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11Alexandr11 [23.1K]
3 years ago
6

How long must a simple pendulum be if it is to make exactly one swing per second? (That is, one complete vibration takes exactly

2.0 s.)
Physics
1 answer:
umka2103 [35]3 years ago
8 0

Oscillation is a type of periodic motion which repeats itself to and from about a point which is called mean position. The period of a Pendulum can be described as a ratio between the length and gravity as,

T = 2\pi \sqrt{\frac{L}{g}}

Here,

L = length

g = Gravity

If we rearrange  to find the length we have that,

L = \frac{T^2g}{4\pi^2}

Our period is 2s and the gravity is 9.8m/s^2, then,

L = \frac{(2)^2(9.8)}{4\pi^2}

L = 0.9929m

The simple required length of the pendulum must be 0.9929m

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What happens to the average kinetic energy of water molecules as water freezes? It decreases as the water absorbs energy from it
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In a double slit experiment, 450 nm light passes through two slits producing an interference pattern where the first bright frin
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Answer:

1.3 x 10⁻⁴ m

Explanation:

\lambda = wavelength of the light = 450 nm = 450 x 10⁻⁹ m

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For bright fringe, Using the equation

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Inserting the values

d Sin0.2° = (1) (450 x 10⁻⁹)

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3 years ago
The diagram below shows a 5.00-kilogram block
bixtya [17]

The name and strength of the force holding the block up is 50 N upward - Normal force.

The given parameters:

  • <em>Mass of the block, m = 5 kg</em>

The weight of the block acting downwards due to gravity is calculated as follows;

W = mg

where;

  • <em>g is acceleration due to gravity = 10 m/s²</em>

W = 5 x 10

W = 50 N <em>(</em><em>downwards</em><em>)</em>

Since the block is at rest, an a force equal to the weight of the block must be acting upwards. This force is known as normal reaction.

Fₙ = 50 N <em>(</em><em>upwards</em><em>)</em>

Thus, the name and strength of the force holding the block up is 50 N upward - Normal force.

Learn more about Normal force here: brainly.com/question/14486416

4 0
2 years ago
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A bucket filled with water has a mass of 54 kg and is hanging from a rope that is wound around a 0.050 m radius stationary cylin
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Answer:

The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

Explanation:

Given that,

Mass of bucket = 54 kg

Radius = 0.050 m

We need to calculate the magnitude of the torque the bucket produces around the center of the cylinder

Using formula of torque

\tau=F\times r

\tau=mg\times r

Where, m = mass

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r = radius

Put the value into the formula

\tau=54\times9.8\times0.050

\tau=26.46\ N-m

Hence, The magnitude of the torque the bucket produces around the center of the cylinder is 26.46 N-m.

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