The acceleration of the 3 kg ball pushed with a 15 N force can be calculated using Newton's second law: F = ma. Rearranging the formula to solve for acceleration, we have a = F/m. Plugging in the values, we get a = 15 N / 3 kg = 5 m/s^2. Hence, the acceleration of the 3 kg ball pushed with a 15 N force is 5 m/s^2.
Using Newton's second law (F=ma), we can rearrange the formula to find acceleration (a=F/m). Plugging in the values, we get a=15N/45kg = 0.33 m/s^2. Therefore, the acceleration of the 45 kg mass pushed with a 15 N force is 0.33 m/s^2.
The net force acting on the ball is 10N (15N - 5N) in the direction of the stronger kick. Using Newton's second law (F = ma), we can calculate the resulting acceleration as 5 m/s^2 (10N / 2kg).
4 N upward
T = R x F T = 0.5m x 15N T = 7.5 N*m
The table provides a support force of 15N to balance the weight of the book, according to Newton's third law. The net force on the book is zero because the support force is equal in magnitude but opposite in direction to the weight of the book, resulting in a balanced force system.
put your hand there to stop it.
Using Newton's second law (F=ma), we can rearrange the formula to find acceleration (a=F/m). Plugging in the values, we get a=15N/45kg = 0.33 m/s^2. Therefore, the acceleration of the 45 kg mass pushed with a 15 N force is 0.33 m/s^2.
The net force acting on the ball is 10N (15N - 5N) in the direction of the stronger kick. Using Newton's second law (F = ma), we can calculate the resulting acceleration as 5 m/s^2 (10N / 2kg).
4 N upward
kiss my butt
T = R x F T = 0.5m x 15N T = 7.5 N*m
-5 -15n = 10add 5 to both sides-15n = 15divide both sides by -15n = -1
The table provides a support force of 15N to balance the weight of the book, according to Newton's third law. The net force on the book is zero because the support force is equal in magnitude but opposite in direction to the weight of the book, resulting in a balanced force system.
He did 5n per m. Joe's ratio of work to distance is 5:1.
The first thing to do is define a positive direction. The right is often defined as being positive, so we shall use that in this problem. We can now define the forces. Let F1= 100lb and F2 = -30lb (It is negative because it is pointing to the left.) Now simply add up the forces: Fnet = F1+F2 Fnet = 100lb + (-30lb) Fnet = 70lb
torque = force * lever length torque = 15 * 55 torque = 825 n-cms
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