Work = force * distance
and newton*meters = Joule
In this case, work = 250N*50m = 12500 J
So the answer is D) 12,500 J
Answer:
Fulcrum
Explanation:
When you use a spoon to prise a lid from a tin you are using a simple machine called a lever. In fact, levers are the basis of many tools in and around your house and work. The way levers operate is by an effort applied at a point, which moves a load at another point through a balance point called the fulcrum.
Answer:
The net force acting on the tennis ball while it is in contact with the racquet is 50.73 N
Explanation:
The impulse-momentum theorem said that the net impulse is equal to the change of the momentum, this is:
![\overrightarrow{I}=\varDelta\overrightarrow{P}\,\,(1)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BI%7D%3D%5CvarDelta%5Coverrightarrow%7BP%7D%5C%2C%5C%2C%281%29)
but the net impulse is too the net force times the change in time:
![\overrightarrow{I}=\overrightarrow{F}\varDelta t\,\,(2)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BI%7D%3D%5Coverrightarrow%7BF%7D%5CvarDelta%20t%5C%2C%5C%2C%282%29)
so using (2) on (1) we have:
![\overrightarrow{F}\varDelta t=\varDelta\overrightarrow{P}\,\,(3)](https://tex.z-dn.net/?f=%5Coverrightarrow%7BF%7D%5CvarDelta%20t%3D%5CvarDelta%5Coverrightarrow%7BP%7D%5C%2C%5C%2C%283%29)
Decomposing that on x and y components:
![F_{x}\varDelta t=\varDelta P_{x}=P_{fx}-P_{ox}\,\,(3)](https://tex.z-dn.net/?f=F_%7Bx%7D%5CvarDelta%20t%3D%5CvarDelta%20P_%7Bx%7D%3DP_%7Bfx%7D-P_%7Box%7D%5C%2C%5C%2C%283%29)
![F_{y}\varDelta t=\varDelta P_{y}=P_{fy}-P_{oy}\,\,(4)](https://tex.z-dn.net/?f=F_%7By%7D%5CvarDelta%20t%3D%5CvarDelta%20P_%7By%7D%3DP_%7Bfy%7D-P_%7Boy%7D%5C%2C%5C%2C%284%29)
(See figure below) with Pfx = m*vfx= m*vf*cos(15°)=(0.058kg)(40m/s)cos(15°),
Pox = -m*vox= m*vo*cos(15°)=-(0.058kg)(30m/s)cos(15°), the same analysis to Pfy and Poy gives
Pfy=(0.058kg)(40m/s)sin(15°), Poy=-(0.058kg)(30m/s)sin(15°), using those values on (3) and (4) and solving for Fy and Fx:
![F_{x}=\frac{(0.058)(40)cos(15\text{\textdegree})-(-(0.058)(30)cos(15\text{\textdegree}))}{0.08}\simeq49N\,\,(5)](https://tex.z-dn.net/?f=F_%7Bx%7D%3D%5Cfrac%7B%280.058%29%2840%29cos%2815%5Ctext%7B%5Ctextdegree%7D%29-%28-%280.058%29%2830%29cos%2815%5Ctext%7B%5Ctextdegree%7D%29%29%7D%7B0.08%7D%5Csimeq49N%5C%2C%5C%2C%285%29)
![F_{y}=\frac{(0.058)(40)sin(15\text{\textdegree})-(-(0.058)(30)sin(15\text{\textdegree}))}{0.08}\simeq13.13N\,\,(6)](https://tex.z-dn.net/?f=F_%7By%7D%3D%5Cfrac%7B%280.058%29%2840%29sin%2815%5Ctext%7B%5Ctextdegree%7D%29-%28-%280.058%29%2830%29sin%2815%5Ctext%7B%5Ctextdegree%7D%29%29%7D%7B0.08%7D%5Csimeq13.13N%5C%2C%5C%2C%286%29)
So the net force acting on the tennis ball while it is in contact with the racquet is:
![F=\sqrt{F_{x}^{2}+F_{y}^{2}}\simeq50.73N](https://tex.z-dn.net/?f=F%3D%5Csqrt%7BF_%7Bx%7D%5E%7B2%7D%2BF_%7By%7D%5E%7B2%7D%7D%5Csimeq50.73N)
Answer:
![a_{c}=v^{2}/R](https://tex.z-dn.net/?f=a_%7Bc%7D%3Dv%5E%7B2%7D%2FR)
The radius of curvature changes so that centripetal acceleration is similar along the entire roller coaster.
Explanation:
We know that the centripetal acceleration is directly proportional to the tangential velocity and inversely proportional to the radius of curvature:
![a_{c}=v^{2}/R](https://tex.z-dn.net/?f=a_%7Bc%7D%3Dv%5E%7B2%7D%2FR)
By energy conservation (and common sense), we know that the speed at the top of the roller coaster is smaller. Therefore if the roller coaster has similar accelerations (therefore also similar normal forces) at the top and at the bottom, it is necessary that the difference in speed be compensated with the radius of curvature, i.e. smaller radius at the top than at the bottom.