To introduce a vector quantity .... (A) it needs magnitude not direction (B) it needs direction not magnitude (C) it needs both magnitude and direction (D) nothing is needed

Short Answer

Expert verified
(C) A vector quantity needs both magnitude and direction.

Step by step solution

01

Understanding Vector Quantity

A vector quantity is a quantity that has both magnitude(complete size or value) and direction. To specify a vector, we must know both the magnitude (how much or the size) and the direction (where it is pointing). Examples of vector quantities are force, velocity, and acceleration. Now, let's analyze the given choices.
02

Choice A: Needs Magnitude, Not Direction

Since we know that a vector quantity requires both magnitude and direction, this choice is incorrect. A quantity with only magnitude, but no direction, is called a scalar quantity.
03

Choice B: Needs Direction, Not Magnitude

This choice is also inaccurate because a vector cannot exist without both magnitude and direction. Without magnitude, the vector would not have any value associated with it.
04

Choice C: Needs Both Magnitude and Direction

This choice is correct since vector quantities require both magnitude and direction. To fully define a vector, we must provide the necessary information about both its size (magnitude) and the direction in which it's pointing or acting.
05

Choice D: Nothing is Needed

This choice is incorrect as vector quantity cannot be defined without specifying its magnitude and direction. A quantity with neither magnitude nor direction does not exist in the context of vector quantities. To conclude, the correct choice is: (C) A vector quantity needs both magnitude and direction.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\mathrm{t}_{1}\) and \(\mathrm{t}_{2}\) are two values of time of a projectile at the same height \(t \mathrm{t}_{1}+\mathrm{t}_{2}=\) (A) Time to reach maximum height (B) flight time for the projectile (C) \((3 / 4)\) time of the flight time. (D) \((3 / 2)\) time of the flight time.

If resultant of $\mathrm{A}^{\rightarrow}=2 \hat{1}+\hat{\jmath}-\mathrm{k}^{\wedge}, \mathrm{B}^{\rightarrow}=\hat{\imath}-2 \hat{\mathrm{j}}+3 \mathrm{k}^{\text {and }} \mathrm{C}^{\rightarrow}$ is unit vector in y direction, then \(\mathrm{C}^{\rightarrow}\) is (A) \(-\hat{j}\) (B) \(3 \hat{i}-2 \hat{j}+2 \mathrm{k}^{\wedge}\) (C) (D) \(2 \hat{i}+3 \mathrm{k}^{\wedge}\)

A body starts its motion with zero velocity and its acceleration is $\left(3 \mathrm{~m} / \mathrm{s}^{2}\right)$. Find the distance travelled by it in fifth second. (A) \(15.5 \mathrm{~m}\) (B) \(17.5 \mathrm{~m}\) (C) \(13.5 \mathrm{~m}\) (D) \(14.5 \mathrm{~m}\)

\(\mathrm{A}^{-}+\mathrm{B}^{-}\) is perpendicular to \(\mathrm{A}^{-}\) and \(\left|\mathrm{B}^{-}\right|=2\left|\mathrm{~A}^{-}+\mathrm{B}^{-}\right|\) What is the angle between \(\mathrm{A}^{-}\) and \(\mathrm{B}^{\rightarrow}\) \((\mathrm{A})(\pi / 6)\) (B) \((5 \pi / 6)\) (C) \((2 \pi / 3)\) (D) \((\pi / 3)\)

Assertion: (A) If both Assertion - Reason are true, reason is correct explanation of Assertion. (B) If both Assertion - Reason are true but reason is not correct explanation of Assertion. (C) Assertion is true but Reason is false. (D) If Reason is true but Assertion is false At the highest point of projectile motion the velocity is not zero. Reason: Only the vertical component of velocity is zero. Where as horizontal component still exists. (A) a (B) \(\mathrm{b}\) (C) \(\mathrm{c}\) (D) \(\mathrm{d}\)

See all solutions

Recommended explanations on English Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free