Answer:
The carpet has a buildup of electrons in it that get released onto your body when you rub across the floor while the hardwood doesn't have electrons built up inside of it.
Explanation:
I study that
Answer:
the center of mass rises because the whole object rises up on the ramp thing so the center of mass rises
Hope it helps!
Explanation:
Answer:
I believe it might be point A since the question ask what will result in the ln a largest increase in potential energy
Explanation:
Answer:
a) No difference
Explanation:
Since the billiard balls are identical , they have the same mass. Also they have the same speed
Since the angular momentum is conserved and the total energy is conserved ( if we assume elastic collision)
1/2 m1 * v i1² +1/2 m2 * v i1² = 1/2 m1 * v f1² +1/2 m2 * v f2²
where m= mass , vi= initial velocity , vf= final velocity
since m1=m2=m , vi1=vi2=vi
1/2 m1 * v i1² +1/2 m2 * v i1² = 1/2 m1 * v f1² +1/2 m2 * v f2²
m * v i² = 1/2 m (v f1² +v f2² )
vi² = 1/2(v f1² +v f2² )
since the 2 balls are indistinguishable from each other (they have identical initial mass and velocity) there is no reason for a preferential speed for one of the balls and therefore its velocities must be equal . Thus vf1=vf2=vf
therefore
v i² = 1/2(v f1² +v f2² ) = v i1² = 1/2* 2vf² = vf²
and thus
vi= vf
in conclusion, there is no difference in speed after the rebound
Answer:
t = 0.55[sg]; v = 0.9[m/s]
Explanation:
In order to solve this problem we must establish the initial conditions with which we can work.
y = initial elevation = - 1.5 [m]
x = landing distance = 0.5 [m]
We set "y" with a negative value, as this height is below the table level.
in the following equation (vy)o is equal to zero because there is no velocity in the y component.
therefore:
![y = (v_{y})_{o}*t - \frac{1}{2} *g*t^{2}\\ where:\\(v_{y})_{o}=0[m/s]\\t = time [sg]\\g = gravity = 9.81[\frac{m}{s^{2}}]\\ -1.5 = 0*t -4.905*t^{2} \\t = \sqrt{\frac{1.5}{4.905} } \\t=0.55[s]](https://tex.z-dn.net/?f=y%20%3D%20%28v_%7By%7D%29_%7Bo%7D%2At%20-%20%5Cfrac%7B1%7D%7B2%7D%20%2Ag%2At%5E%7B2%7D%5C%5C%20%20%20where%3A%5C%5C%28v_%7By%7D%29_%7Bo%7D%3D0%5Bm%2Fs%5D%5C%5Ct%20%3D%20time%20%5Bsg%5D%5C%5Cg%20%3D%20gravity%20%3D%209.81%5B%5Cfrac%7Bm%7D%7Bs%5E%7B2%7D%7D%5D%5C%5C%20-1.5%20%3D%200%2At%20-4.905%2At%5E%7B2%7D%20%5C%5Ct%20%3D%20%5Csqrt%7B%5Cfrac%7B1.5%7D%7B4.905%7D%20%7D%20%5C%5Ct%3D0.55%5Bs%5D)
Now we can find the initial velocity, It is important to note that the initial velocity has velocity components only in the x-axis.
![(v_{x} )_{o} = \frac{x}{t} \\(v_{x} )_{o} = \frac{0.5}{0.55} \\(v_{x} )_{o} =0.9[m/s]](https://tex.z-dn.net/?f=%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7Bx%7D%7Bt%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D%20%5Cfrac%7B0.5%7D%7B0.55%7D%20%5C%5C%28v_%7Bx%7D%20%29_%7Bo%7D%20%3D0.9%5Bm%2Fs%5D)
