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Bezzdna [24]
2 years ago
5

Compare the time period of two simple pendulums of length 4m and 16m at a place.

Physics
1 answer:
Vlad1618 [11]2 years ago
5 0

Answer:

the period of the 16 m pendulum is twice the period of the 4 m pendulum

Explanation:

Recall that the period (T) of a pendulum of length (L)  is defined as:

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

where "g" is the local acceleration of gravity.

SInce both pendulums are at the same place, "g" is the same for both, and when we compare the two periods, we get:

T_1=2\,\pi\,\sqrt{\frac{4}{g} } \\T_2=2\,\pi\,\sqrt{\frac{16}{g} } \\ \\\frac{T_2}{T_1} =\sqrt{\frac{16}{4} } =2

therefore the period of the 16 m pendulum is twice the period of the 4 m pendulum.

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true

Explanation:

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You go to the hardware store to buy a new 50 ft garden hose. You find you can choose between hoses of ½ inch and 5/8 inch inner
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To solve this problem it is necessary to consider two concepts. The first of these is the flow rate that can be defined as the volumetric quantity that a channel travels in a given time. The flow rate can also be calculated from the Area and speed, that is,

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The second concept related to the calculation of this problem is continuity, which is defined as the proportion that exists between the input channel and the output channel. It is understood as well as the geometric section of entry and exit, defined as,

Q_1 = Q_2

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Our values are given as,

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Re-arrange the equation to find the first ratio of rates we have:

\frac{V_1}{V_2}=\frac{A_2}{A_1}

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The second ratio of rates is

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A person weighing 785 newtons on the surface of Earth would weigh 298 newtons on the surface of Mars.
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Answer:

The gravitational field strength on the surface of Mars = 3.72 m/s²

Explanation:

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Mathematically, Gravitational field is represented as,

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m = F/g ..................... Equation 2.

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From the question,

Note: That The mass of the object is constant both on the surface of the earth and on the surface of Mars.

On the Surface of the earth,

Given: F = 785 N, g = 9.8 m/s²

Substituting this values into equation 2,

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m = 80.10 kg.

On the surface of Mars.

Given: m = 80.10 kg, F = 298.

Substituting into equation 2

g = 298/80.1

g = 3.72 m/s²

Thus the gravitational field strength on the surface of Mars = 3.72 m/s²

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