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2 years ago

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A very long straight wire carries a 12 A current eastward and a second very long straight wire carries a 14 A current westward.

The wires are parallel to each other and are 42 cm apart. Calculate the force on a 6.4 m length of one of the wires.

1 answer:

s344n2d4d5 [400]2 years ago

7 0

Answer: 5.12x10∧-4N

Explanation:

Force = I B L

L = 6.4m

Let Current (I) I₁ = I₂= 14A

Distance of the wire = 42cm = 0.42m

BUT

B = μ₀I / 2πr

=(2X10∧-7 X 12) / 0.42

B =5.714×10∧-6T

Force = I B L

Force = 14x [5.714×10-6]×6.4

Force = 5.12x10∧-4N

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I need to choose a theme for my physics assignment My experiment is finding g

Kobotan [32]

<h3>Question:</h3>

How to find g (acceleration due to gravity)

<h3>Solution:</h3>

We know,

Acceleration due to gravity (g)

$= \frac{GM}{ {R}^{2} }$

where, G = Gravitational constant

$= 6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \\$

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R = Radius of the earth

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Putting these values of G, M and R in the above formula, we get

$g \: = \: \frac{6.67 \times {10}^{11} N {m}^{2}/k {g}^{2} \times \: 6 \times {10}^{24} \: kg }{(6.4 \times {10}^{6}m {)}^{2} } \\ = 9.8m/ {s}^{2}$

So, the value of acceleration due to gravity is

$9.8m/s ^{2}$

Hope it helps.

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2 years ago

Given a particle that has the velocity v(t) = 3 cos(mt) = 3 cos (0.5t) meters, a. Find the acceleration at 3 seconds. b. Find th

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Answer:

a. $\rm -1.49\ m/s^2.$

b. $\rm 50.49\ m.$

Explanation:

<u>Given:</u>

Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .

<h2>(a):</h2>

The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.

$\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).$

At time t = 3 seconds,

$\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.$

<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>

<h2>(b):</h2>

The velocity of the particle at some is defined as the rate of change of the position of the particle.

$\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.$

For the time interval of 2 seconds,

$\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt$

The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,

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3 years ago

The force of gravity acting<br> on a child's mass on earth is 490 newtons. what's the child's mass?

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Assoli18 [71]

Answer:

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Part b)

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Part b)

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