If you wish to take a picture of a bullet traveling at 530 m/s, then a very brief flash of light produced by an RC discharge through a flash tube can limit blurring. Assuming 1.00 mm of motion during one RC constant is acceptable, and given that the flash is driven by a 530 µF capacitor, what is the resistance (in Ω) in the flash tube?

3.56*10^-3 Ω

Explanation:

Given

Speed of bullet, v = 530 m/s

Distance during one RC time constant, d = 1 mm = 1*10^-3 m

Capacitance of the capacitor, C = 530 µF

v = d / t = d / τ, so that

τ = d / v

The time constant for the RC circuit is

τ = RC, if we make R subject of formula we have

R = τ / C. Substituting the relation for τ

R = d / vC

R = 1*10^-3 / (530 * 530*10^-6)

R = 1*10^-3 / 0.2809

R = 3.56*10^-3 Ω

Therefore, the resistance in the flash tube is 3.56*10^-3 mΩ

gpe = mass x gravity constant x height or… gpe = mgh

where… mass: measured in kilograms (kg)

gravity: gravity constant on earth = 10.00 m/s2

height: measured in meters (m)

gpe: measured in joules (j) or kilojoules (kj)

how do you solve gpe?

gravitational potential energy (gpe) is the energy an object has because of its position above the. ground.

gpe = mass x gravity constant x height or… gpe = mgh.

mass:

note: to earn full marks when solving science word problems, you must show your work:

formula.

example:

given values:

mass = 0.3 kg.