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
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
This is happened because "the air" above "moves faster" and "the pressure" is "lower"
.
Option: 1
<u>Explanation</u>:
Air movement take place from the region where air pressure is more than the region where the pressure is low. When we "blow" air above the "paper strip" paper rises because "low pressure" is created above the strip as high speed winds always travel with reduced air pressure. Hence due to higher air pressure below the strip, it is pushed upwards. This difference in pressure results into fast air moves. This happen because "speed" of the wind depends on "the difference between the pressures" of the air in the two regions.
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(0.15 x g)/100 = 0.0147N. per cm. length.
<span>Torque short end = (0.0147 x 9) = 0.1323N/cm. </span>
<span>Torque long end = ((100 - 18)/2) x 0.0147 = 0.6027N/cm. </span>
<span>Difference = (0.6027 - 0.1323) = 0.4704N/cm. </span>
<span>(0.4704/3.2) = 0.147cm. from the 18cm. mark, = 17.853cm. from the 0 end of the stick.</span>
