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4 years ago

9

Which statement correctly describes the relationship between the energy of a wave and the wave’s amplitude? High energy waves ha

ve high amplitudes. Low energy waves have high amplitudes. High and low energy waves have the same amplitudes.

2 answers:

schepotkina [342]4 years ago

7 0

Explanation :

The energy of a wave is given by:

$E=\dfrac{hc}{\lambda}$

Where,

h is the planck's constant

c is the speed of light

$\lambda$ is the wavlength

One more factor on which the energy of the wave depends is the amplitude of the wave.

The maximum displacement of the particle is called the amplitude of the wave.

The particles transports energy from one location to another.

Hence, the correct option is (a) " High energy waves have high amplitudes".

creativ13 [48]4 years ago

6 0

Answer:

High energy waves have high amplitudes.

Explanation:

Energy of a wave is dependent on its amplitude. Larger amplitude means that more energy is transferred from one particle to another. Other factors on which energy of a wave depends are :

$E=\dfrac{hc}{\lambda}=hf$

f is frequency

$\lambda$ is wavelength

So, the correct option is (A) " High energy waves have high amplitudes".

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3 years ago

8. An unpowered flywheel is slowed by a constant frictional torque. At time t = 0 it has an angular velocity of 200 rad/s. Ten s

allsm [11]

Answer:

a) $\omega = 50\,\frac{rad}{s}$ , b) $\omega = 0\,\frac{rad}{s}$

Explanation:

The magnitude of torque is a form of moment, that is, a product of force and lever arm (distance), and force is the product of mass and acceleration for rotating systems with constant mass. That is:

$\tau = F \cdot r$

$\tau = m\cdot a \cdot r$

$\tau = m \cdot \alpha \cdot r^{2}$

Where $\alpha$ is the angular acceleration, which is constant as torque is constant. Angular deceleration experimented by the unpowered flywheel is:

$\alpha = \frac{170\,\frac{rad}{s} - 200\,\frac{rad}{s} }{10\,s}$

$\alpha = -3\,\frac{rad}{s^{2}}$

Now, angular velocities of the unpowered flywheel at 50 seconds and 100 seconds are, respectively:

a) t = 50 s.

$\omega = 200\,\frac{rad}{s} - \left(3\,\frac{rad}{s^{2}} \right) \cdot (50\,s)$

$\omega = 50\,\frac{rad}{s}$

b) t = 100 s.

Given that friction is of reactive nature. Frictional torque works on the unpowered flywheel until angular velocity is reduced to zero, whose instant is:

$t = \frac{0\,\frac{rad}{s}-200\,\frac{rad}{s} }{\left(-3\,\frac{rad}{s^{2}} \right)}$

$t = 66.667\,s$

Since $t > 66.667\,s$ , then the angular velocity is equal to zero. Therefore:

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Daniel [21]

Answer:

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Explanation:

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Thus,

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Using integrated rate law for first order kinetics as:

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Where,

$[A_t]$ is the concentration at time t

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So,

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t = ?

$\frac {[A_t]}{[A_0]}=e^{-k\times t}$

$0.8333=e^{-0.1824\times t}$

t ≅ 1 day

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