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Dennis_Churaev [7]

3 years ago

12

A rope has one end tied to a vertical support. You hold the other end so that the rope is horizontal. If you move the end of the

rope back and forth with a frequency of 4 Hz, the transverse wave you produce has a wavelength of 0.5 m. What is the speed of the wave in the rope?a. 0.13 m/sb. 8 m/sc. 2 m/sd. 4 m/s

1 answer:

Strike441 [17]3 years ago

7 0

Answer:

c. 2 m/s

Explanation:

The relationship between speed, frequency and wavelength of a wave is given by:

$v=f \lambda$

where

v is the speed of the wave

f is its frequency

$\lambda$ is the wavelength

For the transverse wave in this problem, we have:

$f = 4 Hz$ is the frequency

$\lambda=0.5 m$ is the wavelength

Substituting these numbers into the equation, we find the speed of the wave:

$v=(4 Hz)(0.5 m)=2 m/s$

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Answer:

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

Explanation:

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$\tau = F \cdot r$

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

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

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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:

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

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

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A 1400 kg car starts from rest on a horizontal road and gains a speed of 61 km/h in 19 s. (a) what is its kinetic energy at the

lana [24]

(a) Let's convert the final speed of the car in m/s:
$v_f = 61 km/h = 16.9 m/s$
The kinetic energy of the car at t=19 s is
$K= \frac{1}{2}mv_f^2= \frac{1}{2}(1400 kg)(16.9 m/s)^2=2.00 \cdot 10^5 J$

(b) The average power delivered by the engine of the car during the 19 s is equal to the work done by the engine divided by the time interval:
$P= \frac{W}{\Delta t}$
But the work done is equal to the increase in kinetic energy of the car, and since its initial kinetic energy is zero (because the car starts from rest), this translates into
$P= \frac{K}{\Delta t}= \frac{2.00 \cdot 10^5 J}{19 s}=1.05 \cdot 10^4 W$

(c) The instantaneous power is given by
$P_i = Fv_f$
where F is the force exerted by the engine, equal to F=ma.

So we need to find the acceleration first:
$a= \frac{v_f-v_i}{\Delta t}= \frac{16.9 m/s}{19 s}=0.89 m/s^2$
And the problem says this acceleration is constant during the motion, so now we can calculate the instantaneous power at t=19 s:
$P_i = Fv=(ma)v=(1400 kg)(0.89 m/s^2)(16.9 m/s)=2.11 \cdot 10^4 W$

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