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ikadub [295]
3 years ago
14

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

Physics
1 answer:
ExtremeBDS [4]3 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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One gallon of paint covers an area of 25 m2. What is the thickness of the paint on the wall?​
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Answer:

Explanation:

1 gal = 231 in³

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3 years ago
A particle, whose acceleration is constant, is moving in the negative x direction at a speed of 4.91 m/s, and 12.9 s later the p
Zinaida [17]

Answer:

The particle’s velocity is -16.9 m/s.

Explanation:

Given that,

Initial velocity of particle in negative x direction= 4.91 m/s

Time = 12.9 s

Final velocity of particle in positive x direction= 7.12 m/s

Before 12.4 sec,

Velocity of particle in negative x direction= 5.32 m/s

We need to calculate the acceleration

Using equation of motion

v = u+at

a=\dfrac{v-u}{t}

Where, v = final velocity

u = initial velocity

t = time

Put the value into the equation

a=\dfrac{7.12-(-4.91)}{12.9}

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We need to calculate the initial speed of the particle

Using equation of motion again

v=u+at

u=v-at

Put the value into the formula

u=-5.321-0.933\times12.4

u=-16.9\ m/s

Hence, The particle’s velocity is -16.9 m/s.

4 0
3 years ago
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