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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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3 years ago
A baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg. What’s the baseballs velocity?
Mashcka [7]

Answer:

<em>v=40 m/s south</em>

Explanation:

<u>Momentum </u>

It's a physical magnitude that measures the product of the mass by the velocity of a particle. Its units in the International System is kg.m/s and the formula is

p=m.v

Where m is the mass and v the velocity of the particle. If we wanted to solve for v, we have

\displaystyle v=\frac{p}{m}

The baseball has a momentum of 6.0 kg.m/s south and mass of 0.15kg, thus

\displaystyle v=\frac{6}{0.15}=40\ m/s

The velocity is directed to the south

3 0
3 years ago
A 4.00 m long, massless beam rests horizontally on a support 3.00 m from the left
Rus_ich [418]

If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Given the data in the question;

  • Length of the massless beam;L = 4.00m
  • Distance of support from the left end; x = 3.00m
  • First mass; m1 = 31.3 kg
  • Distance of beam from  the left end( m₁ is attached to ); x_1 = ?
  • Second mass; m_2 = 61.7 kg
  • Distance of beam from  the right of the support( m₂ is attached to ); x_1 = 0.273m

Now, since it is mentioned that the beam is in static equilibrium, the Net Torque on it about the support must be zero.

Hence, m_1g( x-x_1) = m_2gx_2

we divide both sides by g

m_1( x-x_1) = m_2x_2

Next, we make x_1, the subject of the formula

x_1 = x - [ \frac{m_2x_2}{m_1} ]

We substitute in our given values

x_1 = 3.00m - [ \frac{61.7kg\ * \ 0.273m}{31.3kg} ]

x_1 = 3.00m - 0.538m

x_1 = 2.46m

Therefore, If the beam is in static equilibrium, meaning the Net Torque on it about the support is zero, the value of x₁ is 2.46m

Learn more; brainly.com/question/3882839

6 0
3 years ago
The amount of heat required to melt lead at it melting temperature is 23.0 kJ/kg. The amount of heat required to melt mercury at
svet-max [94.6K]

The addition of 24 kJ of energy will allow all of the mercury and lead to change from solid to liquid. The temperature of each substance will also increase.

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