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alekssr [168]
3 years ago
7

Neutron stars consist only of neutrons and have unbelievably high densities. a typical mass and radius for a neutron star might

be 9.9 x 1028 kg and 1.9 x 103 m. (a) find the density of such a star. (b) if a dime (v = 2.0 x 10-7 m3) were made from this material, how much would it weight (in pounds)?
Physics
1 answer:
Tanya [424]3 years ago
4 0
<span>Density is 3.4x10^18 kg/m^3 Dime weighs 1.5x10^12 pounds The definition of density is simply mass per volume. So let's divide the mass of the neutron star by its volume. First, we need to determine the volume. Assuming the neutron star is a sphere, the volume will be 4/3 pi r^3, so 4/3 pi 1.9x10^3 = 4/3 pi 6.859x10^3 m^3 = 2.873x10^10 m^3 Now divide the mass by the volume 9.9x10^28 kg / 2.873x10^10 m^3 = 3.44588x10^18 kg/m^3 Since we only have 2 significant digits in our data, round to 2 significant digits, giving 3.4x10^18 kg/m^3 Now to figure out how much the dime weighs, just multiply by the volume of the dime. 3.4x10^18 kg/m^3 * 2.0x10^-7 m^3 = 6.8x10^11 kg And to convert from kg to lbs, multiply by 2.20462, so 6.8x10^11 kg * 2.20462 lb/kg = 1.5x10^12 lb</span>
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Answer:

Coulomb's law states that the electrical force between two charged objects is directly proportional to the product of the quantity of charge on the objects and inversely proportional to the square of the separation distance between the two objects.

Explanation:

4 0
3 years ago
A bullet with a mass m b = 11.9 mb=11.9 g is fired into a block of wood at velocity v b = 261 m/s. vb=261 m/s. The block is atta
fredd [130]

Answer:

0.372 kg

Explanation:

The collision between the bullet and the block is inelastic, so only the total momentum of the system is conserved. So we can write:

mu=(M+m)v (1)

where

m=11.9 g = 11.9\cdot 10^{-3}kg is the mass of the bullet

u=261 m/s is the initial velocity of the bullet

M is the mass of the block

v is the velocity at which the bullet and the block travels after the collision

We also know that the block is attached to a spring, and that the surface over which the block slides after the collision is frictionless. This means that the energy is conserved: so, the total kinetic energy of the block+bullet system just after the collision will entirely convert into elastic potential energy of the spring when the system comes to rest. So we can write

\frac{1}{2}(M+m)v^2 = \frac{1}{2}kx^2 (2)

where

k = 205 N/m is the spring constant

x = 35.0 cm = 0.35 m is the compression of the spring

From eq(1) we get

v=\frac{mu}{M+m}

And substituting into eq(2), we can solve to find the mass of the block:

(M+m) \frac{(mu)^2}{(M+m)^2}=kx^2\\\frac{(mu)^2}{M+m}=kx^2\\M+m=\frac{(mu)^2}{kx^2}\\M=\frac{(mu)^2}{kx^2}-m=\frac{(11.9\cdot 10^{-3}\cdot 261)^2}{(205)(0.35)^2}-11.9\cdot 10^{-3}=0.372 kg

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8_murik_8 [283]
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The period of a carrier wave is T=0.01 seconds. Determine the frequency and wavelength of the carrier wave. a. f=10 Hz, λ=3E8 me
makvit [3.9K]

Explanation:

It is given that,

The period of the carrier wave, T = 0.01 s

Let f and \lambda are frequency and the wavelength of the wave respectively. The relationship between the time period and the frequency is given by :

f=\dfrac{1}{T}

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f = 100 Hz

The wavelength of a wave is given by :

\lambda=\dfrac{c}{f}

\lambda=\dfrac{3\times 10^8}{100}

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