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2 years ago

How is energy transformation related to molecular activity.

Physics

1 answer:

rodikova [14]2 years ago

6 0

Explanation:

The law of conservation of energy states that when one form of energy is transformed to another, no energy is destroyed in the process. According to the law of conservation of energy, energy cannot be created or destroyed. So the total amount of energy is the same before and after any transformation.

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In the final situation below, the 8.0 kg box has been launched with a speed of 10.0 m/s across a frictionless surface. Find the

Murljashka [212]

Answer:

the energy of the spring at the start is 400 J.

Explanation:

Given;

mass of the box, m = 8.0 kg

final speed of the box, v = 10 m/s

Apply the principle of conservation of energy to determine the energy of the spring at the start;

Final Kinetic energy of the box = initial elastic potential energy of the spring

K.E = Ux

¹/₂mv² = Ux

¹/₂ x 8 x 10² = Ux

400 J = Ux

Therefore, the energy of the spring at the start is 400 J.

8 0

3 years ago

A cord of mass 0.65 kg is stretched between two supports 8.0 m apart. If the tension in the cord is 120 N, how long will it take

DedPeter [7]

The time taken by the pulse to travel from one support to the other is 0.208 s.

<h3>Given:</h3>

The mass of the cord is m = 0.65 kg.

The distance between the supports is, d = 8.0 m.

The tension in the cord is T = 120 N.

The time taken by the pulse to travel from one support to the other is given as,

$v=\frac{d}{t}$

$t=\frac{d}{v}$

Here, v is the linear velocity of a pulse. Its value is,

$v=\sqrt{\frac{T d }{m} }$

$v=\sqrt{\frac{120 * 8}{0.65} }$

$v= 38.43 m/s$

Then,

$t=\frac{8}{38.43}$

$t=0.208 s$

Thus, the time taken by the pulse to travel from one support to the other is 0.208 s.

Learn more about tension here:

brainly.com/question/24994188

#SPJ4

7 0

1 year ago

What is the highest and lowest frequency a human being can hear

nata0808 [166]

Every person is different. But for a planet-wide overall average that roughly represents all human beings on Earth, the figures usually used are:

from 20 Hz to 20,000 Hz .

6 0

3 years ago

g initial angular velocity of 39.1 rad/s. It starts to slow down uniformly and comes to rest, making 76.8 revolutions during the

MrRa [10]

Answer:

Approximately $-1.58\; \rm rad \cdot s^{-2}$ .

Explanation:

This question suggests that the rotation of this object slows down "uniformly". Therefore, the angular acceleration of this object should be constant and smaller than zero.

This question does not provide any information about the time required for the rotation of this object to come to a stop. In linear motions with a constant acceleration, there's an SUVAT equation that does not involve time:

$v^2 - u^2 = 2\, a\, x$ ,

where

$v$ is the final velocity of the moving object,
$u$ is the initial velocity of the moving object,
$a$ is the (linear) acceleration of the moving object, and
$x$ is the (linear) displacement of the object while its velocity changed from $u$ to $v$ .

The angular analogue of that equation will be:

$(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ , where

$\omega(\text{final})$ and $\omega(\text{initial})$ are the initial and final angular velocity of the rotating object,
$\alpha$ is the angular acceleration of the moving object, and
$\theta$ is the angular displacement of the object while its angular velocity changed from $\omega(\text{initial})$ to $\omega(\text{final})$ .

For this object:

$\omega(\text{final}) = 0\; \rm rad\cdot s^{-1}$ , whereas
$\omega(\text{initial}) = 39.1\; \rm rad\cdot s^{-1}$ .

The question is asking for an angular acceleration with the unit $\rm rad \cdot s^{-1}$ . However, the angular displacement from the question is described with the number of revolutions. Convert that to radians:

$\begin{aligned}\theta &= 76.8\; \rm \text{revolution} \\ &= 76.8\;\text{revolution} \times 2\pi\; \rm rad \cdot \text{revolution}^{-1} \\ &= 153.6\pi\; \rm rad\end{aligned}$ .

Rearrange the equation $(\omega(\text{final}))^2 - (\omega(\text{initial}))^2 = 2\, \alpha\, \theta$ and solve for $\alpha$ :

$\begin{aligned}\alpha &= \frac{(\omega(\text{final}))^2 - (\omega(\text{initial}))^2}{2\, \theta} \\ &= \frac{-\left(39.1\; \rm rad \cdot s^{-1}\right)^2}{2\times 153.6\pi\; \rm rad} \approx -1.58\; \rm rad \cdot s^{-1}\end{aligned}$ .

7 0

3 years ago

15. In the diagram shown, the emf of the secondary winding is______ the emf of the primary winding.

sp2606 [1]

Answer:

Greater than

Explanation:

3 0

3 years ago