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ANEK [815]
3 years ago
12

What are (a) the lowest frequency, (b) the second lowest frequency, and (c) the third lowest frequency for standing waves on a w

ire that is 10.9 m long, has a mass of 55.8 g, and is stretched under a tension of 253 n?
Physics
1 answer:
Delvig [45]3 years ago
8 0
(a) The lowest frequency (called fundamental frequency) of a wire stretched under a tension T is given by
f_1 =  \frac{1}{2L} \sqrt{ \frac{T}{m/L} }
where
L is the wire length
T is the tension
m is the wire mass

In our problem, L=10.9 m, m=55.8 g=0.0558 kg and T=253 N, therefore the fundamental frequency of the wire is
T= \frac{1}{2L} \sqrt{ \frac{T}{m/L} }= \frac{1}{2 \cdot 10.9 m} \sqrt{ \frac{253 N}{0.0558 kg/10.9 m} }=    10.2 Hz

b) The frequency of the nth-harmonic for a standing wave in a wire is given by
f_n = n f_1
where n is the order of the harmonic and f1 is the fundamental frequency. If we use n=2, we find the second lowest frequency of the wire:
f_2 = 2 f_1 = 2 \cdot 10.2 Hz=20.4 Hz

c) Similarly, the third lowest frequency (third harmonic) is given by
f_3 = 3 f_1 = 3 \cdot 10.2 Hz = 30.6 Hz

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Answer:

A.) The arrow`s range is 624,996 m

B.) The arrow`s range is 846.887 m, when the horse is galloping

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.

By the way, the equations that we use for the X axis will be from URM, and those for the Y axis wiil be from UAM.

<u>Equations</u>

X axis:

X=v_{ox}*t

v_{0x} =v_0cos(\alpha)

Y axis:

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

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

Caculate the segcond degree equation to obtain the two possible values for t:

t_1= 10.18 \\t_2= -0.04046

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

Caculate now:

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

Using the same procedure that item A, caculate X

First, we need to know the new time

0=2.4m+92m/s*sin(39)t-(1/2*9.81m/s^2)t^2

And we obtain:

t_1=11.845s\\t_2=-0.041s

One more time, we take the positive time: t_1=11.845s

Finally:

X=92m/s *cos(39)*11.845s=846.887 m

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3 years ago
Which of the following statements is always true?
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Explanation:

An unbalanced force (net force) acting on an object changes its speed and/or direction of motion. An unbalanced force is an unopposed force that causes a change in motion.

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A rocket, initially at rest on the ground, accelerates straight upward from rest with constant (net) acceleration 29.4 m/s2 m /
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Answer:

The maximum height is 2881.2 m.

Explanation:

Given that,

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Time = 7.00 s

We need to calculate the distance

Using equation of motion

s=ut+\dfrac{1}{2}at^2

Put the value into the formula

s=0+\dfrac{1}{2}\times29.4\times7^2

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Using formula of velocity

v=a\times t

Put the value into the formula

v=29.4\times7

v=205.8\ m/s

We need to calculate the height

Using formula of height

H=\dfrac{v^2}{2g}

Put the value into the formula

H=\dfrac{(205.8)^2}{2\times9.8}

H=2160.9\ m

We need to calculate the maximum height

Using formula for maximum height

H'=H+s

Put the value into the formula

H'=2160.9+720.3

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Answer:

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Explanation:

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v=v0 - gt

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in 2 seconds:

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5 0
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