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pogonyaev
2 years ago
14

A student performs a reaction twice. In the second trial, she raises the temperature by 20¨¬C and notices that the reaction take

s place more quickly. She concludes that the reaction must be endothermic. Explain why the student is incorrect.
Physics
2 answers:
Ivan2 years ago
6 0

Explanation:

An endothermic reaction is a reaction in which the reactants absorb energy and the energy of products is more than that of reactants.

Whereas an exothermic reaction is a reaction in which the reactants have more energy as compared to products. Hence, there will be release of energy.

Therefore, when temperature is increased by 20^{o}C then it means we are providing more energy to the reactants. As a result, there will be release of energy. Hence, the reaction will be exothermic in nature and not endothermic.

Thus, we can conclude that the student is incorrect because the reaction is exothermic and not endothermic.

Lerok [7]2 years ago
3 0

Answer: She is incorrect to conclude that the reaction is endothermic.

As in the second trial, the temperature has increased by 20 °C , that means the heat has been released and energy is released in exothermic reactions.

Exothermic reactions: The reactions in which the energy of the products is less than the energy of the reactants, and the excess energy is released as heat.

Endothermic reactions: The reactions in which the energy of the products is more than the energy of the reactants, and the excess energy is absorbed as heat.

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An unstrained horizontal spring has a length of 0.31 m and a spring constant of 220 N/m. Two small charged objects are attached
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The solution you should use is Hooke's law: F=-kx

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8 0
3 years ago
Consider a spring mass system (mass m1, spring constant k) with period T1. Now consider a spring mass system with the same sprin
tatuchka [14]

Answer:

Assuming that both mass here move horizontally on a frictionless surface, and that this spring follows Hooke's Law, then the mass of m_2 would be four times that of m_1.

Explanation:

In general, if the mass in a spring-mass system moves horizontally on a frictionless surface, and that the spring follows Hooke's Law, then

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

Here's how this statement can be concluded from the equations for a simple harmonic motion (SHM.)

In an SHM, if the period is T, then the angular velocity of the SHM would be

\displaystyle \omega = \frac{2\pi}{T}.

Assume that the mass starts with a zero displacement and a positive velocity. If A represent the amplitude of the SHM, then the displacement of the mass at time t would be:

\mathbf{x}(t) = A\sin(\omega\cdot t).

The velocity of the mass at time t would be:

\mathbf{v}(t) = A\,\omega \, \cos(\omega\, t).

The acceleration of the mass at time t would be:

\mathbf{a}(t) = -A\,\omega^2\, \sin(\omega \, t).

Let m represent the size of the mass attached to the spring. By Newton's Second Law, the net force on the mass at time t would be:

\mathbf{F}(t) = m\, \mathbf{a}(t) = -m\, A\, \omega^2 \, \cos(\omega\cdot t),

Since it is assumed that the mass here moves on a horizontal frictionless surface, only the spring could supply the net force on the mass. Therefore, the force that the spring exerts on the mass will be equal to the net force on the mass. If the spring satisfies Hooke's Law, then the spring constant k will be equal to:

\begin{aligned} k &= -\frac{\mathbf{F}(t)}{\mathbf{x}(t)} \\ &= \frac{m\, A\, \omega^2\, \cos(\omega\cdot t)}{A \cos(\omega \cdot t)} \\ &= m \, \omega^2\end{aligned}.

Since \displaystyle \omega = \frac{2\pi}{T}, it can be concluded that:

\begin{aligned} k &= m \, \omega^2 = m \left(\frac{2\pi}{T}\right)^2\end{aligned}.

For the first mass m_1, if the time period is T_1, then the spring constant would be:

\displaystyle k = m_1\, \left(\frac{2\pi}{T_1}\right)^2.

Similarly, for the second mass m_2, if the time period is T_2, then the spring constant would be:

\displaystyle k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Since the two springs are the same, the two spring constants should be equal to each other. That is:

\displaystyle m_1\, \left(\frac{2\pi}{T_1}\right)^2 = k = m_2\, \left(\frac{2\pi}{T_2}\right)^2.

Simplify to obtain:

\displaystyle \frac{m_2}{m_1} = \left(\frac{T_2}{T_1}\right)^2.

6 0
3 years ago
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