The answer is D, it falls between the infrared and ultraviolet.
Answer:
1.82 rad/s².
Explanation:
Applying,
α = (ω₂-ω₁)/t..................... Equation 1
Where α = angular acceleration of the fan blades, ω₂ = final angular velocity of the fan blades, ω₁ = initial angular velocity of the fan blades, t = time.
Given: ω₂ = 350 rpm = (350×0.1047) rad/s = 36.645 rad/s. ω₁ = 250 rpm = (250×0.1047) rad/s = 26.175 rad/s, t = 5.75 s.
Substitute into equation 1
α = (36.645-26.175)/5.75
α = 10.47/5.75
α = 1.82 rad/s².
Hence the magnitude of the angular acceleration of the fan blades = 1.82 rad/s²
Answer:
λ
=8.57 μ m
Explanation:
Given that
Ey = 375 cos [kx − (2.20 × 10¹⁴ rad/s)t] N/C
Standard form
Ey=Eo cos[k x-ωt] N/C
By comparing the given equation with the standard wave equation
Eo = 375 N/C
ω = 2.20 × 10¹⁴ rad/s
We know that ω = 2 π f
f=3.50×10¹³ Hz
We know that the velocity given as
V = f λ
λ
=Wavelength
V=Speed = 3 x 10⁸ m/s
λ
=0.00000857 m ( 1 μ m = 10⁶ m)
λ
=8.57 μ m
Work is force times distance, so W = 40 N * 10 m = 400 J
Answer:
I only know how to answer 22
which is 250 J
Explanation:
KE: 1/2 mv^2
m- 5kg
v- 10 m/s
KE: 1/2 × 5 × 10^2
= 250 J
