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inessss [21]
1 year ago
12

An electric current in a conductor varies with time according to the expression I(t) = 100 sin (120πt) , where I is in amperes a

nd t is in seconds. What is the total charge passing a given point in the conductor from t = 0 to t = 1/240 s ?
Physics
1 answer:
Mice21 [21]1 year ago
7 0

The  total charge passing a given point in the conductor from t = 0 to t = 1/240 s is 12000π Coulombs

<h3>Total charge through a conductor</h3>

A conductor is a substance that allow current and electricity to pass through it.

The expression that will be used to calculate the current is expressed s:

I(t) = 100 sin (120πt)

where

I is in amperes and t is in seconds.

Since dQ = Idt =  I(t) = 100 sin (120πt)

On integrating

Q = 100*120πcos(120πt) |¹/¹²⁴⁰₀

Substitute the limits

Q = 100 * 120π

Q = 12000π Coulombs

Hence the  total charge passing a given point in the conductor from t = 0 to t = 1/240 s is 12000π Coulombs

Learn more on total charge through a conductor here: brainly.com/question/15083960

#SPJ4

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S_A_V [24]

The question is incomplete, the complete question is;

The compound magnesium phosphate has the chemical formula Mg3(PO4)2. In this compound, phosphorus and oxygen act together as one charged particle, which is connected to magnesium, the other charged particle. What does the 2 mean in the formula 5Mg3(PO4)2? A. There are two elements in magnesium phosphate. B. There are two molecules of magnesium phosphate. C. There are two magnesium ions in a molecule of magnesium phosphate. D. There are two phosphate ions in a molecule of magnesium phosphate.

Answer:

There are two phosphate ions in a molecule of magnesium phosphate.

Explanation:

The compound magnesium phosphate is an ionic compound. Ionic compounds always consists of two ions, a positive ion and a negative ion.

In this case, the positive ion is Mg^2+ while the negative ion is PO4^3-.

The subscript, 2 after the formula of the phosphate ion means that there are two phosphate ions in each formula unit of magnesium phosphate.

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2 years ago
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You are a member of an alpine rescue team and must get a box of supplies, with mass 2.20 kg , up an incline of constant slope an
a_sh-v [17]

Answer:

The minimum speed of the box bottom of the incline so that it will reach the skier is 8.19 m/s.

Explanation:

It is given that,

Mass of the box, m = 2.2 kg

The box is inclined at an angle of 30 degrees

Vertical distance, d = 3.1 m

The coefficient of friction, \mu=6\times 10^{-2}

Using the work energy theorem, the loss of kinetic energy is equal to the sum of gain in potential energy and the work done against friction.

KE=PE+W

\dfrac{1}{2}mv^2=mgh+W

W is the work done by the friction.

W=f\times d

f=\mu mg\ cos\theta

W=\mu mg\ cos\theta\times \dfrac{d}{sin\theta}

W=\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}mv^2=mgh+\dfrac{\mu mgh}{tan\theta}

\dfrac{1}{2}v^2=gh+\dfrac{\mu gh}{tan\theta}

\dfrac{1}{2}v^2=9.8\times 3.1+\dfrac{6\times 10^{-2}\times 9.8\times 3.1}{tan(30)}

v = 8.19 m/s

So, the speed of the box is 8.19 m/s. Hence, this is the required solution.

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2 years ago
At a race car driving event, a staff member notices that the skid marks left by the race car are
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A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

A car experiences a deceleration (a) of -41.62 m/s² and comes to a stop (final velocity = v = 0 m/s) after 10.99 m (s).

We can calculate the initial velocity of the car (u) using the following kinematic equation.

v^{2} = u^{2} + 2as\\\\u = \sqrt[]{v^{2}-2as} = \sqrt[]{(0m/s)^{2}-2(-42.61m/s^{2} )(10.99m)} = 30.60m/s

A car that experiences a deceleration of -41.62 m/s² and comes to a stop after 10.99 m has an initial velocity of 30.60 m/s.

Learn more: brainly.com/question/14851168

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