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Goryan [66]
3 years ago
15

Please answer these!

Physics
1 answer:
solong [7]3 years ago
8 0

6. f = 128.62 s^{-1} , T= 0.0077775 s

7. f  = 2.2 * 10^{4} s^{-1}, T = 4.545 * 10^{-5} s

8. 32.64 s^{-1}

9. 3.29 * 10^{14} s^{-1}

Explanation:

Step 1:

6.

For light and sound v = fλ

where v represents the velocity

f represents the frequency

λ represents the wavelength

λ = 2.69 m

v = 346 m/s

f = v/λ = 346/2.69 = 128.62 s^{-1}

Time period is the reciprocal of frequency

T = 1/128.62 = 0.0077775 s

Step 2:

7.

λ = 110 cm = 1.1 m

v = 2.42*10^{4} m/s

f = 2.42*10^{4}/1.1 = 2.2 * 10^{4} s^{-1}

T = 1/(2.2*10^{4}) = 4.545 * 10^{-5} s

Step 3:

8.

λ = 10.6 m

v = 346 m/s

f = v/λ = 346/10.6 = 32.64 s^{-1}

Step 4:

9.

λ = 5.89 * 10^{-7} m

v = 1.94 * 10^{8} m/s

f = v/λ = 3.29 * 10^{14} s^{-1}

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Find the acceleration of an 800-kg car that has a net force of 4,000 N acting upon it.
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While traveling along a highway a driver slows from 24 m/s to 15 m/s in 12 seconds. What is the automobile's acceleration? (Reme
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A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm
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A) 2.75 m/s  B) 0.1911 m    C) 0.109 s

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mass of block = M =0.1 kg

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amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = \sqrt\frac{k}{M}

                                                       = \sqrt\frac{21}{0.1}

                                                       = 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic  Motion is given as:

                                x = A sin (\omega t + \phi)\\

Differentiating this to find speed of the block immediately before the collision:

                    v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\

As bullet strikes at equilibrium position so,

                                  φ = 0

                                   t= 2nπ

                             ⇒ cos (ω₀t + φ) = 1

                             ⇒ v= A\omega_{o}

                                       v=(.19)(14.49)\\v= 2.75 ms^{-1}

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

                              x= Bsin(\omega t + \phi)

To find B, consider law of conservation of energy

K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2}  \\P.E = \frac{1}{2} kB^{2}

\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}

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T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec

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