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Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.
36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.
For every second of acceleration the velocity is increased by that acceleration.
You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.
Not enough information. You also need to know:* The initial speed * How long it takes to speed up If you divide the difference in speed by the time it takes to speed up, you get the average acceleration for that time period.
"Meters per second" is not a unit of acceleration. "Meters per second squared" is one.I must assume that's what you meant in the question.Force = (mass) x (acceleration) = (0.15) x (12) = 1.8 newtons
It is accelerating at 1.2m/s per second.
Acceleration is not measured in meters/second. Meters/second is a unit of speed. Since acceleration is defined as change of speed divided by time, the units are meters/second/second, usually written as meters/second2.
Force = mass * acceleration ( acceleration's unit is m/s2 ) Force = (10 kg)(4 m/s2) = 40 Newtons ==========
F = ma Force (in Newtons) equals mass (in kilograms) times acceleration (in meters per second squared) In this case, 450 = 30a, so the accelerating is 15 meters per second squared
The mass of an object can be determined by taking (the net force in Newtons) divided by (the acceleration in meters per second per second).
Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.Yes. When it moves around the Sun, there is circular acceleration, that can be calculated via the formula a = v2 / r. Velocity should be converted to meters / second, radius (of the orbit) to meters - in this case, the result is in meters per second squared.
36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.36 meters is not a "rate".If you have an acceleration (in meters per second square), use Newton's Second Law:Net force = mass x acceleration.
You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.You divide the given acceleration by the standard acceleration due to Earth's gravity. If the acceleration is in meters per second square, you divide by 9.8.
Not enough information. You also need to know:* The initial speed * How long it takes to speed up If you divide the difference in speed by the time it takes to speed up, you get the average acceleration for that time period.
For every second of acceleration the velocity is increased by that acceleration.
Assuming you want the international units: time: second velocity: meters / second distance: meters acceleration: meters / second2