Alex needs to determine the power dissipated in a resistor. She measures the current through the resistor to be I=10.0±0.6 A, and its resistance to be R=45.0±2.0 Ω. Assuming her calculations are correct, Alex should find that the uncertainty in power dissipated in the resistor due only to the uncertainty in the current is:

Explanation:

Given that,

Current measure is

i=10±0.6 Amps

And also,

R=45.0±2.0 Ω

Power dissipated by

P=i²R

Then

P=(10±0.6)²(45.0±2.0)

P=10²×45

P=450Watts

Now, calculating the uncertainty

∆P=|P| • √(2(∆i/i)²+(∆R/R)²)

∆P=450√ (2×(0.6/10)²+(2/45)²)

∆P=450√(0.0072+0.001975)

∆P=450√0.009175

∆P=43.1

The uncertainty in power is 43.1

Then,

P=450 ± 43.1 Watts

i think the answer is a and b

i wish you luck

you'd use the equation that establishes a relationship between  energy  and  wavelength, which looks like this

e=h⋅cλ, where

λ  - the wavelength of the photon; c  - the speed of light, usually given as  3⋅108m/s; h  -  planck's constant, equal to  6.626⋅10−34j s

plug the value you were given for the energy of the photon in the above equation to get the wavelength

e=h⋅cλ⇒λ=h⋅ce

λ=6.626⋅10−34js⋅3⋅108ms−1.257⋅10−24j=1.581⋅10−1m