A low-power laser used in a physics lab might have a power of 0.50 mW and a beam diameter of 3.0 mm. Calculate:a. The average light intensity of the laser beam. b. The intensity of a lightbulb producing 100-W light viewed from 2.0 m. c.Compare the intensity of the laser to the intensity of the lightbulb. Is it advisable to look directly at a laser

A) I_laser = 70.74 W/m²

B) I_bulb = 1.989 W/m²

C) it is not advisable to look at the laser beam directly.

Explanation:

We are given;

Power; P = 0.50 mW = 0.5 × 10^(-3) W

Diameter; d = 3 mm = 0.003 m

Radius; r = d/2 = 0.003/2 = 0.0015 m

A) Area of beam; A = πr²

A = 0.0015²π

Now, formula for average intensity is;

I = P/A

I = (0.5 × 10^(-3))/0.0015²π

I = 70.74 W/m²

B) We are told to find the intensity of a lightbulb producing 100-W.

Thus, P = 100 W

A light bulb is spherical in shape. Thus;

Area; A = 4πr²

We are told it's 2 m away.

Thus; r = 2 m

A = 4π(2)²

A = 16π

Thus, I = P/A = 100/16π

I = 1.989 W/m²

C) The intensity of the laser beam is far greater than that of the light bulb. Thus, it is not advisable to look at the laser beam directly.

33: i think it’s b

answer: 802.3 kilo joules of energy represents 191.75 kilo calorie of energy.

explanation:

energy released during combustion of one mole methane = 802.3 kilo joules

if 1 kilo joule = 0.239006   kilo calorie

then ,802.3 kilo joule will be = 802.30.239006 kilo calorie

802.3 kilo joule = 191.75 kilo calorie

hence, the energy in kilo calorie is 191.75 kilo calorie.