7.5 × 10¹⁴ Hz is the highest frequency of visible light when wavelengths of visible light range from 400 nm to 700 nm.
The distance a wave travels in one unit of time is known as the wave speed (v).Taking into account that the wave travels one wavelength in one interval,
v=λ/T
Given that T = 1/f, we can write the equation above as,
V = f λ
Given data:
Minimum wavelength of visible light = 400 nm = 4 × 10⁻⁷ m
Speed of light = 3 × 10⁸ m/s
Frequency = c/λ = 3 × 10⁸ / 4 × 10⁻⁷
= 7.5 × 10¹⁴ Hz
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I think its [B]
Personally i would say [B] only because If you are looking beyond the car in front of you..... then what if the car in front of you throws on breaks... you would hit them in the butt because you weren't paying attention to the car.
And majority of the time if your looking in the lanes beside you then you are most likely trying to get in that lane.
Answer:
40 cm
Explanation:
We are given that
Load=800 N
Effort=200 N
Load distance=10 cm
We have to find the effort distance.
We know that
Using the formula
Effort distance=
Effort distance=
Effort distance=40 cm
Hence, the effort distance will be 40 cm.
The answer is B - Current Y has a greater potential difference, and the charges flow at a slower rate.
I just took the quiz
Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
