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3 years ago

9

In a longitudinal wave, particles of the medium vibrate in a direction that is perpendicular that the wave travels? True or Fals

e plzz answer right will give brainliest

2 answers:

Korolek [52]3 years ago

7 0

It’s false they travel parallel to the direction of the wave

vodomira [7]3 years ago

3 0

A transverse wave is a wave in which particles of the medium vibrate perpendicular to the direction that the wave travels.

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If a pickup is placed 16.25 cm from one of the fixed ends of a 65.00-cm-long string, which of the harmonics from n=1 to n=12 wil

Lina20 [59]

Answer:

The answer to this question can be defined as follows:

Explanation:

Therefore the 4th harmonicas its node is right and over the pickup so, can not be captured from 16.25, which is 1:4 out of 65. Normally, it's only conceptual for the certain harmonic, this will be low, would still be heard by the catcher.

Instead, every harmonic node has maximum fractions along its string; the very first node is the complete string length and the second node is half a mile to the third node, which is one-third up and so on.

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3 years ago

When performing the spike the ball should be hit with the ?

VMariaS [17]

Answer:

When you are performing spike it's most effective to strike the ball from the right or left side at a sharp downward angle. Whether you are spiking the ball from the right or left front position, position yourself behind the 10-foot line (attack line), which is the line that is about four steps away from the net.

5 0

3 years ago

A slit has a width of W1 = 4.4 × 10-6 m. When light with a wavelength of λ1 = 487 nm passes through this slit, the width of the

Vitek1552 [10]

Answer:

The width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

Explanation:

To solve the problem it is necessary to apply the concepts related to interference from two sources. Destructive interference produces the dark fringes. Dark fringes in the diffraction pattern of a single slit are found at angles θ for which

$w sin\theta = m\lambda$

Where,

w = width

$\lambda =$ wavelength

m is an integer, m = 1, 2, 3...

We here know that as $sin\theta$ as w are constant, then

$\frac{w_1}{\lambda_1} = \frac{w_2}{\lambda_2}$

We need to find $w_2$ , then

$w_2 = \frac{w_1}{\lambda_1}\lambda_2$

Replacing with our values:

$w_2 = \frac{4.4*10^{-6}}{487}496$

$w_2 = 4.48*10^{-6}m$

Therefore the width of the central bright fringe on the screen is observed to be unchanged is $4.48*10^{-6}m$

3 0

3 years ago

A beaker of vegetable oil contains a beam of light that is aimed at a surface at an angle of 34 degrees as shown. If the index o

OverLord2011 [107]

Answer:

Angle of reflection of light is 34 degree

Explanation:

As per law of reflection of light we know that

angle of incidence of light = angle of reflection of light

So here we know that

angle of incidence on the surface of oil is given as

$\theta_i = 34 degree$

so we know that

$\theta_i = \theta_r$

so here we can say that reflection angle of light will be same as angle of incidence

$\theta_r = 34 degree$

8 0

3 years ago

You are traveling on an airplane. The velocity of the plane with respect to the air is 110.0 m/s due east. The velocity of the a

Mice21 [21]

1. Vpa = 180m/s. @ 0 deg.
Vag = 40m/s @ 120 deg,CCW.

<span> Vpg = Vpa + Vag,
Vpg = (180 + 40cos120) + i40sin120,
Vpg = 160 + i34.64,
Vpg=sqrt((160)^2 + (34.64)^2)=163.7m/s.
</span>
<span>2. tanA = Y / X = 34.64 / 160 = 0.2165,
A = 12.2 deg,CCW. = 12.2deg. North of East. </span>

3. 1 hr = 3600s. <span>d = Vt = 163.7m/s * 3600s = 589,320m.

hope this helps</span>

8 0

3 years ago