Answer:
(a) 1.21 m/s
(b) 2303.33 J, 152.27 J
Explanation:
m1 = 95 kg, u1 = - 3.750 m/s, m2 = 113 kg, u2 = 5.38 m/s
(a) Let their velocity after striking is v.
By use of conservation of momentum
Momentum before collision = momentum after collision
m1 x u1 + m2 x u2 = (m1 + m2) x v
- 95 x 3.75 + 113 x 5.38 = (95 + 113) x v
v = ( - 356.25 + 607.94) / 208 = 1.21 m /s
(b) Kinetic energy before collision = 1/2 m1 x u1^2 + 1/2 m2 x u2^2
= 0.5 ( 95 x 3.750 x 3.750 + 113 x 5.38 x 5.38)
= 0.5 (1335.94 + 3270.7) = 2303.33 J
Kinetic energy after collision = 1/2 (m1 + m2) v^2
= 0.5 (95 + 113) x 1.21 x 1.21 = 152.27 J
Answer:
Explanation:
Given
charge of first body 
charge of second body 
Particle 1 is at origin and particle 2 is at 
third Particle which charge +q must be placed left of 
because it will repel the q charge while 
will attract it
suppose it is placed at a distance of x m
Answer:
The electrical potential energy is 0.027 Joules.
Explanation:
The values from the question are
charge (q) = 
Electric Field strength (E) = 
Distance from source (d) = 0.030 m
Now the formula for the electrical potential energy (U) is given by
So now insert the values to find the answer
On further solving
Explanation:
Lasers produce a narrow beam of light in which all of the light waves have very similar wavelengths. The laser's light waves travel together with their peaks all lined up, or in phase. This is why laser beams are very narrow, very bright, and can be focused into a very tiny spot.
M=2.45 because you multiply out the equation on the right and divide by 10
