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ikadub [295]
2 years ago
14

Can a vector have a component equal to zero and still have nonzero magnitude?

Physics
1 answer:
ExtremeBDS [4]2 years ago
8 0

Answer:

Yes

Explanation:

in a 2-D plane a vector can be resolved into its rectangular components. These are rectangular components because these are at right angles to each other. One component is directed along x-axis and is the termed as horizontal component and the other one is directed along y-axis and is termed as vertical component.

Sometimes, the vector is directed entirely along x-axis or y-axis. When the vector is along x-axis or parallel to x-axis all of its magnitude will be directed along horizontal component while its vertical component will be zero. Similarly, for a vector along y-axis, the horizontal component will be zero and the entire magnitude will be along y-axis.

Consider a vector V of magnitude 5 Newtons along x-axis. Since, V is along axis the angle made with the positive x-axis is 0.

The components of a vector are calculated as:

Horizontal component = V cos θ

Vertical component = V sin θ

θ is the angle made with positive x-axis in anti-clockwise direction which is 0 in this case.

So, the components of our vector will be:

Horizontal component = 5 cos(0) = 5

Vertical component = 5 sin(0) = 0

We have just showed, even if one of the components of a vector is 0, it still can have a non-zero magnitude.

The magnitude of a vector will be zero only when all of its components are equal to 0.

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Answer:

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3 years ago
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Which state has the most fixed shape?<br> O A. Gas<br> O B. Solid<br> O C. Liquid<br> O D. Plasma
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Answer: Liquid

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3 years ago
A scientist directs monochromatic light toward a single slit in an opaque barrier. The light has a wavelength of 580 nm and the
vredina [299]

Answer:

a) 9.72 mm

b) 4.86 mm

Explanation:

wave length of light  λ is  580 nm = 580 \times 10⁻⁹ m

Width of slit d = 0.215\times 10⁻³ m

Distance of screen D  = 1.8 m.

Width of one fringe = \frac{\lambda\times D}{d}

Putting the values we get fringe width

= \frac{580\times10^{-9}\times1.8}{.000315}

=4.86 mm.

a) Width of central maxima = 2 times width of one fringe

= 2 times 4.86

=9.72 mm

b) width of each fringe except central fringe  is same , no matter what the order is.Only brightness changes .

So width of either of the two first order bright fringe will be same and it will be  

= 4.86 mm.

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3 years ago
A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The
qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0
2 years ago
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