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VLD [36.1K]
3 years ago
12

Which diagram best represents the electric field around a negatively charged conducting sphere? (See pic)

Physics
1 answer:
dalvyx [7]3 years ago
3 0
The answer is D !!!!!!!
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Define :density٬archimedes principle
olga2289 [7]

Answer:

density is defined as the amount of mass contained in unit volume of a body .its si unit is kg/m*3

5 0
3 years ago
Whiteout friction you could not, write, drive or fly and airplane. Why not?
aleksklad [387]

friction is the resistance that one surface or object encounters when moving over another. Due to gravity pulling everything down things need to friction in order to move

i hope this helps :/


7 0
3 years ago
In which region of the ear does resonance allow the brain to interpret sound answer
lakkis [162]
I'm not too sure what your asking but here are two answers that may help.
The ear drum amplifies the vibrations.
The cochlea changes vibrations into electric signals.
7 0
3 years ago
What is the approximate mass of air in a living room 4.5m×3.4m×2.9m? the density of air is 1.29 kg/m3?
topjm [15]
First we have to calculate the volume of the living room:
V = L x W x H = 4.5 m * 3.4 m * 2.9 m
V = 44.37 m³
We know that Density = 1.29 kg/m²
D = m / V
m = D · V
m = 1.29 kg/m³ · 44.37 m³
m = 57.2373 kg ≈ 57.2 kg
Answer: The approximate mass of air in living room is 57.2 kg.
6 0
3 years ago
Starting from zero, the electric current takes 2 seconds to reach half its maximum possible value in an RL circuit with a resist
Leno4ka [110]

Answer:

time=4s

Explanation:

we know that in a RL circuit with a resistance R, an inductance L and a battery of emf E, the current (i) will vary in following fashion

i(t)=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}}), where imax=\frac{E}{R}

Given that, at i(2)=\frac{imax}{2} =\frac{E}{2R}

⇒\frac{E}{2R}=\frac{E}{R}(1-e^\frac{-2}{\frac{L}{R}})

⇒\frac{1}{2}=1-e^\frac{-2}{\frac{L}{R}}

⇒\frac{1}{2}=e^\frac{-2}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{2})=\frac{-2}{\frac{L}{R}}

⇒log(2)=\frac{2}{\frac{L}{R}}

⇒\frac{L}{R}=\frac{2}{log2}

Now substitute i(t)=\frac{3}{4}imax=\frac{3E}{4R}

⇒\frac{3E}{4R}=\frac{E}{R}(1-e^\frac{-t}{\frac{L}{R}})

⇒\frac{3}{4}=1-e^\frac{-t}{\frac{L}{R}}

⇒\frac{1}{4}=e^\frac{-t}{\frac{L}{R}}

Applying logarithm on both sides,

⇒log(\frac{1}{4})=\frac{-t}{\frac{L}{R}}

⇒log(4)=\frac{t}{\frac{L}{R}}

⇒t=log4\frac{L}{R}

now subs. \frac{L}{R}=\frac{2}{log2}

⇒t=log4\frac{2}{log2}

also log4=log2^{2}=2log2

⇒t=2log2\frac{2}{log2}

⇒t=4

5 0
3 years ago
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