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grigory [225]
2 years ago
8

If a current of 2. 4 a is flowing in a wire of diameter 2. 0 mm, what is the average current density?

Physics
1 answer:
HACTEHA [7]2 years ago
5 0

The average current density is 7.6 × 10⁵ A/m².

To calculate the current density current will be 2.4 A.

Diameter of a wire = 2mm.

The cross-sectional area of the wire is given by r = d/2

where r is the radius of the wire.  

Then, the cross-sectional area is = 0.00000314159265

                                                                   = 3.1 × 10⁻⁶ m².

<h3>What is average current density?</h3>

         Consider a current carrying conductor, the current density depends upon  the current flow in the conductor. If the current flow in the conductor will be high then the current density will also be high. Using the average current flowing through the conductor, the average current density will be found.

Average current density j = I / A Ampere/ meter².

By substituting the values in the formula,

             j = 2.4 / ( 3.1 × 10⁻⁶)

               = 7.6 × 10⁵ A/m².

Hence, the current density can be calculated.

Learn more about average current density,

brainly.com/question/3981451

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I have multiple questions!
UNO [17]

I'm not sure about the rest but for question 2:

A theory is an attempt to come up with a big picture of all we know so far. It also drives future research as people do experiments to see if what the theory predicts actually happens. When experiments don’t support the theory, you have to change the theory and try again. That’s how science works. We come up with a “best guess” (theory), and then do research to test it’s accuracy. As we discover contradictions, we adjust the theory to take those into account, and then start testing the validity of the new theory.

3 0
3 years ago
The number of bacteria in a certain population increases according to a continuous exponential growth model, with a growth rate
sveta [45]

Answer:

It would take approximately 289 hours for the population to double

Explanation:

Recall the expression for the continuous exponential growth of a population:

N(t)=N_0\,e^{kt}

where N(t) measures the number of individuals, No is the original population, "k" is the percent rate of growth, and "t" is the time elapsed.

In our case, we don't know No (original population, but know that we want it to double in a certain elapsed "t". We also have in mind that the percent rate "k" would be expressed in mathematical form as: 0.0024 (mathematical form of the given percent growth rate).

So we need to solve for "t" in the following equation:

2\,N_0=N_0\,e^{0.0024\,t}\\\frac{2\,N_0}{N_0} =e^{0.0024\,t}\\2=e^{0.0024\,t}\\ln(2)=0.0024\,t\\t=\frac{ln(2)}{0.0024} \\t=288.811\,\, hours

Which can be rounded to about 289 hours

6 0
3 years ago
Tech A says that a vacuum modulator converts manifold vacuum into an engine load signal. Tech B says that the manual valve is mo
alexandr1967 [171]

Answer:

The correct option for the answer is A.) Tech A

Explanation:

Tech A is correct

8 0
3 years ago
The coefficient of static friction between a book and the level surface it slides on is 0.65. If the mass of the book is 0.2 kg,
just olya [345]

Answer: 1.274 N minimum force required to slide the book across the surface

Explanation:

Mass of the book = m = 0.2 kg

Normal force acting on the book,N = mg

Coefficient of friction =mu_s = 0.65

Minimum force required to slide the book across the surface= F

F=\mu_s\times N

F=0.65\times 0.2 kg\times 9.8 m/s^2=1.274 N

1.274 N minimum force required to slide the book across the surface

7 0
3 years ago
Read 2 more answers
Suppose that the moment of inertia of a skater with arms out and one leg extended is 3.0 kg⋅m2 and for arms and legs in is 0.90
babymother [125]

Answer:

Her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

Explanation:

Initial moment of inertia when arms and legs in is I_i=0.90 kg.m^{2}

Final moment of inertia when her arms and on leg open outward, I_f=3.0 kg.m^{2}

Initial angular speed w_i=5.2\frac{rev}{s}

Let the final angular speed be w_f

Since external torque on her is zero so we can apply conservation of angular momentum

\therefore L_f=L_i

=>I_fw_f=I_iw_i

=>w_f=\frac{I_iw_i}{I_f}=\frac{0.9\times5.2 }{3.0}\frac{rev}{s}=1.56\frac{rev}{s}

Thus her angular speed (in rev/s) when her arms and one leg open outward is 1.56\frac{rev}{s}

7 0
3 years ago
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