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levacccp [35]
2 years ago
14

Need help with this! just scienceeeeeee

Physics
1 answer:
LiRa [457]2 years ago
8 0

Answer:

7)  λ = 0.5 m,  8)  f = 4.8 10¹⁴ Hz

Explanation:

The speed of an electromagnetic wave is

          c = λ f

where c is the speed of light in vacuum c = 3 10⁸ m / s

7) indicate the frequency f = 6.0 10⁸ Hz

we do not know the wavelength

         λ = c / f

       

we calculate

        λ = 3 10⁸ / 6.0 10⁸

        λ = 0.5 m

8) indicate the wavelength  λ = 6.25 10-7 m

we do not know the frequency

         f = c / λ

we calculate

        f = 3 10⁸ / 6.25 10⁻⁷

        f = 0.48 10¹⁵ Hz

        f = 4.8 10¹⁴ Hz

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A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that
astraxan [27]

Answer:

A) k=34867.3384\ N.m^{-1}

B) \omega'\approx84\ Hz

Explanation:

Given:

mass of car, m=1380\ kg

A)

frequency of spring oscillation, f=1.6\ Hz

We knkow the formula for spring oscillation frequency:

\omega=2\pi.f

\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f

\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6

k_{eq}=139469.3537\ N.m^{-1}

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

k=\frac{k_{eq}}{4}

k=\frac{139469.3537}{4}

k=34867.3384\ N.m^{-1}

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }

\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }

\omega'\approx84\ Hz

7 0
3 years ago
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