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stiks02 [169]
3 years ago
9

The gravitational force between two objects (mass1 = 10kg, mass2 = 6kg) is measured when the objects are 12 centimeters apart. I

f the distance between them is increased to 36 centimeters, how does the new gravitational attraction compare to the first one that was measured? A. The new gravitational force is 3 times larger than the old one B. The new gravitational force is 9 times larger than the old one C. The new gravitational force is 3 times smaller than the old one D. The new gravitational force is 9 times smaller than the old one
Physics
1 answer:
Yanka [14]3 years ago
8 0
Since the new distance is 3 times the old distance,
the new force is (1/3²) = 1/9th of the old force.

That's kind-of Choice-D, but I really don't like the way choice-D is worded.
"9 times smaller" is really pretty meaningless.  
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How do you find the oscillation period in seconds for different pendulum lengths?
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To find:

The equation to find the period of oscillation.

Explanation:

The period of oscillation of a pendulum is directly proportional to the square root of the length of the pendulum and inversely proportional to the square root of the acceleration due to gravity.

Thus the period of a pendulum is given by the equation,

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

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On substituting the values of the length of the pendulum and the acceleration due to gravity at the point where the period of the pendulum is being measured, the above equation yields the value of the period of the pendulum.

Final answer:

The period of oscillation of a pendulum can be calculated using the equation,

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

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8 months ago
Automobiles must be able to sustain a frontal impacl The automobile design must allow low speed impacts with little sustained da
valentinak56 [21]

Answer: the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

Explanation:

Given that;

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First we determine the energy of the vehicle just prior to impact;

W_v = 1/2mv²

we substitute

W_v = 1/2 × 1000 × (2.5)²

W_v = 3125 J

now, the the effective design stiffness k will be:

at the impact point, energy of the vehicle converts to elastic potential energy of the bumper;

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W_v = 1/2kx²

we substitute

3125 = 1/2 × k (0.04)²

3125 = 0.0008k

k = 3125 / 0.0008

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Therefore, the effective design stiffness required to limit the bumper maximum deflection during impact to 4 cm is 3906250 N/m

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