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1 year ago

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]1 year 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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Answer:

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2 years ago

How much voltage is required to run 0.64 A of current through a 240 resistor? Use V= IR.

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2 years ago

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Two parallel wires separated by a distance of 0.6 m each carry current in the same direction. One wire is carrying a current of

mart [117]

Answer:

The value is $B = 3.33 *10^{-6} \ T$

Explanation:

From the question we are told that

The distance of separation is $d = 0.6 \ m$

The current on the one wire is $I_1 = 9 \ A$

The current on the second wire is $I_2 = 4 \ A$

Generally the magnitude of the field exerted between the current carrying wire is

$B = B_1 - B_2$

Here $B_1$ is the magnetic field due to the first wire which is mathematically represented as

$B_1 = \frac{\mu_o * I_1 }{2 \pi * d_1}$

Here $d_1$ is the distance to the half way point of the separation and the value is

$d_1 = 0.3 \ m$

$B_2$ is the magnetic field due to the first wire which is mathematically represented as

$B_2 = \frac{\mu_o * I_2 }{2 \pi * d_2}$

Here $d_2$ is the distance to the half way point of the separation and the value is

$d_2 = 0.3 \ m$

This means that $d_1 = d_2 = a = 0.3$

$B = \frac{\mu_o * I_1 }{2 \pi * d_1} - \frac{\mu_o * I_2 }{2 \pi * d_2}$

=> $B = \frac{\mu_o * (I_1 - I_2)}{2 \pi *0.3 }$

=> $B = \frac{ 4\pi * 10^{-7} * (9- 4)}{2 * 3.142 *0.3 }$

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2 years ago

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You might want to try calling it

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