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3 years ago

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In an electric shaver, the blade moves back and forth over a distance of 2.0 mm in simple harmonic motion, with frequency 124 Hz

. Find (a) the amplitude, (b) the maximum blade speed, and (c) the magnitude of the maximum blade acceleration.

1 answer:

bagirrra123 [75]3 years ago

8 0

Answer:

(a) A = 1 mm

(b) $V_{max}=0.77872 m/s$

(c) $a_{max}=606.4 m/s^{2}/tex]Explanation:Distance moved back and forth = 2 mm Frequency, f = 124 HzSo, amplitude is the half of the distance traveled back and forth. (a) So, amplitude, A = 1 mm(b) Angular frequency, ω = 2 π f = 2 x 3.14 x 124 = 778.72 rad/s The formula for the maximum speed is given by [tex]V_{max}=\omega \times A$

$V_{max}=778.72 \times 0.001$

$V_{max}=0.77872 m/s$

(c) The formula for the maximum acceleration is given by

$a_{max}=\omega ^{2}A$

$a_{max}=778.72 ^{2}\times 0.001$

[tex]a_{max}=606.4 m/s^{2}/tex]

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Answer:

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Explanation:

We have a case of oblique movement. In these cases the movement in the X axis is a Uniform Rectelinear Movement (URM), and a Uniform Accelerated Movement (UAM) in the Y axis.

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A.) First, it is necessary to know t, total time.

To figure out t value, we use UAM, since time is determined by this movement.

Now, at the end of the movement, $Y=0$ , then

$0= Y_0 +v_{y0} t - \frac{g}{2} t^2$

$0=2.4m+79m/s*sin(39)t-(1/2*9.81m/s^2)t^2$

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But, in physics, time it could not be negative, so we take $t_1= 10.18$

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$X=79m/s*cos(\39)*10.18s= 624.996 m$

B.) Now, the narrow has an additional speed, that could be sum to the speed due to the bow.

$v_0= 79m/s+13m/s= 92m/s$

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And we obtain:

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