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SpyIntel [72]
3 years ago
8

In which type of bone does ossification occur in the membranes?

Physics
1 answer:
Afina-wow [57]3 years ago
3 0
In the formation of flat bones such as the skull the mandibles and the clavicles  
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One of the waste products of a nuclear reactor is plutonium-239 . This nucleus is radioactive and decays by splitting into a hel
Gekata [30.6K]

Answer:

a) v_{U-235} = 2.68 \cdot 10^{5} m/s

v_{He-4} = -1.57 \cdot 10^{7} m/s  

b) E_{He-4} = 8.23 \cdot 10^{-13} J

E_{U-235} = 1.41 \cdot 10^{-14} J

 

Explanation:

Searching the missed information we have:                                        

E: is the energy emitted in the plutonium decay = 8.40x10⁻¹³ J

m(⁴He): is the mass of the helium nucleus = 6.68x10⁻²⁷ kg  

m(²³⁵U): is the mass of the helium U-235 nucleus = 3.92x10⁻²⁵ kg            

a) We can find the velocities of the two nuclei by conservation of linear momentum and kinetic energy:

Linear momentum:

p_{i} = p_{f}

m_{Pu-239}v_{Pu-239} = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}

Since the plutonium nucleus is originally at rest, v_{Pu-239} = 0:

0 = m_{He-4}v_{He-4} + m_{U-235}v_{U-235}  

v_{He-4} = -\frac{m_{U-235}v_{U-235}}{m_{He-4}}    (1)

Kinetic Energy:

E_{Pu-239} = \frac{1}{2}m_{He-4}v_{He-4}^{2} + \frac{1}{2}m_{U-235}v_{U-235}^{2}

2*8.40 \cdot 10^{-13} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}    

1.68\cdot 10^{-12} J = m_{He-4}v_{He-4}^{2} + m_{U-235}v_{U-235}^{2}   (2)    

By entering equation (1) into (2) we have:

1.68\cdot 10^{-12} J = m_{He-4}(-\frac{m_{U-235}v_{U-235}}{m_{He-4}})^{2} + m_{U-235}v_{U-235}^{2}  

1.68\cdot 10^{-12} J = 6.68 \cdot 10^{-27} kg*(-\frac{3.92 \cdot 10^{-25} kg*v_{U-235}}{6.68 \cdot 10^{-27} kg})^{2} +3.92 \cdot 10^{-25} kg*v_{U-235}^{2}  

Solving the above equation for v_{U-235} we have:

v_{U-235} = 2.68 \cdot 10^{5} m/s

And by entering that value into equation (1):

v_{He-4} = -\frac{3.92 \cdot 10^{-25} kg*2.68 \cdot 10^{5} m/s}{6.68 \cdot 10^{-27} kg} = -1.57 \cdot 10^{7} m/s                        

The minus sign means that the helium-4 nucleus is moving in the opposite direction to the uranium-235 nucleus.

b) Now, the kinetic energy of each nucleus is:

For He-4:

E_{He-4} = \frac{1}{2}m_{He-4}v_{He-4}^{2} = \frac{1}{2} 6.68 \cdot 10^{-27} kg*(-1.57 \cdot 10^{7} m/s)^{2} = 8.23 \cdot 10^{-13} J

For U-235:

E_{U-235} = \frac{1}{2}m_{U-235}v_{U-235}^{2} = \frac{1}{2} 3.92 \cdot 10^{-25} kg*(2.68 \cdot 10^{5} m/s)^{2} = 1.41 \cdot 10^{-14} J

 

I hope it helps you!                                                                                    

3 0
3 years ago
a car was traveling at 30.0 m/s when a small child darted onto the road 60.0 m away. Reacting instantly the driver slams the bra
Galina-37 [17]
2 seconds,,,,,,,,,,,,,,,,,,,,,,,
5 0
2 years ago
On the surface of the earth the weight of an object is 200 lb. Determine the height of the
siniylev [52]

Answer:

The height of the  object is 5007.4 miles.

Explanation:

Given that,

Weight of object = 200 lb

We need to calculate the value of Gmm_{e}

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

200=\dfrac{Gmm_{e}}{(3958.756)^2}

200\times(3958.756)^2=Gmm_{e}

Gmm_{e}=3.134\times10^{9}

We need to calculate the height of the  object

Using formula of gravitational force

F=\dfrac{Gmm_{e}}{r^2}

Put the value into the formula

125=\dfrac{200\times(3958.756)^2}{r^2}

r^2=\dfrac{200\times(3958.756)^2}{125}

r^2=25074798.5

r=\sqrt{25074798.5}

r=5007.4\ miles

Hence. The height of the  object is 5007.4 miles.

7 0
3 years ago
which of the following is true? a)speed is velocity w a vector b)speed is a scalar w direction c)speed is velocity w direction d
Aleksandr [31]

The answer is D velocity is speed w a direction

8 0
3 years ago
What would occur if the index of refraction of a material was less than n=1?
Bingel [31]

The correct answer is (A). The speed of light would increase to a speed larger than the maximum speed of light in vacuum.

The index of refraction is the ratio of speed of light in vacuum to the speed of light in a medium.

n=C/V

here, n is the index of refraction, c the speed of light in vacuum, v is speed of light in any medium.

Now if the value of index of refraction is less than one, than the value of speed of light would be greater than the speed of light in the vacuum.


3 0
3 years ago
Read 2 more answers
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