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3 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]3 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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A ski jumper has a mass of 59.6 kg. She is moving with a speed of 23.4 m/s at a height of 44.6 meters above the ground. Determin

Daniel [21]

Hello!

Use the formula:

M = k * p

Data:

M = Mechanic energy

k = Kinetic energy

p = Potencial energy

Descomposing:

M = (0,5*mv²) + (mgh)

Replacing:

M = (0,5 * 59,6 kg * (23,4 m/s)²) + (59,6 kg * 9,81 m/s² * 44,6 m)

M = 16317,28 J + 26076,54 J

M = 42393,82 J

The mechanic energy is <u>42393,82 Joules.</u>

7 0

3 years ago

I need an answer for this plzz!!<br>number 2 <br>anybody can help ??

Anuta_ua [19.1K]

2.1) (i) W = mg downwards
(ii) N = R = Normal Reaction from the ground upwards
(iii) Fe = Force of engine towards the right
(iv) f = friction towards the left
(v) ma = Constant acceleration towards right.
2.2.1)
v = 25 m/s
u = 0 m/s
∆v = v - u = (25 - 0) m/s = 25 m/s
x = X
∆t = 50 s
$a \: = \: \frac{dv}{dt} \: = \: \frac{25 \: \frac{m}{s} }{50 \: s} \: = \: 0.5 \: \frac{m}{ {s}^{2} }$
a = 0.5 m/s².
2.2.2)
F = ma = 900 kg × 0.5 m/s² = 450 N.
2.2.3)
$2ax \: = \: {v}^{2} \: - \: {u}^{2}$
$x \: = \: \frac{ {v}^{2} \: - \: {u}^{2} }{2a} \: = \: \frac{{(25 \: \frac{m}{s})}^{2} \: - \: {(0 \: \frac{m}{s} )}^{2} }{2 \: \times \: 0.5 \: \frac{m}{ {s}^{2} } } \: = \: 625 \: m$
2.3)
Fe = f + ma
Fe - f = ma
For velocity to be constant,
a should be 0, or, a = 0,
Fe = f = 270 N
2.4.1)
v = 0
u = 25 m/s
a = -0.5 m/s²
v = u + at
t = -u/a = -(25)/(-0.5) = 50 s.
2.4.2)
x = -625/(2×(-0.5)) = 625 m.

8 0

3 years ago

Violet light (410 nm) and red light

Leokris [45]

Answer:10.03

Explanation:acellus

7 0

2 years ago

1.Ben is pushing a block up a ramp that is 3 meters high and 6 meters long. The block has a force of 49N. It is taking Ben 37 Ne

inna [77]

Idrk sorry... i’m doing this bc it’s asking me to. wish i could help you guys . i really do !

5 0

3 years ago

If a nearsighted person has a far point df that is 3.50m from the eye, what is the focal length f1 of the contact lenses that th

olga55 [171]

Answer:

f1= -350cm or -3.5m

f2= 22.1cm or 0.221m

Explanation:

A person is nearsighted when the person's far point is less than infinity. A diverging lens is normally used to correct this eye defect. A diverging lens has a negative focal length as seen in the solution attached.

Farsightedness is when a person's near point is farther than 25cm. This eye defect is corrected using a converging lens. The focal length of a converging lens is positive. This is evident in the solution attached. The near point is also referred to as the least distance of distinct vision.

3 0

3 years ago