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xxMikexx [17]
3 years ago
14

The mass of 2 cm3 of gold is 38.6 grams. find the density of the gold. Could you please help me??

Physics
1 answer:
Alex Ar [27]3 years ago
6 0

Density = mass/ volume (so here this is how you would solve the problem)

<span>D = 38.6 g/ 2 cm3 (first step)</span>

<span>D= 19.3 g/cm3  ( Do math and then you would get this)</span>

<span>
</span>

<span>Hope this helps!! :) </span>


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A positively charged objectwith a mass of 0.114 kg oscillates at the end of a spring, generating ELF (extremely low frequency) r
katen-ka-za [31]

Answer:

  • k = 167.33 N/m

Explanation:

  • The radio waves have a fixed relationship between the propagation speed (the speed of light in vacuum), the frequency and the wavelength, as follows:
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        where c= speed of light in vacuum = 3*10⁸ m/s, λ = wavelength =  

        4.92*10⁷ m.

        Solving for f, we get the frequency of the radio waves:

        f = 6.1 Hz

  • Now, from the Hooke's law, we know that the mass attached at the end of the spring oscillates with an angular frequency defined by  a fixed relationship between the spring constant k and the mass m, as follows:

       \omega_{o}^{2} =\frac{k}{m}  (1)

  • Now, we know that there exists a fixed relationship between the angular frequency and the frequency, as follows:

       \omega = 2*\pi *f (2)    

  • We also know that f in (2) is the same that we got for the radio waves, so replacing (2) in (1), and rearranging terms, we can solve for k, as follows:
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3 0
3 years ago
Two particles each of mass m and charge q are suspended by strings of length / from a common point. Find the angle e that each s
ozzi

Answer:

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

Explanation:

Let the length of the string is L.

Let T be the tension in the string.

Resolve the components of T.

As the charge q is in equilibrium.

T Sinθ = Fe       ..... (1)

T Cosθ = mg     .......(2)

Divide equation (1) by equation (2), we get

tan θ = Fe / mg

tan\theta =\frac{\frac{kq^{2}}{AB^{2}}}{mg}

tan\theta =\frac{\frac{kq^{2}}{4L^{2}Sin^{\theta }}}}{mg}

tan\theta =\frac{kq^{2}}{4L^{2}Sin^{2}\theta \times mg}

tan\theta\times Sin^{2}\theta =\frac{kq^{2}}{4L^{2}\times mg}

As θ is very small, so tanθ and Sinθ is equal to θ.

\theta ^{3} =\frac{kq^{2}}{4L^{2}\times mg}

\theta =\left (\frac{kq^{2}}{4L^{2}\times mg}  \right )^{\frac{1}{3}}

7 0
3 years ago
If we increase the force applied to an object and all other factors remain the same that amount of work will
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<span>If we increase the force applied to an object and all other factors remain the same that amount of work will

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Answer:

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Hence, The electric field strength is 6.6\times10^{5}\ V/m

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Answer:

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