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3 years ago

Can anyone plzzzz answer this question... Cuz it's really urgent....

Physics

2 answers:

aev [14]3 years ago

8 0

The measurements in the table allow for the following conclusion: Stiff and dense materials like steel and granite allow sound to propagate extremely fast (5-6 km/s) while lower density ones, like the fluids, reduce this speed to about 1.5km/s, with the lowest-density media (gases) allowing for only order of hundreds of m/s.

The measurements are consistent with a model of energy propagation consisting of spheres (modeling the molecules) and springs connecting them (modeling the molecular bonds). It can be shown that the mass of the spheres and the stiffness of the springs are factors in speed of energy propagation.

nlexa [21]3 years ago

4 0

The fastest in speed of sound is steel. The slowest in speed of sound is oxygen. Hope this helped! I don't really know how to answer this but by looking at the table it looks like this can be one of the conclusions.

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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

Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are

adelina 88 [10]

Answer:

The final temperature of the two objects is the same.

Explanation:

The expression for the heat energy in terms of mass, specific heat and the change in the temperature is as follows:

$Q=mc(T_{f} -T_{i})$

Here, Q is the heat energy, m is the mass of the object, c is the specific heat and $T_{f},T_{i}$ are the final temperature and initial temperature.

According to the given question, Two objects of the same mass, but made of different materials, are initially at the same temperature. Equal amounts of heat are added to each object.

$Q=mc(T_{f}-T_{i})$ ............(1)

$Q=mc(T'_{f}-T_{i})$ .............(2)

From (1) and (2),

$mc(T_{f}-T_{i})=mc(T'_{f}-T_{i})$

$T_{f}-T_{i}=T'_{f}-T_{i}$

$T_{f}=T'_{f}$

Therefore, the final temperature of the two objects is the same.

6 0

3 years ago

As a box slides down a ramp, friction does 23.0 joules of work. At the bottom of the ramp, the box has 3.8 joules of kinetic ene

tensa zangetsu [6.8K]

Answer:

The high of the ramp is 2.81[m]

Explanation:

This is a problem where it applies energy conservation, that is part of the potential energy as it descends the block is transformed into kinetic energy.

If the bottom of the ramp is taken as a potential energy reference point, this point will have a potential energy value equal to zero.

We can find the mass of the box using the kinetic energy and the speed of the box at the bottom of the ramp.

$E_{k}=0.5*m*v^{2}\\\\where:\\E_{k}=3.8[J]\\v = 2.8[m/s]\\m=\frac{E_{k}}{0.5*v^{2} } \\m=\frac{3.8}{0.5*2.8^{2} } \\m=0.969[kg]$

Now applying the energy conservation theorem which tells us that the initial kinetic energy plus the work done and the potential energy is equal to the final kinetic energy of the body, we propose the following equation.

$E_{p}+W_{f}=E_{k}\\where:\\E_{p}= potential energy [J]\\W_{f}=23[J]\\E_{k}=3.8[J]\\$