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3 years ago

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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Calculate the acceleration of a 1000 kg car if the motor provides a small thrust of 1000 N and the static and dynamic friction c

grin007 [14]

Explanation :

It is given that,

Mass of the car, m = 1000 kg

Force applied by the motor, $F_A=1000\ N$

The static and dynamic friction coefficient is, $\mu=0.5$

Let a is the acceleration of the car. Since, the car is in motion, the coefficient of sliding friction can be used. At equilibrium,

$F_A-\mu mg=ma$

$\dfrac{F_A-\mu mg}{m}=a$

$a=\dfrac{1000-0.5(1000)(9.81)}{1000}$

$a=-3.905\ m/s^2$

So, the acceleration of the car is $-3.905\ m/s^2$ . Hence, this is the required solution.

6 0

3 years ago

A girl stands on the edge of a merry-go-round of radius 1.71 m. If the merry go round uniformly accerlerates from rest to 20 rpm

Mashutka [201]

Answer:

$a = 0.53 m/s^2$

Explanation:

initially the merry go round is at rest

after 6.73 s the merry go round will accelerates to 20 rpm

so final angular speed is given as

$\omega = 2\pi f$

$\omega = 2\pi ( \frac{20}{60})$

$\omega = 2.10 rad/s$

so final tangential speed is given as

$v = r\omega$

$v = 1.71 (2.10) = 3.58 m/s$

now average acceleration of the girl is given as

$a = \frac{v_f - v_i}{\Delta t}$

$a = \frac{3.58 - 0}{6.73}$

$a = 0.53 m/s^2$

8 0

3 years ago

Name of a body that changes Chemical energy to electric energy.

Ratling [72]

Answer:

battery

Explanation:

A battery contains stored chemical energy and converts it to electrical energy. (cK.12)

6 0

3 years ago

A pendulum has a length of 2 m and a 30 kg mass hanging on the end. What is the period of the

anastassius [24]

Answer:

T = 2.83701481512 seconds

Explanation:

Hi!

The formula that you will want to use to solve this question is:

$T = 2\pi *\sqrt{\frac{L}{g} }$

T--> period

L --> length of the pendulum

g --> acceleration due to gravity (9.8m/s^2)

since we know that the mass of the bob at the end of the pendulum does not affect the period of the pendulum, we can go ahead and ignore that bit of information (unless, of course, the weight causes the pendulum to stretch)

so now we can plug in our given info into the formula above and solve!

T = 2*pi * sqrt(2/9.8)

T = 2.83701481512 seconds

*Note*

- I used 3.14 to pi, if you need to use a different value for pi (a longer version, etc) your answer will be slightly different

I hope this helped!

7 0

3 years ago

(a) State Hook's law. [2]

murzikaleks [220]

Hookes law state that provided that the elastic limit is not exceeded, the extension is directly proportional to the force

3 0

3 years ago