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KengaRu [80]
3 years ago
13

A 5 cm length of wire carries a current of 3 A in the positive z direction. The force on this wire due to a magnetic field vecto

r is vector = (-0.2 ihatbold + 0.2 jhatbold) N. If this wire is rotated so that the current flows in the positive x direction, the force on the wire is vector = 0.2 khatbold N. Find the magnetic field vector.
Physics
1 answer:
ioda3 years ago
6 0

Answer:

B = 4/3 ( i + j + k )

Explanation:

I = 3 k

L = 5 x 10⁻² m

F = - 0.2 i + 0.2 j

Let magnetic field be B

B = B₁ i + B₂ j + B₃ k

F = L ( I x B )

= 5 x 10⁻² 3 k x ( B₁ i + B₂ j + B₃ k )

=   15 x 10⁻² B₁ j  -   15 x 10⁻² B₂ i

Given

F = - 0.2 i + 0.2 j

equating equal terms

15 x 10⁻² B₂ = .2

15 x 10⁻² B₁ =  .2

B₁ = .02 / 15 x 10⁻² = 20 / 15 N

B₂ = .02 / 15 x 10⁻² = 20 / 15 N

Now wire is rotated so that current flows in positive x direction

I = 3 i

F = . 2 k

F = L ( I x B )

= 5 x 10⁻² x 3 i  x ( B₁ i + B₂ j + B₃ k )

= 15 x 10⁻² B₃ k

15 x 10⁻² B₃ = .2

B₃ = .2 /15 x 10⁻²

B₃ = 20 /15

B = B₁ i + B₂ j + B₃ k

20 /15 9 ( i + j + k )

B = 4/3 ( i + j + k )

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1 m/s² down

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"Calculate the acceleration of the elevator, and find the direction of acceleration."

Sum of forces in the +y direction:

∑F = ma

N − mg = ma

0.9 mg − mg = ma

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How do you calculate a bearing angle and its equivalent angle?
denis-greek [22]

Explanation:

A bearing if an angle is measured clockwise from north direction.

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How much does the gravitational force of attraction change between two asteroids if the two asteroids drift three times closer t
Katen [24]

Answer:

Increase 9 times

Explanation:

We have Newton formula for attraction force between 2 objects with mass and a distance between them:

F_G = G\frac{M_1M_2}{R^2}

where G =6.67408 \times 10^{-11} m^3/kgs^2 is the gravitational constant. M_1, M_2 is the masses of the 2 objects. and R is the distance between them.

Since the force is inversely proportional to the distance squared, if it is reduced by 3 times, the gravitational force between them would increase by 3^2 = 9 times

6 0
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A world-class sprinter running a 100 m dash was clocked at 5.4 m/s 1.0 s after starting running and at 9.8 m/s 1.5 s later. In w
cupoosta [38]

Answer:

<em>The output power is greater in the interval from 1.0 s to 2.5 s</em>

Explanation:

<u>Physical Power </u>

It measures the amount of work W an object does in certain time t. The formula needed to compute power is

\displaystyle P=\frac{W}{t}

Work can be computed in several ways since we are given the motion conditions, we'll use this formula, for F= applied force, x=distance parallel to F

W=F.x

The second Newton's law gives us the net force as

F=m.a

being m the mass of the object and a the acceleration it has for a given period of time. In our problem, we have two different behaviors for each interval and we must calculate this force since the acceleration is changing. Let's calculate the acceleration in the first interval. We can use the formula for the final speed vf knowing the initial speed vo (which is 0 because the sprinter starts from rest), the acceleration a, and the time t:

v_f=v_o+at

v_f=at

Solving for a

\displaystyle a=\frac{v_f}{t}={5.4}{1}

a=5.4\ m/s^2

The distance traveled in the interval is given by

\displaystyle x=v_o.t+\frac{a.t^2}{2}

Since vo=0

\displaystyle x=\frac{a.t^2}{2}=\frac{5.4(1)^2}{2}

x=2.7\ m

The force is given by

F=m.a

We don't know the value of m, so the force is

F=2.7m

Computing the work done by the sprinter

W=F.x=2.7m(5.4)

W=14.58m

The power is finally computed

\displaystyle P=\frac{W}{t}=\frac{14.58m}{1}

P=14.58m

During the second interval, from t=1 sec to 1.5 sec, the speed changes from 5.4 m/s to 9.8 m/s. This allows us to compute the second acceleration

\displaystyle a=\frac{v_f-v_o}{t}=\frac{9.8-5.4}{0.5}

a=8.8\ m/s^2

The distance is

\displaystyle x=(5.4).(0.5)+\frac{8.8(0.5)^2}{2}

x=3.8\ m

The net force is

F=m(8.8)=8.8m

The work done by the sprinter is now computed as

W=8.8m(3.8)=33.44m

At last, the output power is

\displaystyle P=\frac{33.44m}{0.5}=66.88m

By comparing both results, and being m the same for both parts, we conclude the output power is greater in the interval from 1.0 s to 2.5 s

6 0
3 years ago
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