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Llana [10]
3 years ago
11

Which particle is commonly used to initiate a fission chain reaction

Physics
1 answer:
Snezhnost [94]3 years ago
5 0
Neutron is commonly used to initiate a fission chain reaction.
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Which happened after the electromagnetic and weak nuclear forces were separated from the unified force?
alukav5142 [94]
-release of first light
-nucleosynthesis
-formation of stars and galaxies
<span>-formation of elementary particles</span>
7 0
3 years ago
The speed of the center of the earth as it orbits the sun is 107257 kmph and the absolute angular velocity of the earth about it
Sindrei [870]

Answer with Explanation:

We are given that

Speed ,v_0=107257 kmph

Angular velocity,\omega=7.292\times 10^{-5} rad/s

Radius of earth,r=6371 km=6371000 m

1 km=1000 m

Linear velocity,v=r\omega=6371000\times 7.292\times 10^{-5}=464.57 m/s

Linear velocity,v=464.57\times \frac{18}{5}=1672.46 km/h

Velocity at point A,v_A=-vi+v_0 j=-1672.46 i+107257j kmph

Velocity at point B,v_B=v_0j-vj=107257j-1672.46j=105584.54j kmph

Velocity at point C,v_C=v_0j+vi=1672.46 i+107257j kmph

Velocity at point  D,v_D=v_0j+vj=107257j+1672.46j=108929.46jkmph

3 0
3 years ago
determine the amount of potential energy of a 5.0-N book that is moved to three different shelves on a bookcase . the height of
klasskru [66]
Potential energy<span> is the </span>energy<span> that is stored in an object due to its position relative to some zero position. It is calculated by the expression PE = mgh where mg is the weight of the book and h is the height. It is calculated as follows:

PE = 50(1) = 50 J
</span>PE = 50(1.5) = 75 J
PE = 50(2) = 100 J
7 0
3 years ago
An automobile with a standard differential turns sharply to the left. The left driving wheel turns on a 20-m radius. Distance be
Inessa05 [86]

Explanation:

The given data is as follows.

    Inner wheel Radius = 20 m,

   Distance between left and right wheel = 1.5m,

Let us assume speed of drive shaft is N rpm.

Formula to calculate angular velocity is as follows.

    Angular velocity of automobile = w = \frac{V}{R}

where,   V = linear velocity of automobile m/min,

              R = turning radius from automobile center in meter

In the given case, angular velocity remains same for inner and outer wheel but there is change in linear velocity of inner wheel and outer wheel.

Now, we assume that

         u = linear velocity of inner wheel

and,   u' = linear velocity of outer wheel.

Formula for angular velocity of inner wheel w = ,

Formula for angular velocity of outer wheel w =

Now, for inner wheels

                   w =

                      = \frac{u}{(R - d)}

                  u = V \times \frac{(R - d)}{R}

                    = V \times (1 - \frac{d}{R})

If radius of wheel is r it will cover  distance in one min.

Since, velocity of wheel is u it will cover distance u in unit time(min)

Thus,             u = 2\pi rn = V \times (1 - \frac{d}{R})

Now, rotation per minute of inner wheel is calculated as follows.

         n = \frac{V}{2 \pi r \times (1 - \frac{d}{R})}

            = \frac{V}{2 \pi r \times (1 - \frac{0.75}{20})} (since 2d = 1.5m given, d = 0.75m),

             = \frac{V}{r} \times 0.1532

So, rotation per minute of outer wheel; n' =  

                   = \frac{V}{2 \pi r \times (1 + \frac{0.75}{20})}

                   = \frac{V}{r} \times 0.1651

5 0
4 years ago
All objects____ energy.<br><br> Eliminate <br><br> Conduct <br><br> Radiate <br><br> Insulate
Trava [24]

the answer is c. radiate

7 0
4 years ago
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