Answer
Hi,
An increase in amplitude from 3m to 6 m increases the energy it transports. The frequency of the wave is not affected
Explanation
Amplitude is the height of a wave where as frequency is the number of waves that pass by each second. A wave with bigger amplitude has more energy than a wave with smaller amplitude. A point where more waves pass contains more energy that is transferred every second. The change in the amplitude of a wave does not change its frequency. However, frequency is inversely related to the wavelength of a wave.
Best Wishes!
Not totally sure but i would say a normal? its not refraction or incidence if its perpendicular and i dont think its a mirror if its an imaginary line so yeah normal (normals are always perpendicular to their surface too i think so)
F=ma=m(change in velocity/change in time)
Number 1
F=ma
F=55kg(1.1ms^-1/1.6s)=37.8N
Number 2
F=ma
F=0.440kg(10ms^-1/0.02s)=220N
Number 3
F=ma
F=1400kg(15ms^-1/0.73s)=2.88*10^3N or 28,767N
Any questions please feel free to ask.
Answer:
2.9 m
Explanation:
First find the time it takes to reach the floor.
y = y₀ + v₀ t + ½ at²
(0 m) = (1.6 m) + (0 m/s) t + ½ (-9.8 m/s²) t²
t = 0.571 s
Next, find the distance it travels in that time.
x = x₀ + v₀ t + ½ at²
x = (0 m) + (5.0 m/s) (0.571 s) + ½ (0 m/s²) (0.571 s)²
x = 2.86 m
Rounded to two significant figures, the marble travels 2.9 meters in the x direction.
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>
