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Verizon [17]
3 years ago
8

1. When the fission of uranium-235 is carried out, about 0.1 percent of the mass of the reactants is lost during the reaction. W

hat happens to this “lost” mass?
Physics
2 answers:
DerKrebs [107]3 years ago
6 0

The correct answer to the question is that the lost mass has been converted into energy.

EXPLANATION:

From Einstein's theory, we know that energy and mass are inter convertible .

When some amount of mass is lost, same amount of energy equivalent to mass is produced.

Let us consider m is the mass lost during any reaction. Hence, the amount of energy produced will be-

                                 Energy E = mc^2                

Here, c is the velocity of light i.e c = 3\times 10^8\ m/s

As per the question, uranium-235 undergoes fission. The amount of mass defect is 0.1 %.

The mass defect is defined as the difference between mass of reactants and products. During the fission, energy is produced.

The energy produced in this reaction is nothing else than the energy equivalent to mass defect.  Approximately 199.5 Mev of energy equivalent to this mass defect is produced in this reaction.

12345 [234]3 years ago
6 0

It is converted into energy.

You might be interested in
You lift a 10-N physics book up in the air a distance of 1.0 m, at a constant velocity
Anarel [89]

The book is lifted upward, but gravity points down, so the work done by gravity must be negative (so you can eliminate options 1 and 3).

The force exerted on the book by gravity has magnitude

<em>F</em> = <em>mg</em> = (10 N) (9.80 m/s^2) = 9.8 N ≈ 10 N

You raise the book 1.0 m in the opposite direction, so the work done is

<em>W</em> = (10 N) (-1.0 m) = -10 J

5 0
3 years ago
An Alaskan rescue plane traveling 41 m/s drops a package of emergency rations from a height of 192 m to a stranded party of expl
svet-max [94.6K]

Answer:

a)The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) Vertical component of velocity = 61.41 m/s

Explanation:

a) Consider the vertical motion of plane,

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 0 m/s

         Displacement, s = 192 m

         Acceleration, a = 9.81 m/s²

         Substituting

                      s = ut + 0.5 at²

                      192 = 0 x t + 0.5 x 9.81 x t²

                         t = 6.26 seconds

         Now we need to find horizontal distance traveled in 6.26 seconds by the package.

         We have equation of motion, s = ut + 0.5 at²

         Initial velocity, u = 41 m/s

        Time, t = 6.26 s

         Acceleration, a = 0 m/s²

         Substituting

                      s = ut + 0.5 at²

                      s = 41 x 6.26 + 0.5 x 0 x 6.26²

                         s = 256.52 m

     The package strikes 256.2 m in the ground relative to the point directly below where it was released

b) The horizontal component will not change it remains same as 41 m/s

c) We have equation of motion, v = u+ at

          Initial velocity, u = 0 m/s

         Time, t = 6.26 s

         Acceleration, a = 9.81 m/s²  

         Substituting

                      v = u+ at

                       v = 0 + 9.81 x 6.26 = 61.41 m/s

   Vertical component of velocity = 61.41 m/s      

4 0
3 years ago
PLZZZ HELP
Doss [256]

Answer/Explanation:

The weight of an object is defined as the force that is exerted due to the gravitational force.

Mathematically, it can be written as :

W = m g

Where

m is the mass of the object

g is the acceleration due to gravity

Also,  

We know that the value of g varies with respect to the location. At the equator, the value of g is less as compared to the poles.

The feature of an object that affects its weight are :

Mass of the object

Location of the object

How much force Earth exerts on the object

4 0
3 years ago
A circular coil 14.0 cm in diameter and containing nine loops lies flat on the ground. The Earth's magnetic field at this locati
sp2606 [1]

Answer:

a)T = 2.9*10^{-5} N-m

b) north edge will rise up

Explanation:

torque on the coil is given as

T = NIABsin\theta

where N is number of loop =  9 loops

i is current = 7.80 A

-B -earth magnetic field = 5.00*10^{-5} T

A- area of circular coil

A = \frac{\pi d^{2}}{4}

A =\frac{\pi .14^{2}}{4}

A =0.015 m2

PUTITNG ALL VALUE TO GET TORQUE

T = 9*7.8*0.015*5*10^{-5} sin{90-56}

T = 2.9*10^{-5} N-m

b) north edge will rise up

3 0
3 years ago
Read 2 more answers
Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp
yuradex [85]

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

7 0
3 years ago
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