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ira [324]
3 years ago
7

The percent by which the fundamental frequency changed if the tension is increased by 30 percent is ? a)-20.04% b)-40.12% c)-30%

d)-14.02%
Physics
1 answer:
nika2105 [10]3 years ago
3 0

Answer:

Percentage increase in the fundamental frequency is

d)-14.02%

Explanation:

As we know that fundamental frequency of the wave in string is given as

f_o = \frac{1}{2L}\sqrt{\frac{T}{\mu}}

now it is given that tension is increased by 30%

so here we will have

T' = T(1 + 0.30)

T' = 1.30T

now new value of fundamental frequency is given as

f_o' = \frac{1}{2L}\sqrt{\frac{1.30T}{\mu}}

now we have

f_o' = \sqrt{1.3}f_o

so here percentage change in the fundamental frequency is given as

change = \frac{f_o' - f_o}{f_o} \times 100

% change = 14.02%

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Given that a 2.0 kg particle moving along the z-axis experiences the  force shown in a given figure.

Force is the product of mass and acceleration. While acceleration is the rate of change of velocity. Both the force and acceleration are vector quantities. They have both magnitude and direction.

If the particle's velocity is  3.0 m/s at x = 0 m, that mean that the particle experience change of velocity at point x = 0. Since the the force of the particle will change from negative sign to positive sign according to the given figure, we can therefore conclude that the particle will have a turning point at point x = 0.

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Why do objects move at constant speed
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An aircraft flying at a steady velocity of 70 m/s eastwards at a height of 800 m drops a package of supplies.
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The motion of the package can be described as the motion of a projectile,

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Reasons:

Velocity of the aircraft = 70 m/s

Direction of flight of the aircraft = Eastward

Height from which the aircraft drops the package, h = 800 m

a) The initial velocity of the package, \vec{v} = 70·i

b) The time it will take the package to reach the ground, <em>t</em>, is given by the formula;

\displaystyle h = \mathbf{\frac{1}{2} \cdot g \cdot t^2}

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

Therefore;

\displaystyle t = \mathbf{\sqrt{ \frac{2 \cdot h}{g} }}

Which gives;

\displaystyle t = \sqrt{ \frac{2 \times 800}{9.81} } \approx \mathbf{12.77}

The time it will take the package to reach the ground, t ≈ <u>12.77 seconds</u>

c) The vertical velocity just before the package reaches the ground, v_y, is given as follows;

v_y^2 = 2·g·h

Therefore;

v_y = √(2·g·h)

Which gives;

v_y = √(2 × 9.81 × 800) ≈ 125.28

v_y ≈ 125.28 m/s

Which gives; \vec{v} = 70·i - 125.28·j

Therefore, |v| = √(70² + (-125.28)²) ≈ 143.51

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d) The motion of the package that includes both horizontal and vertical motion is a projectile motion.

Therefore;

The path of the package is the path of a projectile, which is a <u>parabolic shape</u>.

e) As seen by someone on the aeroplane, the horizontal velocity will be

zero, therefore, the package will appear as accelerating <u>directly vertical</u>

downwards.

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Monochromatic light is incident on a pair of slits that are separated by 0.220 mm. The screen is 2.60 m away from the slits. (As
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Answer:

a

   \lambda = 1.667 nm

b

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Explanation:

From the question we are told that

   The distance of separation is d  =  0.220 \ mm  =  0.00022 \ m

    The  is distance of the screen from the slit is  D   =  2.60 \ m

    The distance between the central bright fringe and either of the adjacent bright   y  =  1.97 cm  =  1.97 *10^{-2}\ m

Generally  the condition for constructive interference is  

      d sin \tha(\theta ) =  n \lambda

From the question we are told that small-angle approximation is valid here.

So    sin (\theta ) = \theta

=>        d \theta  =  n \lambda

=>        \theta =  \frac{n *  \lambda }{d }

Here n is the order of maxima and the value is  n =  1 because we are considering the central bright fringe and either of the adjacent bright fringes

Generally the distance between the central bright fringe and either of the adjacent bright  is mathematically represented as

         y  =  D * sin (\theta )

From the question we are told that small-angle approximation is valid here.

So

       y  =  D * \theta

=>   \theta  =  \frac{ y}{D}

So

     \frac{n *  \lambda }{d } = \frac{y}{D}

     \lambda =\frac{d * y }{n * D}

substituting values

       \lambda =  \frac{0.00022 * 1.97*10^{-2} }{1 * 2.60 }

        \lambda = 1.667 *10^{-6}

        \lambda = 1.667 nm

In the b part of the question we are considering the next set of bright fringe so  n=  2

    Hence

     dsin (\theta ) =  n \lambda

    \theta  =  sin^{-1}[\frac{ n  *  \lambda }{d} ]

    \theta  =  sin^{-1}[\frac{ 2  *  1667 *10^{-9}}{ 0.00022} ]

    \theta  =  0.8681^o

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4 years ago
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