What is diffraction of light

Answer:

the answer is false.

Explanation:

i took the test and it is false trust me!!!!!!!!!

The gravitational potential energy added when the velociraptor and cage is lifted from the ground to a height of 9 m is approximately 15,998.95 joules.

What is Potential Energy?

Potential energy is a form of energy that is stored in an object due to its position or configuration in a system. It is the energy that an object has the potential to possess, or the ability to do work, as a result of its position or state.

The gravitational potential energy (GPE) added when the velociraptor and cage is lifted from the ground to a height of 9 m can be calculated using the formula:

GPE = mgh

Where m is the mass of the velociraptor and cage, g is the acceleration due to gravity (approximately 9.81 m/\(s^{2}\)), and h is the height lifted.

Given that the mass of the velociraptor and cage together is 175 kg, and the height lifted is 9 m, we can substitute these values into the formula:

GPE = mgh

GPE = (175 kg) x (9.81 m/\(s^{2}\)) x (9 m)

GPE = 15,998.95 J (joules)

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The coefficient of friction between the two blocks is 0.194.

The problem involves finding the coefficient of friction between two blocks of equal mass attached by a string that moves over a sloped ramp with an angle of 30 degrees. The blocks start from rest and are allowed to move freely. The distance traveled by the blocks is 0.50 m, and the time taken to travel this distance is 0.935 s. To solve the problem, we need to consider the forces acting on the system of blocks. The forces acting on the blocks are the gravitational force, the tension in the string, and the frictional force. As the blocks are moving up the ramp, the force of gravity is pulling them down. The tension in the string is pulling the blocks up the ramp. The frictional force is opposing the motion of the blocks and is acting in the opposite direction to the tension in the string.

To determine the coefficient of friction, we can use the equations of motion to find the acceleration of the blocks. Once we have the acceleration, we can use Newton's Second Law to find the net force acting on the blocks. We can then use the force of friction to find the coefficient of friction. Using the equations of motion, we can find the acceleration of the blocks:

a = 2d/t^2

where d is the distance traveled by the blocks and t is the time taken to travel the distance. Plugging in the given values, we get:

a = 2(0.50 m)/(0.935 s)^2 = 1.15 m/s^2

Next, we can use Newton's Second Law to find the net force acting on the blocks:

ΣF = ma

where ΣF is the sum of the forces acting on the blocks. Plugging in the known forces, we get:

T - mg sin θ - μmg cos θ = ma

where T is the tension in the string, m is the mass of the blocks, g is the acceleration due to gravity, θ is the angle of the ramp, and μ is the coefficient of friction.

We can simplify this equation by substituting mg sin θ for the component of the weight of the blocks that is acting down the ramp and mg cos θ for the component of the weight that is acting perpendicular to the ramp:

T - mg sin θ - μmg cos θ = ma

T - mg(1/2) - μmg(√3/2) = ma

We can also use the equation for the tension in the string:

T = 2mg sin θ

Substituting this into the equation for net force, we get:

2mg sin θ - mg(1/2) - μmg(√3/2) = ma

Simplifying and solving for μ, we get:

μ = (2gsinθ - a)/(2gcosθ)

Substituting the given values, we get:

μ = (2(9.81 m/s^2)sin 30° - 1.15 m/s^2)/(2(9.81 m/s^2)cos 30°) = 0.194

Therefore, the coefficient of friction between the two blocks is 0.194.

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Answer:

The charges under study are of the same sign

The calculation of the electric field for each charge separately, there is no relationship between the charges

Explanation:

Let's start by writing the equation for the electric field

          E = k q / r²

where q is the charge under analysis and r the distance from this charge to a positive test charge.

When analyzing the statement the student has some problems.

* The charges under study are of the same sign, it does not matter if positive or negative.

* The calculation of the electric field for each charge separately, there is no relationship between the charges for the calculation of the electric field.

* What is added is the interaction of the electric field with the positive test charge, in this case each field has the opposite direction to the other, so the vector sum gives zero

Answer:

The acceleration due to gravity on the planet Fergie is 0.0123 m/s^2.

Explanation:

We want to find the acceleration due to gravity on the planet Fregie. Let it be g m/s^2.

Now, the acceleration due to gravity is defined through the following equation:

\(mg = GMm/R^2\)

where m is the mass of an object on the surface of the planet, M is the mass of the planet, R is the radius of the planet, and G is the universal Gravitational constant.

Subsituting values for M = 6.23*10^23, R = 5.79*10^7, G = 6.67*10^(-11), we get

g = 0.0123 m/s^2.

Thus the acceleration due to gravity on the planet Fergie is 0.0123 m/s^2.

