A certain mountain road is inclined 3.1° with respect to the horizon. what is the change in altitude of the car as a result of i

Will the positioning of your Slinky along the east-west, north-south, or some other orientation affect your readings? Why?

Oxana [17]

Positioning your Slinky along any direction different from its initial position will affect your reading, because there will be change in the magnetic field.

<h3>Effect of magnet on Slinky</h3>

If the Slinky is made of an iron alloy, it can be magnetized by itself. Moving the Slinky around can cause a change in the magnetic field, even if no current is flowing.

When there is a change in the magnetic field, the reading changes.

At any point, you change the orientation of the Slinky, you will need to zero the reading or adjust the Slinky back to its initial position, even if the sensor does not move.

Thus, Positioning your Slinky along any direction that is different to its initial position will affect your reading because there will be change in the magnetic field.

Learn more about magnetic field here: brainly.com/question/7802337

5 0

2 years ago

What would happened if there is no invented of machine?

yuradex [85]

if there were no invention of machines then life would have been more difficult and simple works could be hard to do. Even now we are using our phones, sitting in a AC room interacting to eachother from different places. without the invention of machines simple things like transportation would have been difficult. There would be horses and donkey for the transportation. There would be no electricity,no internet, no transportation, not even c computers or mobile etc. The market for business will be smaller, the knowledge and news about world would be less.

so the problem would have been bigger than we can imagine. But one thing is that nature could survive lot more compared to what we have done till now by destroying nature.

4 0

1 year ago

5. Tại sao khi lặn ta luôn cảm thấy tức ngực và càng lặn sâu thì cảm giác tức ngực càng tăng? A. Ap suất của nước giảm B. Ap s

natali 33 [55]

Answer:

c and d

Explanation:

obviously kksxsxksxkskxkskxksxksxsxsxsxsxsxsxs

7 0

2 years ago

A uniform disk with mass 35.2 kg and radius 0.200 m is pivoted at its center about a horizontal, frictionless axle that is stati

Sergio [31]

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:

$\tau = I * \alpha (1)$

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:

$\alpha = \frac{2*F}{m*r} = \frac{2*34.5N}{35.2kg*0.2m} = 9.8 rad/s2 (4)$

Since the angular acceleration is constant, we can use the following kinematic equation:

$\omega_{f}^{2} - \omega_{o}^{2} = 2*\Delta \theta * \alpha (5)$

Prior to solve it, we need to convert the angle rotated from revs to radians, as follows:

$0.2 rev*\frac{2*\pi rad}{1 rev} = 1.3 rad (6)$

Replacing (6) in (5), taking into account that ω₀ = 0 (due to the disk starts from rest), we can solve for ωf, as follows:

$\omega_{f} = \sqrt{2*\alpha *\Delta\theta} = \sqrt{2*1.3rad*9.8rad/s2} = 5.1 rad/sec (7)$

Now, we know that there exists a fixed relationship the tangential speed and the angular speed, as follows:

$v = \omega * r (8)$

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:

$v= 5.1 rad/sec* 0.2 m = 1.01 m/s (9)$

b)

There exists a fixed relationship between the tangential and the angular acceleration in a circular movement, as follows:

$a_{t} = \alpha * r (9)$

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:

$a_{t} = 9.8 rad/s2 * 0.200 m = 1.96 m/s2 (10)$

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:

$a_{c} = \omega^{2} * r (11)$

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:

$a_{c} = \omega^{2} * r = (5.1 rad/sec)^{2} * 0.200 m = 5.2 m/s2 (12)$

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:

$a = \sqrt{a_{t} ^{2} + a_{c} ^{2} } = \sqrt{(1.96m/s2)^{2} +(5.2m/s2)^{2} } = 5.6 m/s2 (13)$

6 0

2 years ago

Need help ASAP please Thanks

Karo-lina-s [1.5K]

Answer: D
Explanation: there is less light at that point.

3 0

3 years ago