If he stops running the tea is still going to be moving so it will spill on him.
The cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
<h3>
Acceleration of the cart</h3>
The acceleration of the cart is determined from the net force acting on the mass-cart system.
Upward force = Downward force
ma = mg
13a = 10(9.8)
13a = 98
a = 98/13
a = 7.54 m/s²
Thus, the cart's acceleration to the right after the mass is released is determined as 7.54 m/s².
Learn more about acceleration here: brainly.com/question/14344386
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The period of oscillation is T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
<h3>What is period of oscillation?</h3>
This is the time in seconds it takes to complete one oscillation. where an oscillation is a repetitive to and fro motion. period if the inverse of frequency and both are basic when calculation motion in simple harmonic motion.
The period of oscillation is given as T
T = 2 * pi * sqrt ( m / k )
where
m = mass on this case mass of the spring will be inclusive to the mass of the block such that we have:
m1 = mass of the block
m2 = mass pf the spring
k = force constant of the spring
including the two masses to the period gives
T = 2 * pi * sqrt ( ( m2/3 + m1) / k )
Read more on period of oscillation here: brainly.com/question/22499336
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Answer: Here this will help you..
Explanation:
1 kg-m/s to kilogram-force meter/second = 1 kilogram-force meter/second
5 kg-m/s to kilogram-force meter/second = 5 kilogram-force meter/second
10 kg-m/s to kilogram-force meter/second = 10 kilogram-force meter/second
20 kg-m/s to kilogram-force meter/second = 20 kilogram-force meter/second
30 kg-m/s to kilogram-force meter/second = 30 kilogram-force meter/second
40 kg-m/s to kilogram-force meter/second = 40 kilogram-force meter/second
50 kg-m/s to kilogram-force meter/second = 50 kilogram-force meter/second
75 kg-m/s to kilogram-force meter/second = 75 kilogram-force meter/second
100 kg-m/s to kilogram-force meter/second = 100 kilogram-force meter/second
