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Assoli18 [71]
3 years ago
12

13) Can two nonzero perpendicular vectors be added together so their sum is zero? Explain.

Physics
2 answers:
statuscvo [17]3 years ago
4 0

Answer:

sum of two perpendicular nonzero vectors can not be ZERO

Explanation:

As we know by the sum or addition of two vectors is given by

R = \sqrt{a^2 + b^2 + 2abcos\theta}

here we know that

a and b are the magnitude of two vectors

\theta = angle between two vectors

so as we know that here two vectors are perpendicular to each other

so we will have

R = \sqrt{a^2 + b^2}

now since the is sum of two positive non zero numbers so it can not be zero

so sum of two perpendicular nonzero vectors can not be ZERO

Neporo4naja [7]3 years ago
3 0

In order for two vectors to add to zero, they must have the same magnitude and point in opposite directions.

Two perpendicular vectors, by definition, make a right angle with each other whereas two vectors pointing in opposite directions form a straight line.

Because of this, two perpendicular vectors with nonzero magnitudes will never add to zero.

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Answer:

Explanation:

Given

\mu =0.14 Pa.s

Volume Flow rate Q=5.3\times 10^{-5} m^3/s

Length L=37 m

radius r=0.58 cm

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According to Hagen-Poiseuille Equation

difference in Pressure is given

\Delta P=\frac{128\mu L\cdot Q}{\pi D^4}

P_{in}-P_{out}=\frac{128\mu L\cdot Q}{\pi D^4}

P_{in}=P_{out}+\frac{128\mu L\cdot Q}{\pi D^4}

P_{in}=1.01325\times 10^5+\frac{128\times 0.14\times 37\times 5.3\times 10^{-5}}{\pi\cdot 1.16^4\times 10^{-8}}

P_{in}=1.01325\times 10^5+6.17\times 10^5

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3 years ago
A pure substance can be a ..................... (a) element (b) compound (c) either element or compound (d)none of these
stiks02 [169]

Answer:

C

Explanation:

An element is a pure substance that can not be broken down into anything simpler

A compound in also a pure substance held together in fixed proportion through chemical bonds

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3 years ago
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A diffraction grating with 600 lines/mmlines/mm is illuminated with light of wavelength 510 nmnm. A very wide viewing screen is
Ksenya-84 [330]

Answer:

A.2.95 m

B.7

Explanation:

We are given that

Diffraction grating=600 lines/mm

d=\frac{1 mm}{600}=\frac{1\times 10^{-3} m}{600}=1.67\times 10^{-6} m

Wavelength of light,\lambda=510 nm=510\times 10^{-9} m

l=4.6 m

A.We have to find the distance between the two m=1 bright fringes

sin\theta=\frac{m\lambda}{d}

For first bright fringe, =1

sin\theta=\frac{1\times 510\times 10^{-9}}{1.67\times 10^{-6}}=0.305

\theta=sin^{-1}(0.305)=17.76^{\circ}

The distance between two m=1 fringes

x=2ltan\theta=2\times 4.6 tan(17.76^{\circ})=2.95 m

Hence, the distance between two m=1 fringes=2.95 m

B.For maximum number of fringes,

sin\theta=1

sin\theta=\frac{m\lambda}{d}

Substitute the values

1=\frac{m\times 510\times 10^{-9}}{1.67\times 10^{-6}}

m=\frac{1.67\times 10^{-6}}{510\times 10^{-9}}=3.3\approx 3

Maximum number of bright fringes on the scree=2m+1=2(3)+1=7

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