Anita and Nick are playing tug-of-war near a mud puddle. They are each holding on to an end of a taut rope that has a knot exact
ly in the middle. Anita's position is 6.2 meters east of the center of the puddle and Nick's position is 3.0 meters west of the center of the puddle. What is the location of the knot relative to the center of the puddle? Treat east as positive and west as negative.
1 answer:
Answer:
1.6 meters east
Explanation:
Note: Image attached.
As we can see on the data, Anita is further from the puddle than Nick. Additionally, the distances relatives to the center of the puddle allow us to now the lenght of the rope which is the sum of these two, 9.2 meters. We also know that the knot is at the middle of the rope, so it means it is at 4.6 meters of both Anita and Nick. Now what is next? Well, we now have everything we know. We have the distance between Anita and the puddle and we also know the distance between Anita and the knot (4.6 meters). We just need to subtract the distance to the knot from the distance to the puddle which is equal to 1,6 meters east direction.
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Answer:
a) v = 1.01 m/s
b) a = 5.6 m/s²
Explanation:
a)
- If the disk is initially at rest, and it is applied a constant force tangential to the rim, we can apply the following expression (that resembles Newton's 2nd law, applying to rigid bodies instead of point masses) as follows:
- Where τ is the external torque applied to the body, I is the rotational inertia of the body regarding the axis of rotation, and α is the angular acceleration as a consequence of the torque.
- Since the force is applied tangentially to the rim of the disk, it's perpendicular to the radius, so the torque can be calculated simply as follows:
- τ = F*r (2)
- For a solid uniform disk, the rotational inertia regarding an axle passing through its center is just I = m*r²/2 (3).
- Replacing (2) and (3) in (1), we can solve for α, as follows:
- Since the angular acceleration is constant, we can use the following kinematic equation:
- Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:
- Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:
- Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:
- where r is the radius of the circular movement. If we want to know the tangential speed of a point located on the rim of the disk, r becomes the radius of the disk, 0.200 m.
- Replacing this value and (7) in (8), we get:
b)
- There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:
- where r is the radius of the circular movement. In this case the point is located on the rim of the disk, so r becomes the radius of the disk.
- Replacing this value and (4), in (9), we get:
- Now, the resultant acceleration of a point of the rim, in magnitude, is the vector sum of the tangential acceleration and the radial acceleration.
- The radial acceleration is just the centripetal acceleration, that can be expressed as follows:
- Since we are asked to get the acceleration after the disk has rotated 0.2 rev, and we have just got the value of the angular speed after rotating this same angle, we can replace (7) in (11).
- Since the point is located on the rim of the disk, r becomes simply the radius of the disk,, 0.200 m.
- Replacing this value and (7) in (11) we get:
- The magnitude of the resultant acceleration will be simply the vector sum of the tangential and the radial acceleration.
- Since both are perpendicular each other, we can find the resultant acceleration applying the Pythagorean Theorem to both perpendicular components, as follows:
Answer:
The current is
Explanation:
From the question we are told that
The length of the segment is
The current is
The force felt is
The distance of the second wire is
Generally the current on the second wire is mathematically represented as
Here is the permeability of free space with value
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