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amm1812
3 years ago
15

A conductor has a net charge of +24 uC. How many electrons are needed to make this conductor neutral

Physics
1 answer:
Arte-miy333 [17]3 years ago
7 0
The charge on the conductor is +24 μC,  since the charge is positive, same amount of charge but it will be negative.

That is - 24 μC.

To know the number of electrons, we know that the charge of an  electron =
-1.6 * 10^-19 C.

=    - 24 μC / (-1.6 * 10^-19 C)

=  (-24 * 10^-6) / (-1.6 * 10^-19 C)      Use your calculator

=  15 * 10^13

=  1.5 * 10 ^14  Electrons.
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Answer:

The answer is "In this information, Enzo will be confident that if any fertilizer is required only for plants".

Explanation:

The Beans plants were members of the group of legumes. Legume flowering plant under which the roots form nodules. These nodules require a specific bacterium called bacteria, which fix nitrogen. Its bacterium transforms nitrogen from the atmosphere into the soil's nitrate. Consequently, for some of its growth, beans do not require fertilizer.

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3 years ago
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Feel better and develop communication skills

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6 0
3 years ago
Water is boiled at sea level in a coffeemaker equipped with an immersion-type electric heating element. The coffee maker contain
Luden [163]

Answer:

P=1362\ W

t'=251.659\ s is time required to heat to boiling point form initial temperature.

Explanation:

Given:

initial temperature of water, T_i=18^{\circ}C

time taken to vapourize half a liter of water, t=18\ min=1080\ s

desity of water, \rho=1\ kg.L^{-1}

So, the givne mass of water, m=1\ kg

enthalpy of vaporization of water, h_{fg}=2256.4\times 10^{-3}\ J.kg^{-1}

specific heat of water, c=4180\ J.kg^{-1}.K^{-1}

Amount of heat required to raise the temperature of given water mass to 100°C:

Q_s=m.c.\Delta T

Q_s=1\times 4180\times (100-18)

Q_s=342760\ J

Now the amount of heat required to vaporize 0.5 kg of water:

Q_v=m'\times h_{fg}

where:

m'=0.5\ kg= mass of water vaporized due to boiling

Q_v=0.5\times 2256.4

Q_v=1.1282\times 10^{6}\ J

Now the power rating of the boiler:

P=\frac{Q_s+Q_v}{t}

P=\frac{342760+1128200}{1080}

P=1362\ W

Now the time required to heat to boiling point form initial temperature:

t'=\frac{Q_s}{P}

t'=\frac{342760}{1362}

t'=251.659\ s

6 0
3 years ago
Sam is pulling a box up to the second story of his apartment via a string. The box weighs 53.3 kg and starts from rest on the gr
castortr0y [4]

Answer:

W = 1222.4 J = 1.22 KJ

Explanation:

The work done on an object is the product of the force applied on it and the displacement it covers as a result of this force. It must be noted that the component of displacement in the direction of force should only be used. Hence, the work can be calculated as:

W = F d Cosθ

where,

W = Work Done = ?

F = Force Applied = 64 N

d = Distance Covered by Box = 19.1 m

θ = Angle between force and displacement = 0°

Therefore,

W = (64 N)(19.1 m)Cos 0°

<u>W = 1222.4 J = 1.22 KJ</u>

7 0
2 years ago
PLEASE HELP THIS IS DUE LIKE HELP ME PLEASE
marysya [2.9K]

Answer:

At the end points of motion (either side) the velocity must be zero because the velocity is changing from - to + (it can't turn around around without passing thru zero,

The velocity will then increase to the midpoint of the motion.

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2 years ago
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