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1 year ago

What is the value of n?

Physics

2 answers:

vekshin11 year ago

6 0

Answer:

THE ANSWE IS C

Explanation:

dusya [7]1 year ago

6 0

81 or 85 I think this because a right angle is 90 degrees and 81 or 85 is a little less

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Hey, what is antimatter

jenyasd209 [6]

A antimatter is particle physics, antimatter is a material composed of the antiparticle "partners" to the corresponding particles of ordinary matter. A particle and its antiparticle have the same mass as one another, but opposite electric charge and other quantum numbers.

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2 years ago

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The Sun delivers an average power of 1.575 W/m2 to the top of Neptune's atmosphere. Find the magnitudes of max and max for the e

FrozenT [24]

Answer:

$1.1486813808\times 10^{-7}\ T$

34.46 V/m

Explanation:

$\mu_0$ = Vacuum permeability = $4\pi \times 10^{-7}\ H/m$

c = Speed of light = $3\times 10^8\ m/s$

I = Intensity = 1.575 W/m²

The maximum magnetic field intensity is given by

$B_m=\sqrt{\dfrac{2\mu_0I}{c}}\\\Rightarrow B_m=\sqrt{\dfrac{2\times 4\pi \times 10^{-7}\times 1.575}{3\times 10^8}}\\\Rightarrow B_m=1.1486813808\times 10^{-7}\ T$

The magnetic field intensity is $1.1486813808\times 10^{-7}\ T$

The maximum electric field intensity is given by

$E_m=B_m\times c\\\Rightarrow E_m=1.1486813808\times 10^{-7}\times 3\times 10^8\\\Rightarrow E_m=34.46\ V/m$

The electric field intensity is 34.46 V/m

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2 years ago

Which ion channels mediate the falling phase of an action potontial?

White raven [17]

Answer:

Voltage-gated K+ channels

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2 years ago

What is the function of the muscles around the lens in the human eye

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Its to capture light or to focus. don't forget to like. :D

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2 years ago

Two forces, F₁ and F₂, act at a point. F₁ has a magnitude of 8.00 N and is directed at an angle of 61.0° above the negative x ax

kirill115 [55]

1) -7.14 N

2) +2.70 N

3) 7.63 N

Explanation:

In order to find the x-component of the resultant force, we have to resolve each force along the x-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: this means that the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its x-component is

$F_{1x}=(8.00)(cos (180^{\circ}-61^{\circ}))=-3.88 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: so, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its x-component is

$F_{2x}=(5.40)(cos (180^{\circ}+52.8^{\circ}))=-3.26 N$

So, the x-component of the resultant force is

$F_x=F_{1x}+F_{2x}=-3.88+(-3.26)=-7.14 N$

In order to find the y-component of the resultant force, we have to resolve each force along the y-axis.

The first force is 8.00 N and is directed at an angle of 61.0° above the negative x axis in the second quadrant: as we said previously, the angle with respect to the positive x axis is

$180^{\circ}-61^{\circ}$

so its y-component is

$F_{1y}=(8.00)(sin (180^{\circ}-61^{\circ}))=7.00 N$

F₂ has a magnitude of 5.40 N and is directed at an angle of 52.8° below the negative x axis in the third quadrant: as we said previously, its angle with respect to the positive x-axis is

$180^{\circ}+52.8^{\circ}$

Therefore its y-component is

$F_{2y}=(5.40)(sin (180^{\circ}+52.8^{\circ}))=-4.30 N$