<span>The primary reason a light bulb emits light is due to the heating of the resistance in the filament of the light bulb. In fact, the power dissipated in a resistor is given by
</span>
<span>where I is the current and R the resistance. The larger the resistance or the current in the resistor, the larger the power dissipated. Due to this dissipation of power, the temperature of the filament becomes very high, and the resistance becomes incandescent, emitting light.</span>
Answer:
Option 10. 169.118 J/KgºC
Explanation:
From the question given above, the following data were obtained:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1.61 KJ
Mass of metal bar = 476 g
Specific heat capacity (C) of metal bar =?
Next, we shall convert 1.61 KJ to joule (J). This can be obtained as follow:
1 kJ = 1000 J
Therefore,
1.61 KJ = 1.61 KJ × 1000 J / 1 kJ
1.61 KJ = 1610 J
Next, we shall convert 476 g to Kg. This can be obtained as follow:
1000 g = 1 Kg
Therefore,
476 g = 476 g × 1 Kg / 1000 g
476 g = 0.476 Kg
Finally, we shall determine the specific heat capacity of the metal bar. This can be obtained as follow:
Change in temperature (ΔT) = 20 °C
Heat (Q) absorbed = 1610 J
Mass of metal bar = 0.476 Kg
Specific heat capacity (C) of metal bar =?
Q = MCΔT
1610 = 0.476 × C × 20
1610 = 9.52 × C
Divide both side by 9.52
C = 1610 / 9.52
C = 169.118 J/KgºC
Thus, the specific heat capacity of the metal bar is 169.118 J/KgºC
Answer:
The frictional torque is 
Explanation:
From the question we are told that
The mass attached to one end the string is 
The mass attached to the other end of the string is 
The radius of the disk is 
At equilibrium the tension on the string due to the first mass is mathematically represented as
substituting values
At equilibrium the tension on the string due to the mass is mathematically represented as
The frictional torque that must be exerted is mathematically represented as
substituting values
The formula to find the kinetic energy is:
Ek= 1/2 × m × v^2
1. Ek= 1/2×15×3^2
= 67.5 J
2.Ek= 1/2×8×4^2
=64 J
3.Ek= 1/2×12×5^2
= 150 J
4.Ek= 1/2×10×6^2
= 180 J
So the fourth dog has the most kinetic energy.
