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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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A fairgrounds ride spins its occupants inside a flying saucer-shaped container. If the horizontal circular path the riders follo
cestrela7 [59]

Answer:

13.37 rev/min

Explanation:

acceleration due to gravity (g) = 9.8 m/s², centripetal acceleration (a_c) = 1.8 * g = 1.8 * 9.8 m/s² = 17.64 m/s².

r = 9 m

Centripetal acceleration (a_c) is given by:

a_c=\frac{v^2}{r} \\\\v=\sqrt{a_c*r} \\\\v=\sqrt{17.64\ m/s^2*9\ m}\\\\v=12.6\ m/s

The velocity (v) is given by:

v = ωr;  where ω is the angular velocity

Hence:

ω = v/r = 12.6 / 9

ω = 1.4 rad/s

ω = 2πN

N = ω/2π = 1.4 / 2π

N = 0.2228 rev/s

N = 13.37 rev/min

4 0
2 years ago
Select the correct answer from each drop-down menu. A device that uses electricity and magnetism to create motion is called a __
Fittoniya [83]
<h3><u>Answer;</u></h3>

1st drop; Motor

2nd drop; Electricity

A device that uses electricity and magnetism to create motion is called a <u>motor</u>. In a reverse process, a device that uses motion and magnesium can be used to create <u>electricity</u>.

<h3><u>Explanation</u>;</h3>
  • <em><u>Motors are device which use electricity and magnetism to create motion.</u></em> They pass alternating current through opposing pairs of magnets to create a rotating magnetic field which creates a magnetic field in the rotor of a motor, making it to spin around.
  • <em><u>Electric motors work in a reverse process by using motion and magnetism to generate electricity. </u></em>When a coil or loops of wire are exposed to a changing magnetic field, an electrical current arises or is induced.
5 0
3 years ago
Change in inherited characteristics over time is called what?
adell [148]

Answer:

characteristics

Explanation:

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\color{blue}{ \sf lololololololol}

6 0
2 years ago
4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3
Kazeer [188]

Expand each vector into their component forms:

\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N

Similarly,

\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N

\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N

Then assuming the resultant vector \vec R is the sum of these three vectors, we have

\vec R=\vec A+\vec B+\vec C

\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N

and so \vec R has magnitude

\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N

and direction \theta_R such that

\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ

5 0
3 years ago
You are watching an archery tournament when you start wondering how fast an arrow is shot from the bow. Remembering your physics
spayn [35]

Answer:

v_0 = 3.53~{\rm m/s}

Explanation:

This is a projectile motion problem. We will first separate the motion into x- and y-components, apply the equations of kinematics separately, then we will combine them to find the initial velocity.

The initial velocity is in the x-direction, and there is no acceleration in the x-direction.

On the other hand, there no initial velocity in the y-component, so the arrow is basically in free-fall.

Applying the equations of kinematics in the x-direction gives

x - x_0 = v_{x_0} t + \frac{1}{2}a_x t^2\\63 \times 10^{-3} = v_0t + 0\\t = \frac{63\times 10^{-3}}{v_0}

For the y-direction gives

v_y = v_{y_0} + a_y t\\v_y = 0 -9.8t\\v_y = -9.8t

Combining both equation yields the y_component of the final velocity

v_y = -9.8(\frac{63\times 10^{-3}}{v_0}) = -\frac{0.61}{v_0}

Since we know the angle between the x- and y-components of the final velocity, which is 180° - 2.8° = 177.2°, we can calculate the initial velocity.

\tan(\theta) = \frac{v_y}{v_x}\\\tan(177.2^\circ) = -0.0489 = \frac{v_y}{v_0} = \frac{-0.61/v_0}{v_0} = -\frac{0.61}{v_0^2}\\v_0 = 3.53~{\rm m/s}

6 0
3 years ago
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