Two resistance r1=30 ω and r2=20 ω are connected in series to a 10v battery. what is the heat dissipated in r2 during a1/2 h?

Heat dissipated, H = 1440 Joules

Explanation:

It is given that,

Resistor 1,

Resistor 2,

Voltage, V = 10 volts

Both resistors are connected in series. The equivalent resistance is given by :

R = 30 + 20

R = 50 ohms

In series combination, the current throughout the circuit remains the same. It can be calculated using ohm's law as :

V = IR

I = 0.2 A

Heat dissipated in R2 during (1/2) h, t = 1800 s is given by :

H = 1440 J

So, the heat dissipated during 1800 seconds is 1440 Joules. Hence, this is the required solution.

found in the nucleus

has mass of one amu

can be answered using momentum.

momentum = p

p1= mass1 * velocity1

p1 = p2

therefore

p1 = 6kg * 4m/s

p1 = 24kgm/s

p2=m2 * v2

24 kgm/s = 7kg * v2

v2 = 24 kgm/s / 7kg

v2= 3.42 m/s