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3 years ago

The distance between adjacent nodes in a standing wave pattern is 25.0 cm. What is the

Physics

1 answer:

Novay_Z [31]3 years ago

4 0

Answer:

Answer:

Speed of the wave in the string will be 3.2 m/sec

Explanation:

We have given frequency in the string fixed at both ends is 80 Hz

Distance between adjacent antipodes is 20 cm

We know that distance between two adjacent anti nodes is equal to half of the wavelength

So \frac{\lambda }{2}=20cm

=20cm

\lambda =40cmλ=40cm

We have to find the speed of the wave in the string

Speed is equal to v=\lambda f=0.04\times 80=3.2m/secv=λf=0.04×80=3.2m/sec

So speed of the wave in the string will be 3.2 m/sec

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When a substance is heated the average kinetic energy of molecules increases and they start moving with an increased speed.As a result between mean separation between the molecules also increases

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The humpback whale migrates across areas of the Pacific Ocean, from Hawaii and California in the winter to Alaska in the summer.

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Read 2 more answers

A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

guapka [62]

<h2>Answer: (a)t=0.553s, (b)x=110.656m</h2>

Explanation:

This situation is a good example of the projectile motion or parabolic motion, in which the travel of the bullet has two components: x-component and y-component. Being their main equations as follows:

x-component:

$x=V_{o}cos\theta t$ (1)

Where:

$V_{o}=200m/s$ is the bullet's initial speed

$\theta=0$ because we are told the bullet is shot horizontally

$t$ is the time since the bullet is shot until it hits the ground

y-component:

$y=y_{o}+V_{o}sin\theta t-\frac{gt^{2}}{2}$ (2)

Where:

$y_{o}=1.5m$ is the initial height of the bullet

$y=0$ is the final height of the bullet (when it finally hits the ground)

$g=9.8m/s^{2}$ is the acceleration due gravity

Now, for the first part of this problem, the time the bullet elapsed traveling, we will use equation (2) with the conditions given above:

$0=1.5m+200m/s.sin(0) t-\frac{9.8m/s^{2}.t^{2}}{2}$ (3)

$0=1.5m-\frac{9.8m/s^{2}.t^{2}}{2}$ (4)

Finding $t$ :

$t=\sqrt{\frac{1.5m(2)}{9.8m/s^{2}}}$ (5)

Then we have the time elapsed before the bullet hits the ground:

$t=0.553s$ (6)

For the second part of this problem, we are asked to find how far does the bullet traveled horizontally. This means we have to use the equation (1) related to the x-component:

$x=V_{o}cos\theta t$ (1)

Substituting the knonw values and the value of $t$ found in (6):

$x=200m/s.cos(0)(0.553s)$ (7)

$x=200m/s(0.553s)$ (8)

Finally:

$x=110.656m$

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4 years ago

How does the size of the sun compare to the sizes of other stars?

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3 years ago

a uniform rod of length 1.5m is placed over a wedge at 0.5m from one end .a force of 100 N is applied at its one end near the we

andreev551 [17]

Explanation:

The rod is uniform, so the center of gravity is at the center, or 0.75 m from the end. The wedge is 0.5 m from the end, so the center is 0.25 m from the wedge.

Sum the torques about the wedge (it may help to draw a diagram first). Take counterclockwise to be positive.

∑τ = Iα

W (0.25 m) − (100 N) (0.50 m) = 0

W = 200 N

Sum the forces in the y direction.

∑F = ma

F − 100 N − 200 N = 0

F = 300 N

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