The range of frequencies of visible light in a vacuum is mathematically given as
Fmin=4.19*10^14Hz to Fmax=1*10^15Hz
<h3>What is the range of frequencies of visible light in a vacuum?</h3>
Question Parameters:
The wavelengths of visible light vary from about 300 nm to 700 nm.
Generally, the equation for the frequency is mathematically given as
F=C/\lambda
Therefore
For Fmax

Fmax=1*10^15Hz
Where

Fmin=4.19*10^14Hz
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Answer:
Useable energy, electricity
Answer:
Option C. Objects 1 and 3 will not move, and objects 2 and 4 will accelerate
upward.
Explanation:
The following data were obtained from the question:
OBJECT >>>>>>>>> WEIGHT (N)
1 >>>>>>>>>>>>>>>> 35
2 >>>>>>>>>>>>>>>> 23
3 >>>>>>>>>>>>>>>> 26
4 >>>>>>>>>>>>>>>> 18
Force (F) applied = 25 N
From the above, the force applied to each object is 25N. Thus the following can be concluded based on the data given above:
For object 1:
Weight = 35 N
Force applied = 25 N
Thus, the object will not move since the weight of the object is greater than the force applied
For object 2:
Weight = 23 N
Force applied = 25 N
Thus, the object will move since the force applied is greater than the weight of the object.
For object 3:
Weight = 26 N
Force applied = 25 N
Thus, the object will not move since the weight of the object is greater than the force applied.
For object 4:
Weight = 18 N
Force applied = 25 N
Thus, the object will move since the force applied is greater than the weight of the object.
From the above illustrations, Object 1 and 3 will not move, and objects 2 and 4 will accelerate i.e move
Answer:
9.01amp
Explanation:
Power = V^2/R
Given that v = 11volts, P = 99watts
99 = 11^2/R
11×11 = 99R
121= 99R
R = 121/99
R= 1.22ohms
From ohms Law; V = IR
11volts = I × 1.22ohms
I = 11/1.23
I = 9.01 amp
Answer:
PE = (|accepted value – experimental value| \ accepted value) x 100%
Explanation: