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2 years ago

How does changing the wavelength alter the frequency of the wave? & How does altering frequency alter pitch?

2 answers:

4 0

For any wave, the product of (wavelength) x (frequency) is always the same number ... it's the speed of the wave. So if you were able to change the wavelength somehow, the frequency would automatically change, in the opposite direction but by the SAME FACTOR, so that their product would stay the same.

If the frequency of a sound changes (and we're close enough to listen to it), our brain interprets that as a change of pitch, in the same direction as the frequency changed.

Rufina [12.5K]2 years ago

3 0

for a given type of wave in a given medium a larger frequency means a smaller wavelength

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The greatest height reported for a jump into an airbag is 99.4 m by stuntman Dan Koko. In 1948 he jumped from rest from the top

leva [86]

Answer:

44.1613858478 m/s

Explanation:

t = Time taken

u = Initial velocity = 0

v = Final velocity

s = Displacement = 99.4

a = Acceleration

g = Acceleration due to gravity = 9.81 m/s² = a

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 99.4+0^2}\\\Rightarrow v=44.1613858478\ m/s$

If air resistance was absent Dan Koko would strike the airbag at 44.1613858478 m/s

6 0

2 years ago

You are riding in a hot air balloon that, relative to the ground, has a velocity of 6.0 m/s due east. You see a hawk moving dire

charle [14.2K]

Answer:Velocity = 6.325m/s

Directional angle= 18.43°

Explanation:

Using Right angle triangle

Let Velocity of ballon&hawk be VHB represent the height of the triangle.

Let Velocity of balloon angle ground be VBG represent adjacent of the triangle.

Let Velocity of hawk and ground BE VHG represent the hypothesis.

Theta = opp/Adj= VHB/VBG

using pythagorean

VHG= SQRT(VHB^2+VBG^2)

VHG= sqrt(2^2+6^2)

VHG= sqrt(4+36)

VHG= 6.325m/s

Tan theta= 2/6

Tan theta =0.3333

Tan^-1 0.3333=18.43°

8 0

3 years ago

How are ocean ridges formed

Basile [38]

The answer is divergent boundaries.
I hope this helps you!

5 0

3 years ago

Read 2 more answers

When performing an.experiment similar to Millikan's oil drop, a student measured the following load magnitudes: 3.26x10 ^-19 C 5

Scilla [17]

Answer:

1.6 x 10⁻¹⁹ C

Explanation:

Let us arrange the charges in the ascending order and round them off as follows :-

1.53 x 10⁻¹⁹ C → 1.6x 10⁻¹⁹ C

3.26 x 10⁻¹⁹C → 3.2 x 10⁻¹⁹ C

4.66 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

5.09 x 10⁻¹⁹C → 4.8 x 10⁻¹⁹ C

6.39 x 10⁻¹⁹C → 6.4 x 10⁻¹⁹ C

The rounding off has been made to facilitate easy calculation to come to a conclusion and to accommodate error in measurement.

Here we observe that

2 nd charge is almost twice the first charge

3 rd and 4 th charges are almost 3 times the first charge

5 th charge is almost 4 times the first charge.

This result implies that 2 nd to 5 th charges are made by combination of the first charge ie if we take e as first charge , 2nd to 5 th charges can be written as 2e, 3e ,3e and 4e. Hence e is the minimum charge existing in nature and on electron this minimum charge of 1.6 x 10⁻¹⁹ C exists.

3 0

2 years ago

A 150 g baseball is traveling horizontally at 50 m/s. If the ball takes 20 ms to stop once it is in contact with the catcher’s g

Sliva [168]

To solve for force, you need to get the product of mass and acceleration.
F = ma

Your given is:
m = 150g
a = ?
v = 50 m/s
t = 20ms

As you can see, you do not have acceleration yet. But if you read the problem you can come up with the formula of acceleration.
Acceleration is the change in velocity over a period of time.

a = change in velocity/time

To get the change in velocity, you get the difference between the initial velocity and final velocity:

$a = \frac{vf-vi}{t}$

The ball was moving initially at a velocity of 50 m/s and it came to a stop. This is your clue. If a ball comes to a stop then that means that the final velocity of the ball is 0 m/s.

So we can put it into our formula now:

$a = \frac{0m/s-50m/s}{20ms}$

WAIT! As you can see, the units do not match. We have ms and s into our equation and that means you cannot proceed till they are the same. First we need to convert ms to s.

$20ms$ x $\frac{1s}{1000ms}$ = $\frac{20s}{1000} = 0.02s$

So your new time is 0.02s. Now we put this time into the formula:

$a = \frac{0m/s-50m/s}{0.02s}$
$a = \frac{-50m/s}{0.02s} = -2,500 m/ s^{2}$

As you can see our acceleration is a negative value, this indicates that it decelerated or slowed down which makes sense because it was brought to a stop.

So now we have our acceleration. Now using this, we can get our force.

$F= ma$

Before we start doing this, you need to take note that the unit of force is N, but when you expand it, it is $kg.m/ s^{2}$ but as you can see our mass given is in grams. So again, before you put them into the equation we need to change it into kg first.

$150g = \frac{1kg}{1,000g} = \frac{150kg}{1,000} = 0.150kg$

Our new mass is 0.150kg.

To make things clearer, let us write down all our new values:

m = 0.150kg
a = -2,500 $m/ s^{2}$

Now that all our units match, we can put that into our formula:

$F= ma$
$F= (0.150kg)(-2,500m/s^{2})$
$F = -375kg.m/ s^{2} or -375N$

The value again is negative because it is going against the initial direction of the ball. But if your instructor just wants to get the value of force or the magnitude of the force, just disregard the sign.

4 0

3 years ago