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Dima020 [189]
3 years ago
15

What becomes of the loss in mass of nuclei when heavy atoms split?

Physics
1 answer:
NeTakaya3 years ago
6 0

Answer:

when heavy atoms split the loss of mass is converted into kinetic energy

Explanation:

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A block is attached to the end of a horizontal ideal spring and rests on a frictionless surface. The block is pulled so that the
bezimeni [28]

Answer:

The answer is "a, c and b"

Explanation:

  • Its total block power is equal to the amount of potential energy and kinetic energy.
  • Because the original block expansion in all situations will be the same, its potential power in all cases is the same.
  • Because the block in the first case has no initial speed, the block has zero film energy.
  • For both the second example, it also has the v_o velocity, but the kinetic energy is higher among the three because its potential and kinetic energy are higher.
  • While over the last case the kinetic speed is greater and lower than in the first case, the total energy is also higher than the first lower than that of the second.
  • The greater the amplitude was its greater the total energy, therefore lower the second, during the first case the higher the amplitude.
4 0
3 years ago
If the original pressure is 5 atm and original volume is 100 L, and the new volume is 20 L, what is the new pressure?
Nana76 [90]
P1V1=P2V2
5*100=P2*20
500=20P
P=25
7 0
3 years ago
1. Any push or pull is called
timama [110]

Answer:

b  ok

Explanation:

7 0
3 years ago
What is something I could represent chloroplasts in a project
IgorC [24]
Nulceus - recipe book/instruction manual 
Mitochondria - Battery 
Endoplasmic reticulum - Printer or a pen? 
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Chloroplasts - green rechargable battery 
Cell membrane (elastic band (2 to represent the phospholipid bilayer) 
Ribosomes - I guess maybe an ink pot as its the material thats used to write 
Cell Wall - the paper bag 
lysosomes - washing up liquid (breaks down wate food on a dirty plate) 
vaculoes - bottle of water
3 0
3 years ago
The electric flux through a square-shaped area of side 5 cm near a very large, thin, uniformly-charged sheet is found to be 3\ti
deff fn [24]

Answer:

Explanation:

Given

side of square shape a=5\ cm

Electric flux \phi =3\times 10^{-5}\ N.m^2/C

Permittivity of free space \epsilon_0=8.85\times 10^{-12} \frac{C^2}{N.m^2}

Flux is given by

\phi =EA\cos \theta

where E=electric field strength

A=area

\theta=Angle between Electric field and area vector

E=\frac{\phi }{A\cos (0)}

E=\frac{3\times 10^{-5}}{25\times 10^{-4}\times \cos(0)}

E=0.012\ N/C

and Electric field  by a uniformly charged sheet is given by

E=\frac{\sigma }{2\epsilon_0}

where \sigma=charge density

=\frac{\sigma }{\epsilon_0}

\sigma =0.012\times 8.85\times 10^{-12}

\sigma =2.12\times 10^{-13}\ C/m^2    

5 0
3 years ago
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