Explanation:
The momentum of the three objects are as follow :
11 kg-m/s, -65 kg-m/s and -100 kg-m/s
Before collision, the momentum of the system is :
After collison, they move together. It means it is a case of inelastic collision. In this type of collision, the momentum of the system remains conserved.
It would mean that, after collision, momentum of the system is equal to the initial momentum.
Hence, final momentum = -154 kg-m/s.
Answer:
Keq = 2k₃
Explanation:
We can solve this exercise using Newton's second one
F = m a
Where F is the eleatic force of the spring F = - k x
Since we have two springs, they are parallel or they are stretched the same distance by the object and the response force Fe is the same for the spring age due to having the same displacement
F + F = m a
k₃ x + k₃ x = m a
a = 2k₃ x / m
To find the effective force constant, suppose we change this spring to what creates the cuddly displacement
Keq = 2k₃
Answer:
(a) = -0.16%
(b) = smaller
Explanation:
given
power = 460 W
potential difference = 120 V
(a) what percentage will its heat output drop if the applied potential difference drops to 110 V ?
we know 
.....................(i)
we need to find change in power
..............(ii)
from equations we get
(b)
if we increase temperature resistance will increase and decrease with decrease in temperature and we know power is inversely proportional to resistance so if potential decrease and it would cause drop in power
and due to this increment of heating power resistance will decrease so actual drop in the power would be smaller
3. Law: Every action has a reaction equal in magnitude and opposite in direction.
Answer:
-4.1μC is the final charge on the third sphere
Explanation:
From the given data, q1 and q2 are brought into contact as they are both conductors, as such there will be evenly distribution of charges.
a) charge on each sphere(Q) = q1 + q2 / 2
= +3.8 μC + (- 2.6 μC) / 2 = 1.2μC/2 = 0.6μC
b) Now, one of those two spheres is brought into contact with the third sphere ; Q is brought into contact with q3 = Q + q3 / 2
= 0.6μC - 8.8 μC /2 = -8.2 μC/2
= -4.1μC is the final charge on the third sphere.
