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The period of a pendulum is given by the equation uppercase t = 2 pi startroot startfraction uppercase l over g endfraction endroot , where l is the length of the string suspending the pendulum in meters, and g is the acceleration due to gravity in m/s2. which of the following domains provide a real-value period? g < 0

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Final answer:

To ensure a period with a real value for a pendulum, the acceleration due to gravity, g, must be positive. The domain for g is g > 0 since gravity is an attractive force, and negative gravity is non-physical. The pendulum formula T = 2π√(L/g) relates the period of a pendulum to the length L and the acceleration due to gravity g.

Step-by-step explanation:

The period of a simple pendulum is given by the formula T = 2π√(L/g), where T is the period in seconds, L is the length of the pendulum in meters, and g is the acceleration due to gravity in meters per second squared (m/s²). To have a real-value period, g must be a positive number as the square root of a negative number is an imaginary number, not a real number. Also, since g is the acceleration due to gravity, it cannot be negative because gravity always causes objects to accelerate towards the Earth.

Regarding the specific domains for g to provide a real-value period, only positive values of g are applicable. Thus, the domain for g is g > 0. If the value of g were less than zero, it would imply a non-physical scenario where gravity has a repulsive effect - not something we experience in the real world.

For example, if a pendulum has a period T of 2 seconds and a length L of 1 meter, using the formula T = 2π√(L/g), we can measure gravity by solving for g.

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