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3 years ago

Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST

Physics

2 answers:

Sauron [17]3 years ago

7 0

Electrons are negatively charged and are found outside in the nucleus in the rings surrounding the nucleus.

oksian1 [2.3K]3 years ago

4 0

Charge: CURRENT is the rate of flow of charge

Electrons: the flow of electrons in a traditional CURRENT flow from the negative(-) to the positive(+)

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4. A cinder block is sitting on a platform 20 m high. It has a mass of 4 kg. The block has energy. Calculate it.

zavuch27 [327]

Answer:

Explanation:

Since the block is at rest in an elevated position, we can assume that it only has potential energy.

U=mgh is the formula for potential energy where U=potential energy, m= mass, g=acceleration due to gravity, and h=height.

Plug in known variables....

U=4kg*9.8m/s^2*20m

U=784 joules of potential energy or letter A.

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2 years ago

1. You place an object 63 cm in front of a converging lens, with a 40 cm focal length.

NikAS [45]

Answer:

109.6 cm , - 1.74 , real

1.5

Explanation:

d₀ = object distance = 63 cm

f = focal length of the lens = 40 cm

d = image distance = ?

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{40} = \frac{1}{63} + \frac{1}{d}$

d = 109.6 cm

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{-109.6}{63}$

m = - 1.74

The image is real

d₀ = object distance = a

d = image distance = - (a + 5)

f = focal length of lens = 30 cm

using the lens equation

$\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}$

$\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}$

a = 10

magnification is given as

$m = \frac{-d}{d_{o}}$

$m = \frac{- (- (a +5))}{a}$

$m = \frac{(5 + 10)}{10}$

m = 1.5

6 0

3 years ago

A very long solid insulating cylinder has radius R = 0.1 m and uniform charge density rho0= 10-3 C/m3. Find the electric field a

Galina-37 [17]

Answer:

$E = (0.56 \times 10^8 ) r \ \ N/c$

Explanation:

Given that:

$\rho_o = (10^{-3} ) \ c/m^3$

R = (0.1) m

To find the electric field for r < R by using Gauss Law

${\oint}E^{\to}* da^{\to} = \dfrac{Q_{enclosed}}{\varepsilon_o} --- (1)$

For r < R

$Q_{enclosed}=(\rho) ( \pi r^2 ) l$

$E*(2 \pi rl)= \dfrac{\rho ( \pi r ^2 l)}{\varepsilon_o}$

$E= \dfrac{\rho ( r)}{2 \varepsilon_o}$

where;

$\varepsilon_o = 8.85 \times 10^{-12}$

$E= \dfrac{10^{-3} ( r)}{2 (8.85 \times 10^{-12})}$

$E = (0.56 \times 10^8 ) r \ \ N/c$

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3 years ago

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Answer:

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3 years ago

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siniylev [52]

B. The voltage is the same across all resistors in the circuit.

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3 years ago

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Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST​

Describe current in terms of charge and electrons.PLEASE HELP I WILL GIVE BRAINLIEST