Two resistors, one with 7.00 Ω of resistance and the other with 11.00 Ω of resistance, are connected in series to a 9.00 V battery. How much power is dissipated by these two resistors?

4.5 W

Explanation:

Applying,

P = V²/(R₁+R₂) Equation 1

Where P = Power, V = Voltage, R₁ and R₂ = values of the two resistor.

From the question,

Given: V = 9.00 V, R₁ = 7.00 Ω, R₂ = 11.00  Ω

Substitute these values into equation 1

P = 9²/(7+11)

P = 81/(18)

P = 4.5 Watt.

Hence the power dessipated by the two resistors is 4.5 watt

snell's law

refractve index = sin incidence / sin refraction

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//