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o-na [289]
3 years ago
8

Because of barriers and obstructions, you can see only a tiny fraction of a standing wave on a string. You do not know, for inst

ance, how long the string is. All you know is that the speed of waves on the string is 100 m/s and that the distance is 10 cm. Can you determine the frequency of the standing wave? 1. Sure, the frequency is 1000 Hz. 2. Sure, the frequency is 500 Hz.
Physics
2 answers:
gregori [183]3 years ago
6 0

Answer:

500 Hz

Explanation:

Given that,

speed of waves on the string is 100 m/s

distance is 10 cm

so given v is 100 m/s and L is 10 cm

frequency = speed (wave on string) / 2L

so the frequency is 100/2*10*10^-2m

= 1000/2

=500 Hz  

so the answer is 500 Hz

frequency = speed (wave on string) / 2L

so given v is 100 m/s and L is 10 cm  

so the frequency is 100/2*10*10^-2m

= 1000/2

=500 Hz  

so the answer is 500 Hz

chubhunter [2.5K]3 years ago
4 0

Answer:

(1) Sure, the frequency is 1000 Hz.

Explanation:

Frequency = wave speed ÷ wave distance

wave speed = 100 m/s

wave distance = 10 cm = 10/100 = 0.1 m

Frequency = 100 ÷ 0.1 = 1000 Hz

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Explanation:

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Difference between inverted and upright microscope
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3 years ago
Identify the breed of goat in the picture.
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Answer:

look it up im not a sheaperd sorry

Explanation:

4 0
2 years ago
Read 2 more answers
Size AA modern NiMH rechargeable batteries have a maximum capacity of 2050 mAh and the emf equal to 1.20 V. A flashlight has thr
Nataly_w [17]

Answer:

Explanation:

The same current flows through each part of a series circuit. The total resistance of a series circuit is equal to the sum of individual resistances. Voltage applied to a series circuit is equal to the sum of the individual voltage drops.

I = 0.33 A

= 330 mA

Capacity, P = I × t

= 2050/330

= 6.21 hours

Time, t = 6.21 hours.

6 0
3 years ago
the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter
BartSMP [9]

Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

        PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I                           0                0.10

C                         +S               +2S

E                          S              0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

4 0
3 years ago
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