D, electron, the nucleus is not a single particle to begin with, the proton has a positive charge, a neutron has a neutral charge or no charge, and an electron has a negative charge
Answer:
required distance is 233.35 m
Explanation:
Given the data in the question;
Sound intensity
= 1.62 × 10⁻⁶ W/m²
distance r = 165 m
at what distance from the explosion is the sound intensity half this value?
we know that;
Sound intensity
is proportional to 1/(distance)²
i.e
∝ 1/r²
Now, let r² be the distance where sound intensity is half, i.e
₂ =
₁/2
Hence,
₂/
₁ = r₁²/r₂²
1/2 = (165)²/ r₂²
r₂² = 2 × (165)²
r₂² = 2 × 27225
r₂² = 54450
r₂ = √54450
r₂ = 233.35 m
Therefore, required distance is 233.35 m
Vi=0m/s
Vf=?
A=9.81
D=44
T=not needed
Vf^2=Vi^2+2ad
Vf=2ad square rooted
Vf=2(9.81)(44) square root it
Vf=29.3m/s
Answer:
The fuse must be connected between the device and the power intake source.
Explanation:
A fuse is a protective component of electrical appliances that is designed to be sensitive to a particular range of electric current
The fuse is made of a thing metal strip with a known melting point. Once current abive its carrying capacity flows through it, large heat is generated in the metal strip which melts it and causes the metal strip to cut int two protecting the device from the power spike.