3. With a<br> you can measure just about anything.

Distinguish between friction force and motion?

Ksju [112]

Friction force is when you rub 2 things together and they get warm. Motion, on the other hand, is if your walking along the sidewalk - you hardly get warmer -------

Unless it's a colder day outside and you're walking SO you decide to rub your hands together to get warm, but if you were just walking , its motion and only motion - no friction :):)

6 0

3 years ago

A mass is oscillating with amplitude A at the end of a spring.

Dmitry_Shevchenko [17]

A) $x=\pm \frac{A}{2\sqrt{2}}$

The total energy of the system is equal to the maximum elastic potential energy, that is achieved when the displacement is equal to the amplitude (x=A):

$E=\frac{1}{2}kA^2$ (1)

where k is the spring constant.

The total energy, which is conserved, at any other point of the motion is the sum of elastic potential energy and kinetic energy:

$E=U+K=\frac{1}{2}kx^2+\frac{1}{2}mv^2$ (2)

where x is the displacement, m the mass, and v the speed.

We want to know the displacement x at which the elastic potential energy is 1/3 of the kinetic energy:

$U=\frac{1}{3}K$

Using (2) we can rewrite this as

$U=\frac{1}{3}(E-U)=\frac{1}{3}E-\frac{1}{3}U\\U=\frac{E}{4}$

And using (1), we find

$U=\frac{E}{4}=\frac{\frac{1}{2}kA^2}{4}=\frac{1}{8}kA^2$

Substituting $U=\frac{1}{2}kx^2$ into the last equation, we find the value of x:

$\frac{1}{2}kx^2=\frac{1}{8}kA^2\\x=\pm \frac{A}{2\sqrt{2}}$

B) $x=\pm \frac{3}{\sqrt{10}}A$

In this case, the kinetic energy is 1/10 of the total energy:

$K=\frac{1}{10}E$

Since we have

$K=E-U$

we can write

$E-U=\frac{1}{10}E\\U=\frac{9}{10}E$

And so we find:

$\frac{1}{2}kx^2 = \frac{9}{10}(\frac{1}{2}kA^2)=\frac{9}{20}kA^2\\x^2 = \frac{9}{10}A^2\\x=\pm \frac{3}{\sqrt{10}}A$

3 0

3 years ago

What is the difference between m/s and m.s.

vfiekz [6]

Answer

m/s rate of change of dispalcement per sec. ie velocity

m/s^2 is (m/s)/s ie rate of change of velocity per sec. ie accelerationplanation:

5 0

3 years ago

Read 2 more answers

Two parallel wires are separated by 5.60 cm, each carrying 2.65 A of current in the same direction. (a) What is the magnitude of

ololo11 [35]

Explanation:

It is given that,

The separation between two parallel wires, r = 5.6 cm = 0.056 m

Current in both the wires is 2.65 A

(a) We need to find the magnitude of the force per unit length between the wires. It can be given by :

$\dfrac{F}{l}=\dfrac{\mu_o I_1I_2}{2\pi r}\\\\\dfrac{F}{l}=\dfrac{4\pi \times 10^{-7}\times 2.65\times 2.65}{2\pi \times 0.056}\\\\\dfrac{F}{l}=2.5\times 10^{-5}\ N/m$