A student's calculation for the work done on an object is shown.

2 answers:

4 0

The statement that describes the error in the work is that the distance must be converted to meters (m).

<h3>FORMULA FOR WORK:</h3>

Work can be calculated by using the following formula:

W = F × d

Where;

W = work done
F = force (N)
d = distance (m)

According to this question, the force is given as 140N and the distance is given as 30cm. The force is calculated as follows:

F = 140N × 30cm = 4200J

This calculation is erroneous because the unit of distance should be converted from cm to meters.

Learn more about work done at: brainly.com/question/3902440

Serjik [45]2 years ago

3 0

This question involves the concepts of the work done, S.I. unit, and Distance.

The statement that describes the error in the work is "The distance must be converted to meters."

Work done is defined as the product of the displacement covered by the object and the force applied on that object. The S.I. unit o work is Joule (J). Joule is obtained from the product of Newton (N) and meter (m).

Joule = Newton x meter

Hence, in order to get the work done in S.I. unit of Joule, the displacement used must also be in S.I. unit of meter (m).

Learn more about S.I. unit here:

brainly.com/question/24688141

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A football player runs 20 meters North of a football field, and then 15 meters East. The total motion lasted 15 seconds. What wa

vlada-n [284]

Given:
1st run: 20 meters North
2nd run: 15 meters East
time: 15 seconds

Average speed = total distance covered / total time taken
Ave. Speed = (20m + 15m) / 15s
Ave. Speed = 35m / 15s
Ave. Speed = 2 1/3 meters per second

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3 years ago

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valentina_108 [34]

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2 years ago

On the way home from school, Taylor's car runs out of gas. He has to walk 25m north and 10m west in order to reach the nearest g

spin [16.1K]

Answer:

<em>The distance is 35 m and the magnitude of the displacement is 26.93 m</em>

Explanation:

<u>Displacement and Distance</u>

These are two related concepts. A moving object constantly travels for some distance at defined periods of time. The total distance is the sum of each individual distance the object traveled. It can be written as:

dtotal=d1+d2+d3+...+dn

This sum is calculated independently of the direction the object moves.

The displacement only takes into consideration the initial and final positions of the object. The displacement, unlike distance, is a vectorial magnitude and can even have magnitude zero if the object starts and ends the movement at the same point.

Taylor walks 25 m north and 10 m west. The total distance is the sum of both numbers:

d = 25 m + 10 m = 35 m

To calculate the displacement, we need to know the final position with respect to the initial position. If we set the coordinates of Taylor's car as the origin (0,0), then his final position is (-10,25), assuming the west direction is negative and the north direction is positive.

The magnitude of the displacement is the distance from (0,0) to (-10,25):

$D=\sqrt{(25-0)^2+(-10-0)^2}$