The metals will start to rust lol. i think. because this messes up how the metals conduct the flow of the electricity.
Answer:
option D
Explanation:
given,
Intensity of sound = 20 dB
distance = 15 m
intensity of sound is increased to = 50 dB
distance between the sound level = ?
Using relation

L₁ = 20 dB L₂ = 50 dB r₁ = 15 m r₂ = ?





r₂ = 0.47 m
r₂ = 47 cm
hence, the correct answer is option D
<h2>
Answer: </h2><h2>
- Jupiter has orbiting moons.</h2><h2>
- The Sun has sunspots and rotates on its axis.</h2><h2>
- The Moon has mountains, valleys, and craters.</h2><h2>
- Venus goes through a full set of phases.</h2>
Explanation:
In 1609 Galileo built a telescope, with which he observed mountains and craters on the Moon, discovered Jupiter’s major satellites and the next year he published these discoveries in his book <em>The Sidereal Messenger</em>.
In addition, Galileo observed that Venus presented phases (such as those of the moon) together with a variation in size; observations that are only compatible with the fact that Venus rotates around the Sun and not around Earth. This is because <u>Venus presented its smaller size when it was in full phase and the largest size when it was in the new one, when it is between the Sun and the Earth. </u>
<u />
On the other hand, <u>although Galileo was not the first to observe sunspots</u>, he gave the correct explanation of their existence, which supported the idea that planets revolve around the Sun.
These observations and discoveries were presented by Galileo to the Catholic Church (which supported the geocentric theory at that time) as a proof that completely refuted Ptolemy's geocentric system and affirmed Copernicus' heliocentric theory.
Answer:84Nm
Explanation:
force=400N
Distance=0.210m
Workdone=force x distance
Workdone=400 x 0.210
Workdone=84Nm
Answer:
The correct answer is B
Explanation:
Let's calculate the electric field using Gauss's law, which states that the electric field flow is equal to the charge faced by the dielectric permittivity
Φ
= ∫ E. dA =
/ ε₀
For this case we create a Gaussian surface that is a sphere. We can see that the two of the sphere and the field lines from the spherical shell grant in the direction whereby the scalar product is reduced to the ordinary product
∫ E dA =
/ ε₀
The area of a sphere is
A = 4π r²
E 4π r² =
/ ε₀
E = (1 /4πε₀
) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀
) Q / (1.25R)²
E = (1 /4πε₀
) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀
) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B