Un móvil que viaja a 40 m/s aplica los frenos y desacelera a razón de -0,25 m/seg 2 , ¿Cuánto tiempo le llevo

A mass m0 is attached to a spring and hung vertically. The mass is raised a short distance in the vertical direction and release

iragen [17]

Answer:

The frequency of the oscillations in terms of fo will be f2=fo/3

E xplanation:

T= $2pie\frac sqrt {m}{k}$

$\frac {{f2}/times {fo}}$ =1:3

⇒f2=fo\3

Here frequency f is inversely poportional to square root of mass m.

so the value of remainder of frequency f2 and fo is equal to 1:3.

⇒ $\frac{f2} {f1}$ = $\frac sqrt{m1}[m2}$

⇒ $\frac{f2}{fo}$ = 1:3

⇒f2= $\frac{fo} {3}$

6 0

3 years ago

Energy Transformation

Serjik [45]

Answer:

Explanation:

<u>Kinetic Energy</u>

Is the energy an object has due to its state of motion. It's proportional to the square of the speed.

The equation for the kinetic energy is:

$\displaystyle K=\frac{1}{2}mv^2$

Where:

m = mass of the object

v = speed at which the object moves

The kinetic energy is expressed in Joules (J)

The car has a mass of m=2000 Kg and travels at v=20 m/s. Calculating the kinetic energy:

$\displaystyle K=\frac{1}{2}2000*20^2$

Calculating:

K =400000 J

3 0

3 years ago

A 2 kg toy car moves at a speed of 5 m/s.<br> What is the kinetic energy of the car?

ANTONII [103]

Answer:

<h2>The answer is 25 J</h2>

Explanation:

The kinetic energy of an object can be found by using the formula

$KE = \frac{1}{2} m {v}^{2} \\$

m is the mass

v is the velocity

From the question we have

$KE = \frac{1}{2} \times 2 \times {5}^{2} \\ = 1 \times {5}^{2}$

We have the final answer as

Hope this helps you

5 0

2 years ago

What is the lupac<br>name<br>H3C – CHT<br>the H₂ - CH - CH2<br>CH₂<br>CH₂<br>CH3

dimulka [17.4K]

Answer:??

Explanation:

5 0

3 years ago

When the displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position, what fraction of the

KonstantinChe [14]

Answer:

The ratio is KE : TM = 0.75

Explanation:

from the question we are told that

The displacement of a mass on a spring in simple harmonic motion is A/2 from the equilibrium position

Generally the total mechanical energy of the mass is mathematically represented as

$TM = \frac{1}{2} * k * A^2$

Here k is the spring constant , A is the total displacement of the the mass from maximum compression to maximum extension of the spring

Generally this total mechanical energy is mathematically represented as

$TM = KE + PE$

=> $KE = TM - PE$

Here the potential energy of the mass is mathematically represented as

$PE = \frac{1}{ 2} * k * [ x ]^2$

Here x is the displacement of the mass from maximum compression or extension of the spring to equilibrium position and the value is

$x = \frac{A}{2}$

So

$PE = \frac{1}{ 2} * k * [ \frac{A}{2} ]^2$

So

$KE = \frac{1}{2} * k * A^2 - \frac{1}{2} * k * [\frac{A}{2} ]^2$

=> $KE = \frac{1}{2} * k * A^2 - \frac{1}{8} * k * A ^2$

=> $KE = 0.375 * k * A^2$

So the ratio of $KE : TM$ is mathematically represented as

$\frac{KE}{TM} = \frac{0.375 k A^2 }{0.5 k A^2}$

=> $\frac{KE}{TM} = 0.75$

3 0

3 years ago