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3 years ago

Why would two poles of different magnets repel one another?

Physics

2 answers:

docker41 [41]3 years ago

5 0

Because the poles have like poles and like poles reple each other

Ghella [55]3 years ago

5 0

Answer:

The answer is because they are the same poles!

Explanation:

Same poles will repel each other

You might be interested in

A fireman is sliding down a fire pole. As he speeds up, he tightens his grip on the pole, thus increasing the vertical frictiona

Andrei [34K]

Answer:

The fireman will continue to descend, but with a constant speed.

Explanation:

In kinetic friction <em>(which is the case discussed here) </em>since the fireman is already in motion because of a certain force, once the frictional force matches the normal force, the fireman will stop accelerating and continue moving at a constant rate with the original speed he had. We will need a force greater than the normal force acting on the fireman to cause a deceleration.

We need to understand the difference between static friction and kinetic friction.

Static friction occurs in objects that are stationary, while kinetic friction occurs in objects that are already in motion.

In static friction, when the frictional force matches the weight or normal force of the object, the object remains stationary.

While in kinetic friction, when the frictional force matches the normal force, the object will stop accelerating. This is the case of the fireman sliding down the pole as discussed above.

8 0

3 years ago

A 1.05 kg block slides with a speed of 0.865 m/s on a frictionless horizontal surface until it encounters a spring with a force

djyliett [7]

Answer:

a) U = 0 J

k = 0.393 J

E = 0.393 J

b) U = 0.0229J

k = 0.370 J

E = 0.393 J

c) U = 0.0914 J

k = 0.302 J

E = 0.393 J

d) U = 0.206 J

k = 0.187 J

E = 0.393 J

e) U = 0.366 J

k = 0.027 J

E = 0.393 J

Explanation:

Hi there!

The equations of kinetic energy and elastic potential energy are as follows:

k = 1/2 · m · v²

U = 1/2 · ks · x²

Where:

m = mass of the block.

v = velocity.

ks = spring constant.

x = displacement of the string.

a) When the spring is not compressed, the spring potential energy will be zero:

U = 1/2 · ks · x²

U = 1/2 · 457 N/m · (0 cm)²

U = 0 J

The kinetic energy of the block will be:

k = 1/2 · m · v²

k = 1/2 · 1.05 kg · (0.865 m/s)²

k = 0.393 J

The mechanical energy will be:

E = k + U = 0.393 J + 0 J = 0.393 J

This energy will be conserved, i.e., it will remain constant because there is no work done by friction nor by any other dissipative force (like air resistance). This means that the kinetic energy will be converted only into spring potential energy (there is no thermal energy due to friction, for example).

b) The spring potential energy will be:

U = 1/2 · 457 N/m · (0.01 m)²

U = 0.0229 J

Since the mechanical energy has to remain constant, we can use the equation of mechanical energy to obtain the kinetic energy:

E = k + U

0.393 J = k + 0.0229 J

0.393 J - 0.0229 J = k

k = 0.370 J

c) The procedure is now the same. Let´s calculate the spring potential energy with x = 0.02 m.

U = 1/2 · 457 N/m · (0.02 m)²

U = 0.0914 J

Using the equation of mechanical energy:

E = k + U

0.393 J = k + 0.0914 J

k = 0.393 J - 0.0914 J = 0.302 J

d) U = 1/2 · 457 N/m · (0.03 m)²

U = 0.206 J

E = 0.393 J

k = E - U = 0.393 J - 0.206 J

k = 0.187 J

e) U = 1/2 · 457 N/m · (0.04 m)²

U = 0.366 J

E = 0.393 J

k = E - U = 0.393 J - 0.366 J = 0.027 J.

4 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

What is the equation describing the motion of a mass on the end of a spring which is stretched 8.8 cm from equilibrium and then

Sedbober [7]

Answer:

$x=(0.088m)\cos(\sqrt{\frac{k}{m} } t)$

Explanation:

We first identify the elements of this simple harmonic motion:

The amplitude A is 8.8cm, because it's the maximum distance the mass can go away from the equilibrium point. In meters, it is equivalent to 0.088m.

The angular frequency ω can be calculated with the formula:

$\omega =\sqrt{\frac{k}{m}}$

Where k is the spring constant and m is the mass of the particle.

Now, since the spring starts stretched at its maximum, the appropriate function to use is the positive cosine in the equation of simple harmonic motion:

$x=A\cos(\omega t)$