The acceleration due to gravity (g) on this planet is 39.44 m/s²
<h3>What is solar system?</h3>
Solar system consists of all the planets and the most importantly the center of the solar system is Sun.
Given is an unknown planet in the outer-reaches of the solar system, a pendulum with a 12 g bob and a string length of 4 m oscillates with a period of 2 seconds.
The time period of the pendulum is
T = 2π √l/g
Squaring both sides, we get
l/g = T² / 4π²
g = 4π²l/ T²
Substitute Time period T = 2s and length l = 4m, we get
g = 4π²x 4/ 2²
g =39.44 m/s²
Thus, the acceleration due to gravity on this planet is 39.44 m/s²
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Answer:
0.12
Explanation:
The acceleration due to gravity of a planet with mass M and radius R is given as:
g = (G*M) / R²
Where G is gravitational constant.
The mass of the planet M = 3 times the mass of earth = 3 * 5.972 * 10^24 kg
The radius of the planet R = 5 times the radius of earth = 5 * 6.371 * 10^6 m
Therefore:
g(planet) = (6.67 * 10^(-11) * 3 * 5.972 * 10^24) / (5 * 6.371 * 10^6)²
g(planet) = 1.18 m/s²
Therefore ratio of acceleration due to gravity on the surface of the planet, g(planet) to acceleration due to gravity on the surface of the planet, g(earth) is:
g(planet)/g(earth) = 1.18/9.8 = 0.12
Answer:
The acceleration of the wagon is 3 m/s².
To calculate the acceleration of the wagon, we use the formula below.
Formula:
F = ma............. Equation 1
Where:
F = horizontal Force
m = mass of the wagon
a = acceleration of the wagon.
make a the subject of the equation
a = F/m.............. Equation 2
From the question,
Given:
F = 30 N
m = 10 kg
Substitute these values into equation 2
a = 30/10
a = 3 m/s²
Hence, the acceleration of the wagon is 3 m/s².
Answer:
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