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wolverine [178]
3 years ago
13

An astronaut lands on another planet and wishes to determine the acceleration due to gravity.the astronaut measure a period of 0

.5m pendulum as being 3.2 seconds what is the acceleration due to gravity
Physics
1 answer:
jarptica [38.1K]3 years ago
5 0

Answer:

1.927  m/s^2

Explanation:

period = 2 pi  sqrt ( l/g)  

3.2   =  2 pi sqrt (.5/g) =1.927 m/s^2

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Answer:

The input force (effort) is the amount of effort used to push down on a rod, or pull on a rope in order to move the weight. In this example, the force the little guy is using to pull the elephant is the input force.

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A child with mass of 25 kg gets into a toy car with mass of 80 kg on a playground, causing it to sink on its springs (with effec
Bess [88]

Answer:

The frequency of the oscillation is 0.9Hz

Explanation:

This problem bothers on simple harmonic motion of a spring

Given data

Mass of the child m= 25kg

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Amplitude a= 31cm

But the period of the motion as a result of the adults sholve is expressed as

T=2π√m/k

T=2*3.142√25/791

T=6.284√0.031

T=6.284*0.176

T=1.11 sec

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3 years ago
Three systems of two charged particles are shown below. All the particles have the same mass,
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3 years ago
Consider the standing wave pattern below created by a string fixed at both ends. The string is under a tension of 0.98 N and has
Shalnov [3]

Answer:

11.07Hz

Explanation:

Check the attachment for diagram of the standing wave in question.

Formula for calculating the fundamental frequency Fo in strings  is V/2L where;

V is the velocity of the wave in string

L is the length of the string which is expressed as a function of its wavelength.

The wavelength of the string given is 1.5λ(one loop is equivalent to 0.5 wavelength)

Therefore L = 1.5λ

If L = 3.0m

1.5λ = 3.0m

λ = 3/1.5

λ = 2m

Also;

V = √T/m where;

T is the tension = 0.98N

m is the mass per unit length = 2.0g = 0.002kg

V = √0.98/0.002

V = √490

V = 22.14m/s

Fo = V/2L (for string)

Fo = 22.14/2(3)

Fo = 22.14/6

Fo = 3.69Hz

Harmonics are multiple integrals of the fundamental frequency. The string in question resonates in 2nd harmonics F2 = 3Fo

Frequency of the wave = 3×3.69

Frequency of the wave = 11.07Hz

3 0
4 years ago
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