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
Learn more about solar system.
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Answer:
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Explanation:
Answer:
424.26 m/s
Explanation:
Given that Two air craft P and Q are flying at the same speed 300m/s. The direction along which P is flying is at right angles to the direction along which Q is flying. Find the magnitude of velocity of the air craft P relative to air craft Q
The relative speed will be calculated by using pythagorean theorem
Relative speed = sqrt(300^2 + 300^2)
Relative speed = sqrt( 180000 )
Relative speed = 424.26 m/s
Therefore, the magnitude of velocity of the air craft P relative to air craft Q is 424.26 m/s
Acceleration = (velocity final-velocity initial)/ time
where
velocity final = 135 km/hr x 1 hr /3600 s x 1000m/1km
= 37.5 m/s
velocity initial = 35 km/hr x 1hr /3600 s x 1000 m/1 km
= 9.72 m/s
a) acceleration = 2.646 m/s^2
b) acceleration in g units = (2.646m/s^2)/(9.8m/s^2)
= 0.27 units