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miskamm [114]
3 years ago
7

The period T of a pendulum of length L is measured to determine g at the surface of Earth. The equation used is T=2π√L/g. The ma

ss of the pendulum bob is 10.0 kg, and the length of the pendulum is 1.00 m. Which of the following would contribute most to the expected error in the result?
a. 10% uncertainty in the measurement of the mass.
b. Approximating the value of t as 3.14.
c. Variation in the value of g as the pendulum bob moves along its arc.
d. Using the average period of 10 timed oscillations instead of the period of 1 timed oscillation.
d. Starting the pendulum swinging by releasing it from a horizontal position D. E.
Physics
1 answer:
saul85 [17]3 years ago
4 0

Answer:

C: Variation in the value of g as the pendulum bob moves along its arc.

Explanation:

The formula for period of a simple pendulum is given by;

T = 2π√(L/g)

Where;

L is length

g is acceleration due to gravity

Now, from this period equation, it is clear that the only thing that can affect the period of a simple pendulum are changes to its length and acceleration due to gravity.

Looking at the options, the only one that talks about either the length or gravity as being potential causes of the error is option C

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4 years ago
you and a friend setup an umbrella and chairs at a beach. your friend goes into the surf zone while you relax on the sand. sever
Dmitrij [34]

Answer:

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Explanation:

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6 0
3 years ago
A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

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Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

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3 years ago
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