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Nesterboy [21]
2 years ago
7

The period of a simple pendulum in a grandfather clock on another planet is 1.80 s. What is the acceleration due to gravity (in

m/s2) on this planet if the length of the pendulum is 1.00?
Physics
1 answer:
weeeeeb [17]2 years ago
4 0

Answer:

12.17 m/s²

Explanation:

The formula of period of a simple pendulum is given as,

T = 2π√(L/g)........................ Equation 1

Where T = period of the simple pendulum, L = length of the simple pendulum, g = acceleration due to gravity of the planet. π = pie

making g the subject of the equation,

g = 4π²L/T²................... Equation 2

Given: T = 1.8 s, l = 1.00 m

Constant: π = 3.14

Substitute into equation 2

g = (4×3.14²×1)/1.8²

g = 12.17 m/s²

Hence the acceleration due to gravity of the planet = 12.17 m/s²

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1. An asphalt block has a mass of 90 kg and a volume of 0.075 m. Determines the density of the asphalt.
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Explanation:

1. We know that the density is defined by the following expression.

Density = \frac{mass}{volume} \\where:\\mass=90[kg]\\volume=0.075[m^{3} ]\\density=\frac{90}{0.075} \\density=1200[\frac{kg}{m^{3} }]

2. First we need to convert the units to meters.

wide = 35[cm] = 35/100 = 0.35[m]

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Now we can find the density using the expression for the density.

density= \frac{mass}{volume} \\where:\\volume = wide*long*thick\\volume=0.35*1.1*0.015 = 0.005775[m^3]\\\\mass= density*volume = 2700*0.005775 = 15.59[kg]

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