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diamong [38]
3 years ago
8

Why does a rock weigh less in water than it does on land?

Physics
1 answer:
Cloud [144]3 years ago
7 0
Water is more dense then air so it sorta holds the rock as it sinks
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What is physical quantity ? Give examples.​
Dahasolnce [82]

Explanation:

physical quantity is any physical property that can be qualified that,is, be measured using numbers e.g mass, amount of substance,time and length

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3 years ago
At one point in the circuit, there is an LED and a resistor in parallel with one another. If you measure the voltage drop across
kifflom [539]

Answer:

C. The voltage drop across the resistor is 2.1V and nothing about the current through the resistor.

Explanation:

When connected in parallel, voltage across the resistances are the same. So if 2.1V was dropped across the LED then 2.1V was also dropped across the resistor. However, this tells us nothing about the current through the resistor. We can find the current across the resistor if we know the resistance of the resistor, but that's about it.

If it were a series connection, then the current would have been the same, but the voltage drop were another story.

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Is the process of introducing a non blood fluid into the blood​
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The lymphatic system
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What would you predict for this elephant's stride frequency? That is, how many steps per minute will the elephant take?
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8 0
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Read 2 more answers
A uniformly charged solid disk of radius R = 0.45 m carries a uniform charge density of σ = 175 μC/m². A point P is located a di
siniylev [52]

Answer:

1408.685 KN/C

Explanation:

Given:

R = 0.45 m

σ = 175 μC/m²

P is located a distance a = 0.75 m

k = 8.99*10^9

  • The Electric Field Strength E of a uniformly solid disk of charge at distance a perpendicular to disk is given by:

                                  E = 2*pi*k*o * (1 - \frac{a}{\sqrt{a^2 + R^2} })\\

part a)

Electric Field strength at point P: a = 0.75 m

E = 2*pi*8.99*10^9*175*10^-6 * (1 - \frac{0.75}{\sqrt{0.75^2 + 0.45^2} })\\\\E = 9885021.285*(0.1425070743)\\\\E = 1408.685 KN/C

part b)

Since, R >> a, we can approximate a / R = 0 ,

Hence, E simplified relation becomes:

E = 2*pi*k*o * (1 - \frac{a/R}{\sqrt{a^2/R^2 + 1} })\\\\E = 2*pi*k*o * (1 - \frac{0}{\sqrt{0 + 1} })\\\\E = 2*pi*k*o

E = σ / 2*e_o

part c)

Since, a >> R, we can approximate. that the uniform disc of charge becomes a single point charge:

Electric Field strength due to point charge is:

E = k*δ*pi*R^2 / a^2  

Since, R << a, Surface area = δ*pi

Hence,

E = (k*δ*pi/a^2)

 

6 0
3 years ago
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