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denpristay [2]
3 years ago
10

Kayla starts at -3, walks 5 blocks right and 3 blocks left. What is her displacement?

Physics
1 answer:
Elanso [62]3 years ago
7 0

Answer: The displacement is 1 block.

Explanation:

Let's define:

The right is the positive side.

The left is the negative side.

Then if you start at position A, and you walk N blocks to the right, the new position is:

A + N

And if you start at position A, and you walk M blocks to the left, the new position is:

A - M.

In this case, we know that Kayla starts at -3 and she walks 5 blocks to the right.

Then her new position is:

-3 + 5 = 2

Now she walks 3 blocks to the left, then her new position is:

2 - 3 = -1

The displacement will be equal to the difference between the final position (-1) and the initial position (-2)

Then the displacement is:

D = -1 - (-2) = -1 +2 = 1

The displacement is 1 block.

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what is the dot product and cross product of of two vectors if the angle is between them is 90 degree?​
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\sf{\pink{\underline{\underline{\blue{GIVEN:-}}}}}

  • The angle between the two vectors is 90° .

\sf{\pink{\underline{\underline{\blue{TO\: FIND:-}}}}}

  1. The dot product of two vectors .
  2. The cross product of two vectors .

\sf{\pink{\underline{\underline{\blue{SOLUTION:-}}}}}

⚡ Let \rm{\vec{a}} and \rm{\vec{b}} are the two vectors .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:.\:\vec{b}\:=\:ab\cos{\theta}\:}}}}

Where,

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\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\cos{90^{\degree}}\:}

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\rm{\implies\:\vec{a}\:.\:\vec{b}\:=\:ab\times{0}\:}

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\rm{\red{\therefore}} [1] The dot product of two vectors is “ <u>0</u> ” .

✍️ We have know that,

\orange\bigstar\:\rm{\pink{\boxed{\green{\vec{a}\:\times\:\vec{b}\:=\:ab\sin{\theta}\:}}}}

Where,

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\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\times{1}\:}

\rm{\implies\:\vec{a}\:\times\:\vec{b}\:=\:ab\:}

\rm{\red{\therefore}} [2] The cross product of two vectors is “ <u>ab</u> ” .

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