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An electric current in a conductor varies with time according to the expression i(t) = 110 sin (120πt), where i is in amperes and t is in seconds. what is the total charge passing a given point in the conductor from t = 0 to t = 1/180 s?

Answers

Answer 1

Answer:

As we know that current is defined as rate of flow of charge

$i = (dq)/(dt)$

so by rearranging the equation we can say

$q = \int i dt$

here we know that

$i(t) = 110 sin(120\pi t)$

here we will substitute it in the above equation

$q = \int 110 sin(120\pi t) dt$

$q = 110 [- (cos(120\pi t))/(120\pi)]$

now here limits of time is from t = 0 to t = 1/180s

so here it will be given as

$q = (110)/(120\pi)( -cos0 + cos((2\pi)/(3)))$

$q = 0.44 C$

so total charge flow will be 0.44 C

Answer 2

Answer:

Answer:

The total charge passing a given point in the conductor is 0.438 C.

Explanation:

Given that,

The expression of current is

$i(t)=110\sin(120\pi t)$

$(dq(t))/(t)=110\sin(120\pi t)$

$dq(t)=110\sin(120\pi t)dt$ ....(I)

We need to calculate the total charge

On integrating both side of equation (I)

$\int_(0)^(q)dq(t)=\int_(0)^{(1)/(180)}110\sin(120\pi t)dt$

$q=110((-\cos(120\pi t))/(120\pi))_(0)^{(1)/(180)}$

$q=-(110)/(120\pi)(cos(120\pi((1)/(180)))-\cos120\pi(0))$

$q=-0.2918(-(1)/(2)-1)$

$q=0.438\ C$

Hence, The total charge passing a given point in the conductor is 0.438 C.

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What allows two or more atoms to be held together? shared protons shared electrons shared energy shared neutrons

Answers

It is shared electrons.

The following information should be considered:

In the case when two or more atoms can be together at the time when they share electrons with each other.
By sharing, they create a covalent bond and that way the atoms can be stable.

learn more: brainly.com/question/2514933?referrer=searchResults

Answer:

try electrons i hope this helps!!

Explanation:

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

Answers

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_(1) +q_(2) =-q_(3)$ -----eqution 1

where,

$q_(1)$ is the heat absorbed by the solid at 0⁰C

$q_(2)$ is the heat absorbed by the liquid at 0⁰C

$q_(3)$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *(1 mole H_(2) O)/(18g) = 13.167 moles of H_(2)O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_(1) = 13.167 moles *6.01(KJ)/(mole) = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_(2) = -q_(3)$

Step 4: calculate the final temperature of the water

$79.13KJ+M_(sample) *C*\delta {T_(sample)} =-M_(water) *C*\delta {T_(water)$

Substitute in the values; we will have,

$79.13KJ + 237*4.18(J)/(g^(o)C)*(T_(f)-218}) = -350*4.18(J)/(g^(o)C)*(T_(f)-100})$

79.13 kJ + 990.66J* $(T_(f)-218})$ = -1463J* $(T_(f)-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_(f)-218})$ = -1.463KJ* $(T_(f)-100})$

$79.13 + 0.99066T_(f) -215.96388= -1.463T_(f)+146.3$

collect like terms,

2.45366 $T_(f)$ = 283.133

∴ $T_(f) =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

The particle with charge q is now released and given a quick push; as a result, it acquires speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m , what was its initial speed v ?.

Answers

The given situation is illustrated below. A particle is released and given a quick push. As a result, it acquires a speed v. Eventually, this particle ends up at the center of the original square and is momentarily at rest. If the mass of this particle is m, the initial speed of the particle is $√((2qV/m).)$

To solve the problem, we need to apply the law of conservation of energy, which states that energy can neither be created nor destroyed; it can only be transformed from one form to another.
Initial potential energy = Final kinetic energy
The initial potential energy of the particle is given by
U = qV
where V is the potential difference between the corner and the center of the square.
At the center of the square, the potential energy is zero.
The final kinetic energy of the particle is given by
K = (1/2) mv^2
where m is the mass of the particle and v is its final velocity.
Since the particle is momentarily at rest at the center of the square, its final kinetic energy is zero.
Therefore, we have
qV = (1/2) mv^2
Solving for v, we get
v = $√((2qV/m).)$

for such more question on speed

brainly.com/question/13943409

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Determine whether or not each of the following statement is true. If a statement is true, prove it. If the statement is false, provide a counterexample and explain how it constitutes a counterexample. Diagrams can be useful in explaining such things. If the electric potential in a certain region of space is constant, then the charge enclosed by any closed surface completely contained within that region is zero.

Answers

Answer:

True

Explanation:

This is a representation of Gauss law.

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