Total energy associated with a spring:
When x = 0.5a:
The ratio:
When x = 0.5a:
The ratio:
Lasers are classified according to the eye-damage danger they pose. Class 2 lasers, including many laser pointers, produce visib
1 answer:
Response:
a) 318.2 W/m^2
b) 2.5 x 10^-4 J
c) 1.55 x 10^-8 v/m
Reasoning:
The laser power P = 1 mW = 1 x 10^-3 W
duration t = 250 ms = 250 x 10^-3 s
Taking a beam diameter of 2 mm = 2 x 10^-3 m
therefore
the beam's cross-sectional area A = = (3.142 x
)/4
A = 3.142 x 10^-6 m^2
a) The intensity I = P/A
where P refers to the laser's power
and A represents the beam's cross-sectional area
I = ( 1 x 10^-3)/(3.142 x 10^-6) = 318.2 W/m^2
b) The total energy delivered E =Pt
where P is the beam's power
and t is the exposure duration
E = 1 x 10^-3 x 250 x 10^-3 = 2.5 x 10^-4 J
c) The peak electric field can be computed as
E =
where I signifies the beam's intensity
and E is the electric field
c is the speed of light = 3 x 10^8 m/s
= 8.85 x 10^9 m kg s^-2 A^-2
E = = 1.55 x 10^-8 v/m
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Response:
The acceleration of car 2 is four times that of car 1.
Rationale:
Centripetal acceleration occurs when an object travels in a circular route. It can be expressed as:
In this scenario, two race cars are moving at consistent speeds around a circular course. Both automobiles are located at an equal distance from the center, but car 2 is operating at twice the speed of car 1.
Thus,
1 and 2 represent the first and second cars, respectively.
Consequently,
Therefore, car 2's acceleration is four times that of car 1.
Answer:
The equivalent distance in kilometers is 4012 × km.
Explanation:
It's known that 1 millimeter converts to meters. Then, 1 meter converts to
kilometers. Therefore, the conversion for 1 millimeter to kilometers can be stated as
1 mm = m
1 m = km
Thus, 1 mm = ×
km =
km.
Given the distance of 4012 mm, the corresponding distance in kilometers will be
4012 mm = 4012 × km.
The distance therefore is 4012 × km.