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Answer:

ITS A repairing a stove instead of buying a new one not D

Explanation:

i dont know why the dude above me has 5 stars its wrong and I got proof ;) for the future edge ppl

Answer:

A) repairing a stove instead of buying a new one

Explanation:

GOT IT RIGHT :DDD

The statement" The direction of the force on the charge placed in the electric field is upward" is false because the direction of the force on a negative charge (-6 C) placed in an upward-directed uniform electric field of magnitude 24 N/C would be downward.

The direction of the force on a charged particle placed in an electric field is determined by the charge of the particle and the direction of the electric field. In this case, a charge of -6 C is placed in an electric field directed upward with a magnitude of 24 N/C.

The force on a charged particle in an electric field can be calculated using the formula:

F = q * E

Where F is the force, q is the charge of the particle, and E is the electric field.

Since the charge q in this case is negative (-6 C) and the electric field E is directed upward, we can substitute the values into the formula:

F = (-6 C) * (24 N/C)

F = -144 N

The negative sign in the force value indicates that the force is in the opposite direction to the electric field. Therefore, the force on the charge placed in the electric field is downward, not upward.

The force on a negative charge is always opposite to the direction of the electric field. This is because negative charges experience an attractive force towards positive charges, and electric fields are directed from positive charges to negative charges.

Therefore, the statement "The direction of the force on the charge placed in the electric field is upward." is false.

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The effective resistance of the combination when two resistors are connected in series is 2+x²/x.

The resistance is a quantity that offers obstruction to the current flow and the unit of resistance is the ohm. The resistors are connected in series and parallel. The effective resistance of series connections, R(eff) = R₁+R₂+...+Rₙ. The effective resistance of parallel connections are, 1/R(eff) = 1/R₁+1/R₂+.....+1/Rₙ.

From the given,

the parallel resistance of two-ohm x, the effective resistance,

1/R(eff) = 1/x+1/x = 2/x.

The effective resistance in a series combination of resistors, 2/x, and x are,

R(eff) = R₁+R₂ = 2/x + x = 2+x²/x.

Hence, the effective resistance is 2+x²/x.

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Slope is rise over run, meaning the ratio of the change in height to the change in horizontal distance. So a slope of 15.80 corresponds to an angle of ascension θ of

tan(θ) = 15.80   →   θ ≈ 86.38°

The order of the blocks (i.e. whether the large one is pulling the smaller one up or vice versa) does not matter, since friction is not a concern. So if we take the connected blocks as a single mass, by Newton's law we have a net force acting parallel to the incline of

∑ F = P - (M + m) g sin(θ) = (M + m) a

(see the attached free body diagram)

where

P = magnitude of the push

g = 9.80 m/s²

a = 2.53 m/s²

and M and m are the given masses.

Then the system requires a push of

P = (M + m) (a + g sin(θ))

P = (14.2 kg + 4.73 kg) (2.53 m/s² + (9.80 m/s²) sin(86.38°))

P ≈ 233 N

If you have to find the tension in the rope, consider the free body diagram for one of the blocks. By Newton's second law, the net parallel force acting on, say, the larger block (if it's being pulled by the rope) is

∑ F = T - M g sin(θ) = M a

where

T = tension in the rope

Then

T = M (a + g sin(θ))

T = (14.2 kg) (2.53 m/s² + (9.80 m/s²) sin(86.38°))

T ≈ 175 N

Answer:

Explanation:

Equilibrium is a state in which the algebraic sum of all forces acting on an object is zero. Thus the object has no force acting on it. The types are: stable, unstable and neutral equilibrium. While a torque is a turning force which are equal but acts in an opposite direction. When applied to on object, it constitute a turning effect. Example is the force applied on a tap, handle wheel of a car etc.

In the given question, the condition stated shows that the stick would experience a torque, thus not in equilibrium. Since the forces at its ends are in opposite directions, then it continues to rotate about its axis.

Answer:

4.41N or 4.5N (check explanation)

Explanation:

450g = 0.45kg

F = ma

Using 10m/s² = 10(0.45) = 4.5N

Using 9.8m/s² = 9.8(0.45) = 4.41N

The value added by the baker is $3. The bread's contribution to GDP is the final sale price of the bread, which is $6

In this scenario, each person involved in the production and sale of the bread adds value to the final product. The concept of value added refers to the increase in the market value of a product at each stage of production.

The farmer grows the wheat and sells it for $1. The value added by the farmer is $1.

The miller processes the wheat into flour, increasing its value. The miller sells the flour to the baker for $3, so the value added by the miller is $3 - $1 = $2.

The baker uses the flour to make bread, further adding value to the product. The baker sells the bread to the engineer for $6, so the value added by the baker is $6 - $3 = $3.

