.M of bob simple pendulum

Physics
1 answer:
jeka57 [31]2 years ago
6 0

Explanation:

The Simple Pendulum. A simple pendulum is defined to have a point mass, also known as the pendulum bob, which is suspended from a string of length L with negligible mass ((Figure)). Here, the only forces acting on the bob are the force of gravity (i.e., the weight of the bob) and tension from the string.

hope it helps you

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<h2>Answer: Toward the center of the circle.</h2>

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However, in this movement the velocity has a constant magnitude, but its direction varies continuously.

Let's say \vec{V} is the velocity vector, whose direction is perpendicular to the radius r of the trajectory, therefore   the acceleration \vec{a} is directed toward the center of the circumference.

 

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Which scientific law describes the relationship between action and reaction force pairs? Newton's third law of motion Kepler's l
Hitman42 [59]

Newton's third law of motion

Explanation:

Newton's third law of motion states that:

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It is important to note that this law is always valid, even when it seems it is not.

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7 0
3 years ago
In Millikan's experiment, an oil drop of radius 1.362 μm and density 0.888 g/cm3 is suspended in chamber C when a downward-point
Misha Larkins [42]

Answer:

The charge on the oil drop is 3e

Explanation:

F = qE

Where;

F is the applied force in Newton

E is the electric field potential N/C

q is charge in C

Given;

Radius, r = 1.362 μm = 1.362 X 10⁻⁶ m

density, ρ = 0.888 g/cm³ = 0.888 X 10³ kg/m³

Electric field potential = 1.92 ✕ 10⁵ N/C

F =mg

mass of the oil drop = density, ρ  X volume of the oil drop

volume of the oil drop (spherical) =  (4/3)πr³ = 1.3333π(1.362 X 10⁻⁶)³

⇒ volume of the oil drop = 10.584 X 10⁻¹⁸ m³

mass of the oil drop = 0.888 X 10³ (kg/m³) X 10.584 X 10⁻¹⁸ (m³)

⇒ mass of the oil drop = 9.399 X 10⁻¹⁵ kg

⇒ F =mg = 9.399 X 10⁻¹⁵ kg X 9.8 = 9.21 X10⁻¹⁴ N

F = qE

q = F/E

q = (9.21 X10⁻¹⁴)/(1.92 ✕ 10⁵) = 4.797 X 10⁻¹⁹ C

In terms of e

1e = 1.6 X10⁻¹⁹ C

=  (4.797 X 10⁻¹⁹ C)/(1.6 X10⁻¹⁹ C) = 3e

Therefore, the charge on the oil drop is 3e

7 0
3 years ago
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