Answer:
At 11.5 m
Explanation:
The power per unit area corresponds to the intensity, which is given by

where
P is the power
is the area irradiated at a distance r from the source (it corresponds to the surface area of a sphere of radius r)
Here we want the intensity of the two light bulbs to be the same, so

where we have
P1 = 100 W is the power of the first light bulb
P2 = 75 W is the power of the second light bulb
r2 = 10 m is the distance from the second light bulb
Solving for r1, we find

<h2>
Answer: U-238</h2>
Explanation:
Let's begin by explaining that for radioactive geological dating (also called radioisotope dating) in which radioactive impurities were selectively incorporated when the fossil materials were formed, it is very useful to compare it with a naturally occurring radioisotope having a known half-life.
Now, taking into account that the <u>fossils are millions and millions of years old, radioisotopes are needed that exceed this measure.
</u>
To understand it better:
The longer the half-life of a radioisotope, the greater its utility for estimating fossil ages or geological formations.
In this sense, uranium-238 (U238) has a half-life of 4,470 million years, therefore, it is among the most commonly used radioisotopes for fossil and geological dating.
Answer:
violet has the highest frequency
Explanation:
Answer
The rate at which the magnetic field is changing is
Explanation
From the question we are told that
The electric field strength is 
The radius is 
The rate of change of the magnetic field is mathematically represented as

Where
is change of a unit length

Where A is the area which is mathematically represented as

So
where L is the circumference of the circle which is mathematically represented as

So
![E (2 \pi r ) = (\pi r^2 ) [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%282%20%5Cpi%20r%20%29%20%3D%20%20%28%5Cpi%20r%5E2%20%29%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![E = \frac{r}{2} [\frac{dB}{dt} ]](https://tex.z-dn.net/?f=E%20%20%3D%20%20%20%5Cfrac%7Br%7D%7B2%7D%20%20%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D)
![[\frac{dB}{dt} ] = \frac{E}{ \frac{r}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7BE%7D%7B%20%5Cfrac%7Br%7D%7B2%7D%20%7D)
substituting values
![[\frac{dB}{dt} ] = \frac{3.5 *10^{-3}}{ \frac{15}{2} }](https://tex.z-dn.net/?f=%5B%5Cfrac%7BdB%7D%7Bdt%7D%20%5D%20%3D%20%5Cfrac%7B3.5%20%2A10%5E%7B-3%7D%7D%7B%20%5Cfrac%7B15%7D%7B2%7D%20%7D)
Answer:
L= 0.059 mH
Explanation:
Given that
R = 855 Ω and C = 6.25 μF
V= 84 V
Frequency
ω = 51900 1/s
We know that

L=Inductance
C=Capacitance
ω =angular Frequency
ω² L C =1
(51900)² x L x 6.25 x 10⁻⁶ = 1
L= 5.99 x 10⁻⁵ H
L= 0.059 mH