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tia_tia [17]
3 years ago
12

A certain car traveling in a straight line path has a velocity of +10.0 m/s at some instant. after 3.00 s, its velocity is +6.00

m/s. what is the average acceleration of the car during this time interval?
Physics
1 answer:
makvit [3.9K]3 years ago
3 0
The formula is:
v = v o + a t
6 = 10 + 3 * a
3 a = 10 - 6
a = 4 : 3
a =  - 1.33 m/s² ( because the car slows down )
Answer: The average acceleration of the car is - 1.33 m/s²
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E fundamental frequency of an open organ pipe corresponds to the middle c (261.6 hz on the chromatic musical scale). the third r
luda_lava [24]

The wavelength of the third resonance of the closed organ pipe is equal to the ratio between the speed of sound and the frequency of the 3rd harmonic:

\lambda_3 = \frac{c}{f_3}=\frac{343 m/s}{261.6 Hz} =1.31 m

The relationship between length of a closed pipe and wavelength of the standing waves inside is:

L=\frac{n}{4}\lambda_n

where n is the number of the harmonic. In this case, n=3, so the length of the pipe is

L=\frac{3}{4}(1.31 m)=0.98 m

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Use the velocity vs time graph to analyze the motion of the object.
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Explanation:

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an airplane travels 2100 km at a speed of 1000 km/h for the next 1300 km. whatis the average speed for this plane trip?
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4 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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