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4 years ago

15

What happens in a tug of war if the net forces are balanced and why?

1 answer:

FinnZ [79.3K]4 years ago

6 0

Answer:

Balanced forces are responsible for unchanging motion. Balanced forces are forces where the effect of one force is cancelled out by another. A tug of war, where each team is pulling equally on the rope, is an example of balanced forces. The forces exerted on the rope are equal in size and opposite in direction.

Explanation:

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1 year ago

A long, rigid conductor, lying along the x-axis, carries a current of 7.0 A in the negative direction. A magnetic field B is pre

Alisiya [41]

Answer:

0.546 $\hat k$

Explanation:

From the given information:

The force on a given current-carrying conductor is:

$F = I ( \L \limits ^ {\to } \times B ^{\to})\\ \\ dF = I(dL\limits ^ {\to } \times B ^{\to})$

where the length usually in negative (x) direction can be computed as

$\L ^ {\to } = -x\hat i \\dL\limits ^ {\to }- dx\hat i$

Now, taking the integral of the force between x = 1.0 m and x = 3.0 m to get the value of the force, we have:

$\int dF = \int ^3_1 I ( dL^{\to} \times B ^{\to})$

$F = I \int^3_1 ( -dx \hat i ) \times ( 4.0 \hat i + 9.0 \ x^2 \hat j)$

$F = I \int^3_1 - 9.0x^2 \ dx \hat k$

$F = I (9.0) \bigg [\dfrac{x^3}{3} \bigg ] ^3_1 \hat k$

$F = I (9.0) \bigg [\dfrac{3^3}{3} - \dfrac{1^3}{3} \bigg ] \hat k$

where;

current I = 7.0 A

$F = (7.0 \ A) (9.0) \bigg [\dfrac{27}{3} - \dfrac{1}{3} \bigg ] \hat k$

$F = (7.0 \ A) (9.0) \bigg [\dfrac{26}{3} \bigg ] \hat k$

F = 546 × 10⁻³ T/mT $\hat k$

F = 0.546 $\hat k$

4 0

3 years ago