law of conservation of energy
aka the first law of thermodynamics
A because it's the smaller the thicker you just can't have 0 gage
Answer:
4.4 seconds
Explanation:
Given:
a = -5.5 m/s²
v₀ = 0 m/s
y₀ = 53 m
y = 0 m
Find: t
y = y₀ + v₀ t + ½ at²
0 = 53 + 0 + ½ (-5.5) t²
0 = 53 − 2.75 t²
t = 4.39
Rounded to two significant figures, it takes 4.4 seconds for the object to land.
A 5.00 A current runs through a 12 gauge copper wire (diameter 2.05 mm) and through a light bulb. Copper has 8.5*10^28 free electrons per cubic metre.
a) How many electrons pass through the light bulb each second?
b) What is the current density in the wire? (answer in A/m^2)
<span>c) At what speed does a typical electron pass by any given point in the wire? (answer in m/s)
</span>a) 5.0 A = 5.0 C/s
. Number of electrons in 5.0C = 5.0 / 1.60^-19 = 3.125^19
. 5.0 A ►= 3.125^19 electrons/s
b) A/m² = 5.0 / π(1.025^-3 m)² .. .. ►= 1.52^6 A/m²
c) Charge density (q/m³) = 8.50^28 e/m³ x 1.60^-19 = 1.36^10 C/m³
(q/m³)(m²)(m/s) = q/s (current i in C/s [A])
(m²) = Area
(m/s) = mean drift speed
(q/m³)(A)(v) = i
v = i.[(q/m³)A]ˉ¹
<span>v = 5.0 [1.36^10 * π(1.025^-3 m)²]ˉ¹.. .. ►v = 1.10^-4 m/s</span>
Answer:
the maximum frequency observed is 2.0044 10⁶ Hz
Explanation:
This is a Doppler effect exercise. Where the emitter is still and the receiver is mobile, therefore the expression that describes the process is
f ’= 
the + sign is used when the observer approaches the source
typical speeds of a baby's heart stop are around 200 m / min
let's reduce to SI units
v₀ = 200 m / min (1 min / 60 s) = 3.33 m / s
let's calculate
f ’= 2 10⁶ (
)
f ’= 2.0044 10⁶ Hz
f ’= 1,9956 10⁶ Hz
therefore the maximum frequency observed is 2.0044 10⁶ Hz
