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3 years ago

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What is the magnitude of the magnetic field created 4.0m away from a 10.0 A current?

1 answer:

kvasek [131]3 years ago

4 0

Answer:

The magnitude of the magnetic field created is 5 x 10⁻⁷ T

Explanation:

Given;

magnitude of current, I = 10.0 A

distance of the magnetic field, r = 4.0m

Apply Biot-Savart Law to determine the magnitude of of the magnetic field created;

$B = \frac{\mu_oI}{2 \pi r}$

where;

B is the magnitude of of the magnetic field

μ₀ is constant

I is current

r is the distance of the field

$B = \frac{4 \pi*10^{-7} *10}{2 \pi*4} = 5*10^{-7} \ T$

Therefore, the magnitude of the magnetic field created is 5 x 10⁻⁷ T

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3 years ago

A car moving with an initial speed of 25 m/s slows down to a speed of 5 m/s in 10 seconds Calculate a) the acceleration of the c

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Answer :

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Explanation :

By the 1st equation of motion,

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v = final velocity = 5 m/s

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By the 2nd equation of motion,

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$s=(25m/s)\times (10s)+\frac{1}{2}\times (-2m/s^2)\times (10s)^2$

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3 years ago

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Answer:

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Explanation:

From the question we are told that

The mass of the uniform disk is $m_d = 40.0kg$

The radius of the uniform disk is $R_d = 0.200m$

The force applied on the disk is $F_d = 30.0N$

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Where $\theta$ is the angular displacement given from the question as

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$\alpha$ is the angular acceleration which is mathematically represented as

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Substituting values

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Substituting values

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Substituting values

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Substituting values

$= (0.200)(4.34)$

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$\alpha_r = \frac{v^2}{R}$

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$\alpha _t = R \alpha$

Substituting values

$\alpha _t = (0.200) (7.5)$

$= 1.5 m/s^2$

The magnitude of resultant acceleration at that point is

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Answer:

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Thus

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Taking log on both the sides:

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