When all group iia elements lose their valence electrons, the remaining electron configurations are the same as for what family
2 answers:
You might be interested in
Refer to the diagram shown below.
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
μ = the coefficient of dynamic friction between the crate and the ramp.
1. The applied force of F acts over a distance, d.
The work done is F*d.
2. The component of the weight of the crate acting down the ramp is
mg sin(30) = 0.5mg.
The work done by this force is 0.5mgd.
3. The normal force is N = mgcos(30) = 0.866mg.
This force is perpendicular to the ramp, therefore the work done is zero.
4. The frictional force is μN = μmgcos(30) = 0.866μmg.
The work done by the frictional force is 0.866μmgd.
5. The total force acting on the crate up the ramp is
F - mgsin(30) - μmgcos(30) = F - mg(0.5 - 0.866μ)
6. The work done on the crate by the total force is
d*(F - 0.5mg - 0.866μmg)
Answer:
Speed of lighter ball is 4 m/s.
Explanation:
Applying the principle of conservation of linear momentum,
momentum before collision = momentum after collision.
+
=
-
= 3 kg,
= 8 m/s,
= 2 kg,
= 0 m/s ( since it is at rest),
= 2 m/s,
= ?
(3 x 8) + (2 x 0) = (8 x 2) - (2 x )
24 + 0 = 16 - 2
2 = 16 - 24
2 = -8
=
= -4 m/s
This implies that the light ball moves at the speed of 4 m/s in the opposite direction of the heavier ball after collision.