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2 years ago

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:

Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?Change iChange in inherited characteristics over time is called what?n inherited characteristics over time is called what?Change in inherited characteristics over time is called what?

Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?Change in inherited characteristics over time is called what?

$\color{blue}{ \sf lololololololol}$

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