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Vesna [10]
3 years ago
14

When a certain string is clamped at both ends, the lowest four resonant frequencies are 50, 100, 150, and 200 Hz. When the strin

g is also clamped at its midpoint, the lowest four resonant frequencies are __________.
Physics
1 answer:
mario62 [17]3 years ago
4 0

Answer:

Explanation:

Given

Lowest four resonance frequencies are given with magnitude

50,100,150 and 200 Hz

The frequency of vibrating string is given by

f=\frac{n}{2L}\sqrt{\frac{T}{\mu }}

where n=1,2,3 or ...n

L=Length of string

T=Tension

\mu =Mass per unit length

When string is clamped at mid-point

Effecting length becomes L'=0.5 L

Thus new Frequency becomes

f' =\frac{n}{L}\sqrt{\frac{T}{\mu }}

i.e. New frequency is double of old

so new lowest four resonant frequencies are 100,200,300 and 400 Hz      

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Explanation:AYEEE

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A type O star is likely to appear _____. yellow, blue, green or red
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A type O star is likely to appear blue. 
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Which of the following is an example of a chemical change?
ss7ja [257]

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Burning Paper

Explanation:

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2 years ago
A projectile is launched diagonally into the air and has a hang time of 24.5 seconds. Approximately how much time is required fo
Rasek [7]

Answer:

t=12.25\ seconds

Explanation:

<u>Diagonal Launch </u>

It's referred to as a situation where an object is thrown in free air forming an angle with the horizontal. The object then describes a known path called a parabola, where there are x and y components of the speed, displacement, and acceleration.

The object will eventually reach its maximum height (apex) and then it will return to the height from which it was launched. The equation for the height at any time t is

x=v_ocos\theta t

\displaystyle y=y_o+v_osin\theta \ t-\frac{gt^2}{2}

Where vo is the magnitude of the initial velocity, \theta is the angle, t is the time and g is the acceleration of gravity

The maximum height the object can reach can be computed as

\displaystyle t=\frac{v_osin\theta}{g}

There are two times where the value of y is y_o when t=0 (at launching time) and when it goes back to the same level. We need to find that time t by making y=y_o

\displaystyle y_o=y_o+v_osin\theta\ t-\frac{gt^2}{2}

Removing y_o and dividing by t (t different of zero)

\displaystyle 0=v_osin\theta-\frac{gt}{2}

Then we find the total flight as

\displaystyle t=\frac{2v_osin\theta}{g}

We can easily note the total time (hang time) is twice the maximum (apex) time, so the required time is

\boxed{t=24.5/2=12.25\ seconds}

4 0
3 years ago
Based on experimental observations, the acceleration of a particle is defined by the relation a = -(0.1 + sin x/b), where a and
Fiesta28 [93]

Answer:

Velocity,v = 0.323 m/s

Explanation:

The acceleration of a particle is given by :

a=-(0.1+sin\dfrac{x}{b})

b = 0.8 m when x = 0

Since, a=v\dfrac{dv}{dx}  

v\dfrac{dv}{dx}=-(0.1+sin\dfrac{x}{b})  

\int{v.dv}=\int{-(0.1+sin\dfrac{x}{b})}.dx

\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]+c

At x = 0, v = 1 m/s

\dfrac{1}{2}=0.8+c

c=-0.3

\dfrac{v^2}{2}=-[0.1x-0.8cos\dfrac{x}{0.8}]-0.3

At x = -1 m

\dfrac{v^2}{2}=-0.1(-1)+0.8cos\dfrac{(-1)}{0.8}-0.3

{v^2}=0.1045

v = 0.323 m/s

So, the velocity of the particle is 0.323 m/s. Hence, this is the required solution.

8 0
3 years ago
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