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artcher [175]
4 years ago
15

Which are base units used in the metric system? Check all that apply.

Physics
2 answers:
Anestetic [448]4 years ago
8 0

Quarts - X

<h3>Liters - √</h3><h3>Grams - √</h3>

Degrees Fahrenheit - X

<h3>Degrees Celsius - √</h3>

Pints - X

I've tried my best with this, sorry if any are wrong.

Hope this helps,

Davinia.

Shtirlitz [24]4 years ago
7 0

Liters

Grams

Degrees Celsius

The other answer choices are from the imperial system

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Two infinite wires 20 cm apart each carry a current of 3 A into the paper. d I I d/2 d/2 At a distance d 2 below their midpoint,
Rzqust [24]

<u>Answer:</u>

<h3>As electric current is carried in a cable, around it, a magnetic field is created. The lines of the magnetic fields form concentric circles around the wire. The direction of the magnetic field hinges on the direction of the current. It can be calculated by pointing the thumb of your right hand in the direction of the moment, using the "right hand law." The position of your curled fingers is in the magnetic field lines. The magnetic field magnitude depends on the sum of current, and the distance from the wire carrying the charge.</h3>

<u></u>

<u>Explanation:</u>

Determine the direction of vector B magnitude B: B: B=\mu_{0} * 1 /(2 \pi r): r=d / 2 * \sqrt{2}

\cos \alpha=1 / 2 \Rightarrow \alpha=450

Resultant magnitude strength: B=2 B^{*} \cos \alpha=B=u_{0}^{*} 1 /(2 \pi r)=4.24^{*} 10^{-6} T=4.24 u T its direction is pointing to the left.

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3 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:

1)

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

2)

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

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F_y=F_{1y}+F_{2y}=7.00+(-4.30)=2.70 N

3)

The two components of the resultant force representent the sides of a right triangle, of which the resultant force corresponds tot he hypothenuse.

Therefore, we can find the magnitude of the resultant force by using Pythagorean's theorem:

F=\sqrt{F_x^2+F_y^2}

Where in this problem, we have:

F_x=-7.14 N is the x-component

F_y=2.70 N is the y-component

And substituting, we find:

F=\sqrt{(-7.14)^2+(2.70)^2}=7.63 N

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Answer:

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