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Stolb23 [73]
3 years ago
11

A long wire carrying I=1.3A of constant current oriented East-to-West (and the current is running eastwards) is placed distance

d=0.8cm over the top of a compass needle. Assuming the Earth magnetic field to be BEarth= 29μT what is the absolute value of the angular deviation θ of the compass needle from the North-to-South direction?
Physics
1 answer:
Nastasia [14]3 years ago
7 0

Answer:

Explanation:

Magnetic field due to a long wire

= 10⁻⁷ x 2i / r where i is current in the wire and r is distance of the point from the wire

In the given case magnetic field

B = 10⁻⁷ x 2i / r

= 10⁻⁷ x 2x 1.3 / .8 x 10⁻²

= 3.25 x 10⁻⁵ T

= 32.5 x 10⁻⁶ T

It will create a deflecting torque on the compass needle

Let after turning by angle θ , equilibrium point is reached

at this point , deflecting torque = restoring torque by earth's magnetic field

MB cosθ = MHsinθ , M is magnetic moment of magnetic needle , H is horizontal component of earth's magnetic field.

Tanθ = B / H

= 32.5 / 29

=1.20

θ = 50 degree approx .

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A scientific theory states that the universe is expanding. Which of these statements is correct about this theory?
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____________occurs when an object travels in a curved path.
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Answer:

Acceleration

Explanation:

can you mark me  brainlies

So, if an object travels in a curved path, it changes velocity, and, thus, accelerates. This acceleration must be tied to a force. ... Therefore, whenever an object travels in a curved path, there must be an unbalanced force acting upon it. It is important to understand that all this may occur without a change in speed.t

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Riders on the Tower of Doom, an amusement park ride, experience 2.0 s of free fall, after which they are slowed to a stop in 0.5
umka2103 [35]

Newton's second law and kinematics allow finding the answers for the changes in weight when the body stops are:

        A) The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

Given parameters

  • Free fall time t1 = 2.0 s
  • The time tot stops at t2 = 0.5 s
  • Mass m = 65 kg

To find

   A)  Apparent weight

   B)  Factor that exceeds the actual weight

Kinematics studies the movement of bodies, looking for relationships between the position, speed and acceleration of bodies.

This exercise we will do them in two parts

1st part. Freefall

We look for the velocity at the end of the trajectory, as part of rest the initial velocity is zero

            v = v₀ - g t

Where v is the velocity, vo the initial velocity, g the acceleration due to gravity and y time

            v = -gt

            v = -9.8 2.0

            v = -19.6 m / s

The negative sign indicates that the velocity is directed downwards

2nd part. Braking movement

We look for the acceleration to stop the body, in this case the final velocity is zero

            v_f = v + a t

             0 = v + at

             a = - \frac{v}{t}- v / t

             a = -\  \frac{-19.6}{0.5}

             a = + 39.2 m / s²

The positive sign indicates that the acceleration is directed upward.

A) Newton's second law states that the force is proportional to the mass and the acceleration of the body

            ∑ F = m a

            F -W = m a

            F = m (a + g)

            F = 65 (39.2 + 9.8)

            F = 3185 N

Body weight is

             W = mg

             W = 65 9.8

             W = 637 N

therefore the apparent weight is

             ΔW = F -W

             ΔW = 3185 - 637

             Δw = 2548 N

B) the weight change factor

             Factor =\frac{F}{W} F / W

             Factor =\frac{3185}{937}

             factor = 5

In conclusion, using Newton's second law and kinematics, we can find the answers for the changes in weight when the body stops are:

       A)  The apparent weight is 3548 N directed upwards

      B)  The factor of change of weights is 5

Learn more about Newton's second law and kinematics here:

brainly.com/question/15378807

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