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11Alexandr11 [23.1K]
4 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]4 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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A satellite of mass M = 270kg is in circular orbit around the Earth at an altitude equal to the earth's mean radius (6370 km). A
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To solve this problem we will apply the concepts related to Orbital Speed as a function of the universal gravitational constant, the mass of the planet and the orbital distance of the satellite. From finding the velocity it will be possible to calculate the period of the body and finally the gravitational force acting on the satellite.

PART A)

V_{orbital} = \sqrt{\frac{GM_E}{R}}

Here,

M = Mass of Earth

R = Distance from center to the satellite

Replacing with our values we have,

V_{orbital} = \sqrt{\frac{(6.67*10^{-11})(5.972*10^{24})}{(6370*10^3)+(6370*10^3)}}

V_{orbital} = 5591.62m/s

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PART B) The period of satellite is given as,

T = 2\pi \sqrt{\frac{r^3}{Gm_E}}

T = \frac{2\pi r}{V_{orbital}}

T = \frac{2\pi (2*6370*10^3)}{5.591*10^3}

T = 238.61min

PART C) The gravitational force on the satellite is given by,

F = ma

F = \frac{1}{4} mg

F = \frac{270*9.8}{4}

F = 661.5N

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This statement is true. The greater the mass is in an object, it is indeed the higher resistance to a change in movement the object will have. That only mean that the mass of an object and its resistance to change of movement is directly proportional.
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