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Arada [10]
3 years ago
9

A simple pendulum is used to determine the acceleration due to gravity at the surface of a planet. The pendulum has a length of

2 m and its period is measured to be 2 s. The value of g obtained in this investigation is most nearly __________.A. 1 m/s²B. 2 m/s²C. 5 m/s²D. 10 m/s²E. 20 m/s²
Physics
1 answer:
SVEN [57.7K]3 years ago
8 0

Answer:

Acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer

Explanation:

We have given length of the pendulum l = 2 m

Time period of the pendulum T = 2 sec

We have to find acceleration due to gravity g

We know that time period of pendulum is given by

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

2=2\times 3.14 \sqrt{\frac{2}{g}}

0.3184= \sqrt{\frac{2}{g}}

Squaring both side

0.1014= {\frac{2}{g}}

g=19.71=20m/sec^2

So acceleration due to gravity is 20 m/sec^2

So option (E) will be correct answer.

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As a moon follows its orbit around a planet, the maximum gravitational force exerted on the moon by the planet exceeds the minim
professor190 [17]

Answer:

rmax/rmin = √1.127

Explanation:

F = GmM / r²

As the masses can be assumed to be constant, the force between the two is proportional to the inverse of the square of the distance between them

(Fmax - Fmin) / Fmin = 0.127

           (Fmax - Fmin) = 0.127Fmin

       1/rmin² - 1/rmax² = 0.127(1/rmax²)

                      1/rmin² = 0.127(1/rmax²) + 1/rmax²

                      1/rmin² = 1.127(1/rmax²)

              rmax²/rmin² = 1.127

                 rmax/rmin = √1.127 ≈ 1.06160256...

8 0
3 years ago
This should be pretty easy i think.
timama [110]

Answer:

its a sure for not but i think its a 192583

3 0
2 years ago
10 points
enot [183]

Answer:

B

Explanation:

I just took the test

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