In a series circuit one of the resistors is replaced with a resistor having a lower resistance value. how does this affect the current in the circuit?

In a series circuit, the equivalent resistance is given by the sum of individual resistances of all the resistors in series in the circuit. As a resistor is replaced by another resistor of lower resistance value, the sum of resistances will decrease. hence total resistance of the circuit decrease.

In a series circuit, the current in the circuit is given using ohm's law as

i = V/R

where V = battery Voltage

R = Total resistance of the resistors in series.

we know that the battery voltage remains same all the time. hence the current depends inversely on the resistance of the circuit. As the resistance of the circuit decreases on replacement of the one resistor, the current in the circuit increases.

Increase.

The velocity of the second mass is -0.8*v

Explanation: This is a perfectly inelastic collision, where two objects that collide move together as if they were stuck to each other.

The final velocity in this type of collision is

Vf = (m1*v1 + m2*v2)/(m1 + m2)

Where the "m"'s are the masses of each object, and the v's are the velocities.

Here we know that m1 = m2 = m, v1 = v, Vf = 0.100*v

we want to find the value of v2

0.100*v = (m*v + m*v2)/(2m) = m(v + v2)/(2m) = (v + v2)/2

0,100*v = v/2 + (v2)/2

v(0.100 - 1/2) = (v2)/2

-v*0.400*2 = v2

v2 = -v*0.8

Then we know that the second mass is moving with a velocity of -0.8*v (the minus sign means that this object is moving in the opposite direction with respect to the other object)

One interesting thing in this type of collision is that the kinetic energy does not conserve