Question

please answer all of them ​

Answer 1

Answer:

here is your answer hope it helps

Answer 2

Solution first! If an object has moved distance then yes, it could have zero displacement. If the initial position on the final position of an object is same then the displacement is zero. For example, according to the figure and object starts moving from an initial position that is A and go to B then the object returned back to A and stop there. Therefore it be it's final position too. Therefore, the answer is supported! (Attachment first)

Solution 2nd

Solution 2ndGiven that: A farmer move along the boundary of a square field of side 10 m in 40 s. What will be the magnitude of displacement of farmer at the end of 2 min 20 seconds

According to statement:

According to statement:Provided that:

• Side of square field = 10 metres

• (Attachment second)

• Time taken to cover 10 metres = 40 seconds

To calculate:

• The magnitude of displacement of farmer at the end of 2 min 20 seconds

Solution:

• The magnitude of displacement of farmer at the end of 1 min 20 seconds = 10√2 m

Full solution:

~ Firstly let us find out perimeter of the given square field!

${\small{\underline{\boxed{\sf{Perimeter \: of \: square \: = 4 \times side}}}}}$

Therefore, according to formula

→ Perimeter of square = 4 × side

→ Perimeter of square = 4 × 10

→ Perimeter of square = 40 m

~ Now let's convert minutes into seconds by applying suitable formula!

${\small{\underline{\boxed{\sf{1 \: minute \: = 60 \: seconds}}}}}$

Therefore, according to formula

→ 1 minute = 60 seconds

→ 2 min 20 sec = 60 × 2 + 20

→ 2 min 20 sec = 120 + 20

→ 140 seconds...henceforth, converted!

~ Now let's see what to do!

Explanation: As it's given that he covers each 10 metres (40 m because as we find the perimeter) in 40 seconds Therefore, he covers

→ 40/40 = 1 m in 1 second

~ Now let's find how many rounds he take in his field that is in square shape!

→ Rounds = Time/Perimeter

→ Rounds = 140/40

→ Rounds = 3.5

• (Attachment third)

~ Now let's find out the magnitude of displacement!

Explanation: As we are able to see in the attachment that how he moved and we observe that at last movement he take half round that is 3.5 rounds he take means 3 and half rounds.

So here we can apply phythagoras theorm as we are able to see that it is looking like a right angle traingle.

Also we can apply formula to find diagonal of square here too and see it's also looking like diagonal and here, there is a square field!

Don't forget! Displacement is said to be the shortest distance!

(Choice may yours!)

By applying diagonal of sq. formula we get the following results!

${\small{\underline{\boxed{\sf{Diagonal \: of \: square \: = a \sqrt{2}}}}}}$

Here, a denotes side.

→ 10√2

→ Displacement = 10√2 metres

• (Attachment fourth)

By applying phythagoras theorm we get the following results!

${\small{\underline{\boxed{\sf{(H)^{2} \: = (P)^{2} \: + (B)^{2}}}}}}$

Here, H is hypotenuse, P is perpendicular, B is the base.

→ H² = P² + B²

→ H² = (10)² + (10)²

→ H² = 100 + 100

→ H² = 200

→ H = √200

→ H = 10√2 metres

→ Displacement = 10√2 metres

• (Attachment fifth)

Solution 3rd! Both the statements are not correct!