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3 years ago

** PLEASE PLEASE HELP A student uses the right-hand rule as shown.

Physics

2 answers:

guapka [62]3 years ago

7 0

As per right hand thumb rule if we put our thumb of right hand along the current in the wire then curl of fingers will show the direction of magnetic field.

So here as we can see the fingers in the above case is curled such that in front position its showing towards right.

So the direction of magnetic field near the student in front position must be towards Right.

elena-14-01-66 [18.8K]3 years ago

5 0

What is the direction of the magnetic field in front of the wire closest to the student?

B) Right

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Point charges of 21.0 μC and 47.0 μC are placed 0.500 m apart. (a) At what point (in m) along the line connecting them is the el

rewona [7]

Answer:

a) x = 0.200 m

b)E = 3.84*10^{-4} N/C

Explanation:

$q_1 = 21.0\mu C$

$q_1 = 47.0\mu C$

DISTANCE BETWEEN BOTH POINT CHARGE = 0.5 m

by relation for electric field we have following relation

$E = \frac{kq}{x}^2$

according to question E = 0

FROM FIGURE

x is the distance from left point charge where electric field is zero

$\frac{k21}{x}^2 = \frac{k47}{0.5-x}^2$

solving for x we get

$\frac{0.5}{x} = 1+ \sqrt{\frac{47}{21}}$

x = 0.200 m

b)electric field at half way mean x =0.25

$E =\frac{k*21*10^{-6}}{0.25^2} -\frac{k*47*10^{-6}}{0.25^2}$

E = 3.84*10^{-4} N/C

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3 years ago

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How many sides and points do a hexagon have

tamaranim1 [39]

A hexagon has 6 sides

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The diameter of the moon is a little less than the distance across the United States. Please select the best answer from the cho

dlinn [17]

Answer:

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Explanation:

The diameter of the Moon is 3474 km. The distance across the United States, from Florida to Washington, is 4509.382 km.

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A bead slides without friction around a loopthe-loop. The bead is released from a height 21.9 m from the bottom of the loop-the-

wariber [46]

Answer:

Part a)

$v = 12.45 m/s$

Part B)

$F_n = 0.05 N$

Explanation:

Part A)

As we know that the point A lies on the top of the loop

so we will have by energy conservation

$mgH = \frac{1}{2}mv^2 + mg(2R)$

so the speed at point A is given as

$mg(H - 2R) = \frac{1}{2}mv^2$

$v = \sqrt{2g(H - 2R)}$

$v = \sqrt{2(9.81)(21.9 - 2\times 7)}$

$v = 12.45 m/s$

Part B)

Now the force equation at point A is given as

$F_n + mg = \frac{mv^2}{R}$

$F_n = \frac{mv^2}{R} - mg$ [/tex]

$F_n = 0.004(\frac{12.45^2}{7} - 9.81)$

$F_n = 0.05 N$

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3 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

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3 years ago