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
2 answers:
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 =
Here, c is the velocity of light i.e c =
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.
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Answer:
q₁= +0.5nC
Explanation:
Theory of electrical forces
Because the particle q3 is close to three other electrically charged particles, it will experience two electrical forces and the solution of the problem is of a vector nature.
To solve this problem we apply Coulomb's law:
Two point charges (q1, q2) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
o solve this problem we apply Coulomb's law:
Two point charges (q₁, q₂) separated by a distance (d) exert a mutual force (F) whose magnitude is determined by the following formula:
F=K*q₁*q₂/d² Formula (1)
F: Electric force in Newtons (N)
K : Coulomb constant in N*m²/C²
q₁,q₂:Charges in Coulombs (C)
d: distance between the charges in meters
Data:
Equivalences
1nC= 10⁻⁹ C
1cm= 10⁻² m
Data
q₃=+5.00 nC =+5* 10⁻⁹ C
q₂= -2.00 nC =-2* 10⁻⁹ C
d₂= 5.00 cm= 5*10⁻² m
d₁= 2.50 cm= 2.5*10⁻² m
k = 8.99*10⁹ N*m²/C²
Calculation of magnitude and sign of q1
Fn₃=0 : net force on q3 equals zero
F₂₃:The force F₂₃ that exerts q₂ on q₃ is attractive because the charges have opposite signs,in direction +x.
F₁₃:The force F₂₃ that exerts q₂ on q₃ must go in the -x direction so that Fn₃ is zero, therefore q₁ must be positive and F₂₃ is repulsive.
We propose the algebraic sum of the forces on q₃
F₂₃ - F₁₃=0
We eliminate k*q₃ of the equation
q₁= +0.5*10⁻⁹ C
q₁= +0.5nC
Answer:
Explanation:
given data:
flow Q = 9 m^{3}/s
velocity = 8 m/s
density of air = 1.18 kg/m^{3}
minimum power required to supplied to the fan is equal to the POWER POTENTIAL of the kinetic energy and it is given as
here is mass flow rate and given as
Putting all value to get minimum power
The question is missing, but I guess the problem is asking for the distance between the cliff and the source of the sound.
First of all, we need to calculate the speed of sound at temperature of
:
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff:
First of all, we need to calculate the speed of sound at temperature of
The sound wave travels from the original point to the cliff and then back again to the original point in a total time of t=4.60 s. If we call L the distance between the source of the sound wave and the cliff, we can write (since the wave moves by uniform motion):
where v is the speed of the wave, 2L is the total distance covered by the wave and t is the time. Re-arranging the formula, we can calculate L, the distance between the source of the sound and the cliff: