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11Alexandr11 [23.1K]
3 years ago
12

Compare to a low wavelength sound wave, a high wavelength sound wave Select all that apply Group of answer choices travels slowe

r has a smaller frequency travels at the same speed has a higher frequency travels faster has the same frequency
Physics
1 answer:
viktelen [127]3 years ago
5 0

Answer:

B. has a smaller frequency

C. travels at the same speed  

Explanation:

The wording of the question is a bit confusing, it should be short/long for wavelength and low/high for frequency. I assume low wavelength mean short wavelength.

All sound wave travel with the same velocity(343m/s) so wavelength doesn't influence its speed at all. It won't be faster or slower, it will have the same speed.

Velocity is a product of wavelength and frequency. So, a long-wavelength sound wave should have a lower frequency.

The option should be:

A. travels slower -->false

B. has a smaller frequency -->true

C. travels at the same speed  --->true

D. has a higher frequency  --->false

E. travels faster has the same frequency  --->false

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A chemical reaction that has the general formula of AB + C → CB + A is best classified as a reaction.
MissTica

Answer:

Single replacement

Explanation:

A reaction in which one element replaces a similar element is called single replacement.  In this case, C is replacing A.

7 0
3 years ago
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In order to increase the amount of work done, we need to:
Nadusha1986 [10]
Option (D) is the correct one.


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8 0
3 years ago
A horizontal force of 100 N is required to push a 50 kg crate across a factory floor at a constant speed. What is the accelerati
luda_lava [24]

Answer:

a = 2m/s^2

Explanation:

Force (F) = 100 N

Mass (m) = 50 kg

Here,

F = m×a

100 = 50 × a

a = 100÷50

a = 2m/s^2

Thus, the acceleration on the cart is a = 2m/s^2

-TheUnknownScientist

6 0
3 years ago
The earth's radius is 6.37×106m; it rotates once every 24 hours.What is the speed of a point on the earth's surface located at 3
bagirrra123 [75]

Answer:

v = 120 m/s

Explanation:

We are given;

earth's radius; r = 6.37 × 10^(6) m

Angular speed; ω = 2π/(24 × 3600) = 7.27 × 10^(-5) rad/s

Now, we want to find the speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator.

The angle will be;

θ = ¾ × 90

θ = 67.5

¾ is multiplied by 90° because the angular distance from the pole is 90 degrees.

The speed of a point on the earth's surface located at 3/4 of the length of the arc between the equator and the pole, measured from equator will be:

v = r(cos θ) × ω

v = 6.37 × 10^(6) × cos 67.5 × 7.27 × 10^(-5)

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8 0
3 years ago
Ml(d^2θ/dt^2) =-mgθ
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The equation of motion of a pendulum is:

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\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

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This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
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