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2 years ago

5

Hilda plans to display in a clear cube the needle betsy ross used to sew the first american flag. the needle will fit exactly as

the diagonal of the cube. if the side length of the clear cube is startroot 3 endroot inches, and the diagonal of a face of the cube is startroot 6 endroot inches, what is the length of the needle? startroot 6 endroot inches 3 inches 3 startroot 2 endroot inches 6 inches

2 answers:

inysia [295]2 years ago

8 0

Based on the length of side of the cube given, the length of the needle is 2.28 inches.

<h3>What is the length of the needle?</h3>

The length of the side of a cube and its diagonal are related by the formula below:

$Length \: of \: Diagonal = \sqrt{3s}$

where s is length of side

$Side \: Length \: of \: the \: Cube = \sqrt{3inches}$

$Diagonal \: of \: a \: face \: of \: the \: cube = \sqrt{6inches}$

$The \: length \: of \: the \: needle = \sqrt{3\sqrt{3}} = 2.28 \: inches \\$

Therefore, the length of the needle is 2.28 inches

Learn more about length of diagonal of cube at: brainly.com/question/800212

#SPJ4

Irina-Kira [14]2 years ago

7 0

Answer:

Its B 3in

Explanation:

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Chris and Jamie are carrying Wayne on a horizontal stretcher. The uniform stretcher is 2.00 m long and weighs 100 N. Wayne weigh

BabaBlast [244]

Answer:

<h2>E. 650N</h2>

Explanation:

step one:

given

length of stretcher= 2m

weight of stretcher=100N

Wayne weighs =800N

distance of Wayne weighs from chris's end= 75cm= 0.75m

The force that Chris is exerting to support the stretcher, with Wayne on it, can be computed by taking moments of the weight of the stretcher and Wayne weighs about Chris's end, the end result is the reaction at Chris's end

Taking moment about Chris's end

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3 years ago

What is called the process of finding exact quantity of a substance?

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2 years ago

Two particles are fixed to an x axis: particle 1 of charge q1 = 2.78 × 10-8 c at x = 15.0 cm and particle 2 of charge q2 = -3.24

Oksi-84 [34.3K]

Refer to the attached figure. Xp may not be between the particles but the reasoning is the same nonetheless.
At xp the electric field is the sum of both electric fields, remember that at a coordinate x for a particle placed at x' we have the electric field of a point charge (all of this on the x-axis of course):
$E=\frac{1}{4\pi\varepsilon_0}\frac{q}{(x-x')^2}$
Now At xp we have:
$\frac{1}{4\pi\varepsilon_0}\frac{q_1}{(x_p-x_1)^2}-\frac{1}{4\pi\varepsilon_0}\frac{3.29q_1}{(x_p-x_2)^2}=0$
$\implies (x_p-x_1)^2=\frac{(x_p-x_2)^2}{3.29}\\
\implies(1-\frac{1}{3.29})x_p^2+2(\frac{x_2}{3.29}-x_1)x_p+x_1^2-\frac{x_2^2}{3.29}=0$
Which is a second order equation, using the quadratic formula to solve for xp would give us:
$xp=\frac{-(\frac{x_2}{3.29}-x_1)-\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
or
$xp=\frac{-(\frac{x_2}{3.29}-x_1)+\sqrt{(\frac{x_2}{3.29}-x_1)^2-(1-\frac{1}{3.29})(x_1^2-\frac{x_2^2}{3.29})}}{(1-\frac{1}{3.29})}$
Plug the relevant values to get both answers.
Now, let's comment on which of those answers is the right answer. It happens that BOTH are correct. This is simply explained by considring the following.

Let's place a possitive test charge on the system This charge feels a repulsive force due to q1 but an attractive force due to q2, if we place the charge somewhere to the left of q2 the attractive force of q2 will cancel the repulsive force of q1, this translates to a zero electric field at this x coordinate. The same could happen if we place the test charge at some point to the right of q1, hence we can have two possible locations in which the electric field is zero. The second image shows two possible locations for xp.

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3 years ago