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3 years ago

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A pendulum is formed by taking a 2kg mass and hanging it from the ceiling using a steel wire with a diameter of 1.1 mm. It is ob

served that the wire stretches by 0.05 mm under the weight of the mass. What is the period of oscillation of the pendulum?

1 answer:

Mekhanik [1.2K]3 years ago

5 0

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To calculate acceleration you must know both the objects velocity and_____

amm1812

You need to know the time as well.

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3 years ago

Define the unit of time and unit of length

garik1379 [7]

This is your perfect answer

The base unit for time is the second (the other SI units are: metre for length, kilogram for mass, ampere for electric current, kelvin for temperature, candela for luminous intensity, and mole for the amount of substance). The second can be abbreviated as s or sec.

7 0

2 years ago

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WOULUJUTUL RECIPECUIUS.

3241004551 [841]

The force between the two objects is 19.73 nN.

<u>Explanation: </u>

Any force acting between two objects tends to be directly proportional to the product of their masses and inversely proportional to the square of the distance between the two objects. And this kind of attraction force between two objects is termed as gravitational force.

So if we consider $M_{1}$ and $M_{2}$ as the masses of both objects and let d be the distance of separation of two objects. Then the force between the two objects can be determined as below:

$\text {Gravitational force}=\frac{G \times M_{1} \times M_{2}}{d^{2}}$

As gravitational constant $G=6.67 \times 10^{-11} \mathrm{m}^{3} \mathrm{kg}^{-1} \mathrm{s}^{-2}$ , $M_{1}$ = 20 kg and $M_{2}$ = 100 kg, while d = 2.6 m, then

$\text {Gravitational force}=\frac{6.67 \times 10^{-11} \times 20 \times 100}{(2.6)^{2}}=\frac{6.67 \times 20 \times 10^{-9}}{6.76}$

Thus, we get finally,

$\text {Gravitational force}=19.73 \times 10^{-9} \mathrm{N}$

As we know, nano denoted by letter 'n' equals to $10^{-9}$

So the force acting between two objects is 19.73 nN.

7 0

3 years ago

Two loudspeakers are placed next to each other and driven by the same source at 500 Hz. A listener is positioned in front of the

stellarik [79]

To solve this problem we will apply the concepts related to wavelength as the rate of change of the speed of the wave over the frequency. Mathematically this is

$\lambda = \frac{v}{f}$

Here,

v = Wave velocity

f = Frequency,

Replacing with our values we have that,

$\lambda = \frac{340}{500}$

\lambda = 0.68m

The distance to move one speaker is half this

$\lambda/2 = 0.34m$

Therefore the minimum distance will be 0.34m

7 0

3 years ago

According to Newton’s first law of motion, what will an object in motion do when no external force acts on it?

KonstantinChe [14]

By definition, we have to:

Newton's first law states that any object will remain in a state of rest or with a uniform rectilinear motion unless an external force acts on it.

Therefore, according to the first law of Newton, if the object is already in motion and has no force acting on it then, it will remain with a uniform rectilinear motion.

Answer:

The object will remain with a uniform rectilinear movement when the external force does not act on it.

4 0

3 years ago

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