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Natalka [10]
3 years ago
6

30) When you press a piano key, a hammer strikes a very tight cable to produce a clear tone. How does the cable that produces a

low note compare to one the produces a high note?
a) (A) The low note cable is thinner or shorter than the high note cable.
b) (B) The low note cable is longer or thicker than the high note cable.
c) (C) The cables are the same, but the hammer that produces the low note is heavier.
d) (D) The cables are the same, but the hammer that produces the low note is lighter.
Physics
1 answer:
skelet666 [1.2K]3 years ago
3 0

Answer:

Remember: in physics and music, frequency equates to pitch, and amplitude equates to loudness. Also, remember frequency is inversely proportional to wavelength- remember this!

Thus, (B) should be the correct answer, because, think about it,  if a note is higher, than the wavelength should be shorter so the length should also be shorter, but in the case of B, if the cable is longer or thicker then the wavelength gets longer. Therefore, going from low to high, the string should be getting shorter, due to the fact that as wavelength gets shorter, frequency gets higher.

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Concept Simulation 20.4 provides background for this problem and gives you the opportunity to verify your answer graphically. Ho
77julia77 [94]

Answer:

The time constant is 1.049.

Explanation:

Given that,

Charge q{t}= 0.65 q_{0}

We need to calculate the time constant

Using expression for charging in a RC circuit

q(t)=q_{0}[1-e^{-(\dfrac{t}{RC})}]

Where, \dfrac{t}{RC} = time constant

Put the value into the formula

0.65q_{0}=q_{0}[1-e^{-(\dfrac{t}{RC})}]

1-e^{-(\dfrac{t}{RC})}=0.65

e^{-(\dfrac{t}{RC})}=0.35

-\dfrac{t}{RC}=ln (0.35)

-\dfrac{t}{RC}=-1.049

\dfrac{t}{RC}=1.049

Hence, The time constant is 1.049.

6 0
3 years ago
It takes 130 J of work to compress a certain spring 0.10m. (a) What is the force constant of this spring? (b) To compress the sp
Ede4ka [16]

Explanation:

Given that,

Work done to stretch the spring, W = 130 J

Distance, x = 0.1 m

(a) We know that work done in stretching the spring is as follows :

W=\dfrac{1}{2}kx^2\\\\k=\dfrac{2W}{x^2}\\\\k=\dfrac{2\times 130}{(0.1)^2}\\\\k=26000\ N/m

(b) If additional distance is 0.1 m i.e. x = 0.1 + 0.1 = 0.2 m

So,

W=\dfrac{1}{2}kx^2\\\\W=\dfrac{1}{2}\times 26000\times 0.2^2\\\\W=520\ J

So, the new work is more than 130 J.

6 0
3 years ago
Brainliest if right
mixas84 [53]
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3 0
3 years ago
Which rule(s) is used for combining velocities?
DIA [1.3K]

Answer: B. Vector rules

Explanation:

4 0
3 years ago
The space shuttle usually orbited Earth at altitudes of around 300.0 km. 1) Determine the time for one orbit of the shuttle abou
777dan777 [17]

Answer:

Explanation:

distance of shuttle from centre of the earth = radius of the orbit

= 6300 + 300 = 6600 km

= 6600 x 10³

Formula of time period of the satellite

T = 2π R /v₀ , v₀ is orbital velocity

v₀ = √gR , ( if height  is small with respect to radius )

T = 2π R /√gR

= 2π√ R /√g

= 2 x 3.14 x √ 6600 x 10³ / √9.8

= 2 x 3.14 x 256.9 x 10 / 3.13

= 5154.41 s

= 5154.41 / 60 minutes

= 85.91 m

85.9 minutes.

2 ) No of sunrise per day = no of rotation per day

= 24 x 60 / 85.9

= 16.76

or 17 sunrises.

3 0
4 years ago
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