The wavelengths of the light are 4.3 * 10^-12 m and 0.2 m respectively.
<h3>What is wavelength?</h3>
The term wavelength has to do with the horizontal distance that is covered by a wave. We know that a long wavelength implies that the wave is able to travel a long distance from one point to another.
Given that;
c = λf
c = speed of light
λ = wavelength of ight
f = frequency of light
Thus;
λ = 3 * 10^8/ 7.00 x 10^19
λ = 4.3 * 10^-12 m
λ = 3 * 10^8/1.50 x 10^9
λ = 2 * 10^-1 or 0.2 m
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Missing parts:
What are the wavelengths of electromagnetic wave in free space that have the following frequencies? (a) 7.00 x 10^19 Hz______ pm (b) 1.50 x 10^9 Hz__________ cm
Answer:
Option B
Explanation:
Magnification of Microscope is
Mo= Magnification of objective lens and
Me= magnification of the eyepiece.
Both magnifications( of objective and eyepiece) are inversely proportional to the focal length.
Magnification,
when the focal length is less magnification will be high and when the magnification is the low focal length of the microscope will be more.
Thus. Magnification will increase by decreasing the focal length.
The correct answer is Option B i.e. using shorter focal length
It will take 21s for yhe alligator to crawl that distance. Reported answer contains 2 Significant Figures.
Answer:
0.739
Explanation:
If we treat the four tire as single body then
W ( weight of the tyre ) = mass × acceleration due to gravity (g)
the body has a tangential acceleration = dv/dt = 5.22 m/s², also the body has centripetal acceleration to the center = v² / r
where v is speed 25.6 m/s and r is the radius of the circle
centripetal acceleration = (25.6 m/s)² / 130 = 5.041 m/s²
net acceleration of the body = √ (tangential acceleration² + centripetal acceleration²) = √ (5.22² + 5.041²) = 7.2567 m/s²
coefficient of static friction between the tires and the road = frictional force / force of normal
frictional force = m × net acceleration / m×g
where force of normal = weight of the body in opposite direction
coefficient of static friction = (7.2567 × m) / (9.81 × m)
coefficient of static friction = 0.739
Answer:
1. sediment layering and compacting on top of each other and solidifying
2. sediment layering and compacting (not as much) on top of each other and solidifying, just not as much
Explanation:
hope this helps! :))
