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3 years ago

6

Which of the following is an obstacle to creating computer-based models for tracking a hurricane?

1 answer:

iren [92.7K]3 years ago

5 0

Answer:

4. All of the above I think, not to sure about 1. but the rest are right so im like 90.99999 percent sure good luck

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Why is a specimen smaller than 200 nm not visible with a light microscope?

kifflom [539]

Because the specimen is very small with a light microscope

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3 years ago

Which officeholders are elected by the electoral college

NikAS [45]

Answer:Electoral Vote

Explanation:i did it before

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3 years ago

Two long, parallel wires separated by 2.00 cm carry currents in opposite directions. The current in one wire is 1.75 A, and the

Natasha_Volkova [10]

The force per unit length between the two wires is $6.0\cdot 10^{-5} N/m$

Explanation:

The magnitude of the force per unit length exerted between two current-carrying wires is given by

$\frac{F}{L}=\frac{\mu_0 I_1 I_2}{2\pi r}$

where

$\mu_0 = 4\pi \cdot 10^{-7} Tm/A$ is the vacuum permeability

$I_1, I_2$ are the currents in the two wires

r is the separation between the two wires

For the wires in this problem, we have

$I_1 = 1.75 A$

$I_2 = 3.45 A$

r = 2.00 cm = 0.02 m

Substituting into the equation, we find

$\frac{F}{L}=\frac{(4\pi \cdot 10^{-7})(1.75)(3.45)}{2\pi (0.02)}=6.0\cdot 10^{-5} N/m$

Learn more about current and magnetic fields:

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3 years ago

A single-turn circular loop of radius 6 cm is to produce a field at its center that will just cancel the earth's field of magnit

djverab [1.8K]

Answer:

The current is $I = 6.68 \ A$

Explanation:

From the question we are told that

The radius of the loop is $r = 6 \ cm = 0.06 \ m$

The earth's magnetic field is $B_e = 0.7G= 0.7 G * \frac{1*10^{-4} T}{1 G} = 0.7 *10^{-4} T$

The number of turns is $N =1$

Generally the magnetic field generated by the current in the loop is mathematically represented as

$B = \frac{\mu_o * N * I}{2 r }$

Now for the earth's magnetic field to be canceled out the magnetic field generated by the loop must be equal to the magnetic field out the earth

$B = B_e$

=> $B_e = \frac{\mu_o * N * I }{ 2 * r}$

Where $\mu$ is the permeability of free space with value $\mu _o = 4\pi * 10^{-7} N/A^2$

$0.7 *10^{-4}= \frac{ 4\pi * 10^{-7} * 1 * I}{2 * 0.06}$

=> $I = \frac{2 * 0.06 * 0.7 *10^{-4}}{ 4\pi * 10^{-7} * 1}$

$I = 6.68 \ A$

3 0

3 years ago

A runner is moving at a constant speed of 8.00 m/s around a circular track. If the distance from the runner to the center of the

Genrish500 [490]

Answer: Last option

2.27 m/s2

Explanation:

As the runner is running at a constant speed then the only acceleration present in the movement is the centripetal acceleration.

If we call a_c to the centripetal acceleration then, by definition

$a_c =w^2r = \frac{v^2}{r}$

in this case we know the speed of the runner

$v =8.00\ m/s$

The radius "r" will be the distance from the runner to the center of the track

$r = 28.2\ m$

$a_c = \frac{8^2}{28.2}\ m/s^2$

$a_c = 2.27\ m/s^2$

The answer is the last option

3 0

4 years ago