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4 years ago

Within a neutron star, the material at the center has a density of 1.0 x 10 18 kg/m 3 . If a small

Physics

1 answer:

umka2103 [35]4 years ago

7 0

Density of the neutron star is given as

$\rho = 1 \time 10^{18} kg/m^3$

Radius of the star is also given as

$R = 1 \times 10^{-5} m$

now in order to find the mass of star we can assume that star is spherical in shape

so here we can use

$V = \frac{4}{3}\pi R^3$

$V = \frac{4}{3}\pi (1 \times 10^{-5})^3$

$V = 4.2 \times 10^{-15} m^3$

now mass of the star will be given as

$M = \rho V = 4.2 \times 10^{-15} \times 1 \times 10^{18}$

$M = 4.2 \times 10^3 kg$

so weight of the star on the earth will be

$W = Mg$

$W = 4.2 \times 10^3 \times 9.8$

$W = 4.1 \times 10^4 N$

so weight on the earth is given by above equation

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4 years ago

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3 years ago

In which medium does light travel faster: one with a critical angle of 27.0° or one with a critical angle of 32.0°? Explain. (Fo

Eddi Din [679]

Answer:

Among those two medium, light would travel faster in the one with a reflection angle of $32^{\circ}$ (when light enters from the air.)

Explanation:

Let $v_{1}$ denote the speed of light in the first medium. Let $v_{\text{air}}$ denote the speed of light in the air. Assume that the light entered the boundary at an angle of $\theta_{1}$ to the normal and exited with an angle of $\theta_{\text{air}}$ . By Snell's Law, the sine of $\theta_{1}\!$ and $\theta_{\text{air}}\!$ would be proportional to the speed of light in the corresponding medium. In other words:

$\displaystyle \frac{v_{1}}{v_{\text{air}}} = \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})}$ .

When light enters a boundary at the critical angle $\theta_{c}$ , total internal reflection would happen. It would appear as if the angle of refraction is now $90^{\circ}$ . (in this case, $\theta_{\text{air}} = 90^{\circ}$ .)

Substitute this value into the Snell's Law equation:

$\begin{aligned}\frac{v_{1}}{v_{\text{air}}} &= \frac{\sin(\theta_{1})}{\sin(\theta_{\text{air}})} \\ &= \frac{\sin(\theta_{c})}{\sin(90^{\circ})} \\ &= \sin(\theta_{c})\end{aligned}$ .

Rearrange to obtain an expression for the speed of light in the first medium:

$v_{1} = v_{\text{air}} \cdot \sin(\theta_{1})$ .

The speed of light in a medium (with the speed of light slower than that in the air) would be proportional to the critical angle at the boundary between this medium and the air.

For $0 < \theta < 90^{\circ}$ , $\sin(\theta)$ is monotonically increasing with respect to $\theta$ . In other words, for $\!\theta$ in that range, the value of $\sin(\theta)\!$ increases as the value of $\theta\!$ increases.

Therefore, compared to the medium in this question with $\theta_{c} = 27^{\circ}$ , the medium with the larger critical angle $\theta_{c} = 32^{\circ}$ would have a larger $\sin(\theta_{c})$ . such that light would travel faster in that medium.