Is it possible for three vectors of different magnitudes to add to zero?.

Mathematics

1 answer:

amid [387]1 year ago

6 0

Answer:

No, it is not possible to obtain zero by adding two vectors of unequal magnitudes.

Explanation:

The Sum of two vectors can only be zero if they are equal in magnitude and opposite in direction. So no two vectors of unequal magnitude cannot be added to give a null vector. Three vectors of equal magnitude and making an angle 120 degrees with each other gives a zero resultant.

You might be interested in

there are approximately 3 million births in the unitad states each year .find the birth rate per minute

juin [17]

Answer:

Approx. 6 b/m

Step-by-step explanation:

Find out How many minutes in a year which is 525600 then divide that by 3,000,000 which makes 5.7077 so rounded is 6.

3 0

1 year ago

Read 2 more answers

THE AP CALCULUS EXAM SCORES WERE RELEASED IN JULY. THE SECONDARY MATH SPECIALIST PREPARED THE BOX PLOT SHOWN TO PRESENT THE SCOR

Ganezh [65]

Answer : option A

To find the range of scores that represents the middle 50 % of the student who took the test , we find inter quartile

Inter quartile range is the middle 50% of the given range of scores.

The difference between the upper quartile and lower quartile is the inter quartile that is middle 50%

From the diagram , we can see that

Upper quartile = 89

lower quartile = 65

So range is 65% to 89%

8 0

2 years ago

Which ordered pair is a solution to the system of inequalities?<br> y>−2 and x+y≤4

Kamila [148]

Answer:-2,-3

Step-by-step explanation:

8 0

1 year ago

Read 2 more answers

1. Simplify. 5–1(3–2)

konstantin123 [22]

Answer with explanation:

Using following law of indices

$1.x^a \times x^b=x^{a+b}\\\\2. (x^m)^n=x^{mn}\\\\ 3. x^{-n}=\frac{1}{x^n}\\\\4. \frac{x^m}{x^n}=x^{m-n}$

$1.5^{-1}\times 3^{-2}=\frac{1}{5}\times\frac{1}{3^2}\\\\=\frac{1}{5\times9}\\\\=\frac{1}{45}\\\\2. \frac{mn^{-4}}{p^0*q^{-2}}=\frac{mq^2}{n^4}\\\\ 3.0.0042=4.2\times 10^3\\\\4.6.12 \times 10^3=6.12 \times 1000=6120\\\\ 5.0.5 \times(8\times10^5})=\frac{1}{2}\times8 \times 10^5=4\times 10^5\\\\6. (9 \times 10^3)^2=9^2 \times (10^3)^2=81 \times 10^6\\\\7.( \frac{1}{2}a^{-4}\times b^2 )_{a=-2,b=4}{\text{or}}( \frac{1}{2}a^{4}\times b^{-2} )_{a=-2,b=4}=\frac{1}{2}\times (-2)^{-4}\times(4)^2{\text{or}}=\frac{1}{2}\times (-2)^{4}\times(4)^{-2}=\frac{16}{2\times16}=\frac{1}{2}$

$9. (4\times x\times y^2)^3\times (xy)^5=4^3\times x^3\times (y^2)^3\times x^5 \times y^5\\\\=64\times x^{3+5}\times y^{5+6}\\\\=64x^8y^{11}\\\\10.(2\times t^3)^3(0.4\times r)^2=2^3 \times (t^3)^3\times (0.4)^2 \times r^2=8\times t^9 \times 0.16 \times r^2=1.28r^2t^9$