Answer:
a) 4.2m/s
b) 5.0m/s
Explanation:
This problem is solved using the principle of conservation of linear momentum which states that in a closed system of colliding bodies, the sum of the total momenta before collision is equal to the sum of the total momenta after collision.
The problem is also an illustration of elastic collision where there is no loss in kinetic energy.
Equation (1) is a mathematical representation of the the principle of conservation of linear momentum for two colliding bodies of masses
and
whose respective velocities before collision are
and
;

where
and
are their respective velocities after collision.
Given;

Note that
=0 because the second mass
was at rest before the collision.
Also, since the two masses are equal, we can say that
so that equation (1) is reduced as follows;

m cancels out of both sides of equation (2), and we obtain the following;

a) When
, we obtain the following by equation(3)

b) As
stops moving
, therefore,

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Answer:
μ = 0.189
Explanation:
The spring potential energy will convert to work of friction
Fd = ½kx²
μmgd = ½kx²
μ = kx²/2mgd
μ = 167(0.132²) / (2(2.77)(9.81)(0.284))
μ = 0.18852424...
Answer:
No. of 100 W bulbs, n = 28,539 bulbs
Given:
Power of a single bulb = 100 W
Time period, T = 1 yr =
= 31,536,000 s
mass of matter, m = 1 g = ![1\times 10^{-3}]](https://tex.z-dn.net/?f=1%5Ctimes%2010%5E%7B-3%7D%5D)
Solution:
According to Eintein's mass-energy equivalence:
E = 
E = 
E = 
Power of a single bulb, P = 
P = 
P = 28,53,881 W
No. of 100 W bulbs, n = 
n = 
n = 28,539