(a) The velocity of the first ball before the collision with the second ball is 11.18 m/s.
(b) The final velocity of the two balls after the collision is determined as 5.59 m/s.
<h3>
Speed of the block when pushed by the spring</h3>
The speed of the block when pushed by the spring is calculated as follows;
K.E = Ux
¹/₂mv² = ¹/₂kx²
mv² = kx²
v² = kx²/m
v² = (25 x 0.5²)/0.05
v² = 125
v = 11.18 m/s
<h3>Final velocity of the two balls after the collision</h3>
The velocity of the two balls after the collision is calculated as follows;
Pi = Pf
where;
- Pi is initial momentum
- Pf is final momentum
m1u1 + m2u2 = v(m1 + m2)
0.05(11.18) + 0.05(0) = v(0.05 + 0.05)
0.559 = 0.1v
v = 5.59 m/s
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Answer:
option C
Explanation:
The correct answer is option C
There is no external force acting in the system hence the momentum will be conserved.
As the milk is leaking out of the tank mass of the tanker is decreasing.
When the mass of the container will decrease to conservation the momentum speed of the container will have to be increased.
So, the car carrying milk will speed up.
Answer:
6010.457N
Explanation:
Centripetal acceleration = a= V²/R
At a radius of 3.6m and velocity of 16.12m/s,
Acceleration is
a = 16.12²/ 3.6 = 72.182 m/s²
Force = Mass (m) * Acceleration (a)
36 = m * 72.182
m = 36/72.182
At breaking point
Radius = 0.468 m and Velocity = 75.1 m/s
a = V²/R = 75.1²/0.468
a = 12051.3 m/s
F = Mass(m) * Acceleration (a)
F = m * 12051.3
m = F/ 12051.3
Settings the ratio of mass equal
m = m
=> 36/72.182 = F/12051.3
F = 12051.3 * 36/72.182
F = 6010.457N
Answer:
For a relative frequency distribution, relative frequency is computed as the class frequency divided by the number of observations.
