According to Newton's second law, Force equals the rate of change of momentum over time. Momentum change is equal to Force times time. So, F=ma can be rearranged to a=F/m, a more recognizable formulation of Newton's second law
Using a relevant kinematic equation for mass m: V=u+at; where initial speed u=0; thus, acceleration a=F/m gives V=(F/m)xt, which translates to t=mV/F. For mass 2m, applying the same formula: V=u+at; u=0; a=F/2m indicates V=(F/2m)xt, leading to t=2mV/F (possibly double the initial time)
I might have erred somewhere along the line, but the fundamental concept seems valid... using another kinematic equation for m: s=ut + (1/2)at²; with s=d; and initial speed u=0; a=F/m; t=1; results in d=(1/2)(F/m) = F/2m. Similarly, for 2m: s=ut + (1/2)at²; s=d; u=0; a=F/2m; and t=1 gives d=(1/2)(F/2m)=F/4m (half the distance perhaps???? WHAT???!)
Response:
The man's speed is 0.144 m/s
Explanation:
This exemplifies conservation of momentum.
The momentum of the ball prior to being caught must equal the momentum of the man-ball system after catching the ball.
Mass of the ball = 0.65 kg
Mass of the man = 54 kg
Speed of the ball = 12.1 m/s
The momentum of the ball before impact can be calculated as mass multiplied by velocity
= 0.65 x 12.1 = 7.865 kg-m/s
After catching the ball, the momentum of the combined system is
(0.65 + 54)Vf = 54.65Vf
Where Vf denotes their final shared velocity.
Setting the initial momentum equal to the final momentum,
7.865 = 54.65Vf
Vf = 7.865/54.65 = 0.144 m/s
1. τbiceps = +(Positive)
2. τforearm = -(Negative)
3. τball = -(Negative)
Explanation:
The attached figure illustrates the following: 1. For the biceps, τbiceps indicates that torque is calculated as Torque = r x F, where r and F are vectors. Here, r corresponds to the vector from the elbow to the biceps. In the figure, the force from the biceps is directed upwards. Applying the right-hand rule from r to F results in counterclockwise torque, which is considered positive (+).
2. The torque related to the weight of the forearm, τforearm, uses the same torque formula, with r being the vector from the elbow to the forearm. The weight acts downward, causing a clockwise torque that is negative (-).
3. Similarly, for the weight of the ball, τball, the downward force from the ball's weight generates a clockwise torque, which also registers as negative (-).
The Pythagorean Theorem can be utilized here: Imagine a car navigating through traffic—when it turns left to travel north, a right angle of 90 degrees is formed. However, the displacement is always the shortest distance connecting the origin and the endpoint, which forms a triangle in this scenario. In a right triangle, the Pythagorean theorem applies: 215^2+45^2=c^2; therefore, v=√(215^2+45^2).