The engineer consumes the bread, but since no further economic value is added, there is no additional value added by the engineer.

The bread's contribution to GDP (Gross Domestic Product) is the final sale price of the bread, which is $6. GDP measures the total value of all goods and services produced within a country's borders, and the sale of the bread represents the final output of the production chain.

Overall, the value added at each stage contributes to the final price of the bread, and the final sale price of the bread represents its contribution to GDP.

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The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The fraction of a wavelength represented by the first harmonic is 1/2.

The higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

1. The first harmonic of a standing wave on a string with a loose end occurs when there is a node near the driver end and an antinode at the loose end. To measure the frequency of the first harmonic, we need to determine the fraction of a wavelength represented by this standing wave.

The first harmonic of the standing wave on a string with a loose end represents half a wavelength.

The first harmonic of a standing wave on a string with a loose end consists of a node near the driver end and an antinode at the loose end. This configuration creates the simplest standing wave pattern.

In a standing wave, a node is a point where the amplitude of the wave is always zero, representing a point of minimum displacement. An antinode, on the other hand, is a point of maximum displacement, where the amplitude is at its highest.

Since the loose end does not invert the phase of the wave upon reflection, the reflected wave and the original wave constructively interfere at the loose end, resulting in an antinode.

In the first harmonic, there is exactly half a wavelength between the node near the driver end and the antinode at the loose end.

Therefore, the fraction of a wavelength represented by the first harmonic is 1/2.

2. To predict the frequencies of higher harmonics, we can use the relationship that the frequency of each harmonic is a multiple of the frequency of the first harmonic. The higher harmonics can be calculated as follows:

Second Harmonic: The second harmonic consists of two nodes and one additional antinode compared to the first harmonic. The fraction of a wavelength for the second harmonic is 1/2 * 2 = 1. Thus, the second harmonic has a frequency that is twice that of the first harmonic.

Third Harmonic: The third harmonic consists of three nodes and two additional antinodes compared to the first harmonic. The fraction of a wavelength for the third harmonic is 1/2 * 3 = 1.5. Thus, the third harmonic has a frequency that is three times that of the first harmonic.

Fourth Harmonic: The fourth harmonic consists of four nodes and three additional antinodes compared to the first harmonic. The fraction of a wavelength for the fourth harmonic is 1/2 * 4 = 2. Thus, the fourth harmonic has a frequency that is four times that of the first harmonic.

In general, the higher harmonics of a standing wave on a string with a loose end are odd-number multiples of the first harmonic.

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Answer:

Therefore, the 10 kg mass should be placed at x = 5.7 m along the x-axis to achieve a center of mass located at y = 4.5 m.

Explanation:

To find the position along the x-axis where a 10 kg mass should be placed such that the center of mass is located at y = 4.5, we can use the formula for the center of mass:

x_cm = (m1 * x1 + m2 * x2) / (m1 + m2)

Here, m1 and x1 represent the mass and position of the 8 kg mass, respectively. m2 is the mass of the 10 kg mass, and we need to find x2, its position.

Given:

m1 = 8 kg

x1 = 3 m

x_cm = unknown (to be found)

m2 = 10 kg

y_cm = 4.5 m

Since the center of mass is at y = 4.5, we only need to consider the y-coordinate when calculating the center of mass position along the x-axis.

To solve for x2, we can rearrange the formula as follows:

x2 = (x_cm * (m1 + m2) - m1 * x1) / m2

Substituting the given values:

x2 = (x_cm * (8 kg + 10 kg) - 8 kg * 3 m) / 10 kg

Simplifying:

x2 = (x_cm * 18 kg - 24 kg*m) / 10 kg

Now, we can set the y-coordinate of the center of mass equal to 4.5 m and solve for x_cm:

4.5 m = (8 kg * 3 m + 10 kg * x2) / (8 kg + 10 kg)

Simplifying:

4.5 m = (24 kg + 10 kg * x2) / 18 kg

Multiplying both sides by 18 kg:

81 kg*m = 24 kg + 10 kg * x2

Subtracting 24 kg from both sides:

10 kg * x2 = 81 kg*m - 24 kg

Dividing both sides by 10 kg:

x2 = (81 kg*m - 24 kg) / 10 kg

Simplifying:

x2 = 8.1 m - 2.4 m

x2 = 5.7 m

(brainlest?) ples:(

Answer:

the 10 kg mass should be placed at x = -2.4 m to achieve a center of mass at y = 4.5 m.

Explanation:

To find the position along the x-axis where the 10 kg mass should be placed so that the center of mass is located at y = 4.5, we can use the principle of the center of mass.

The center of mass of a system is given by the equation:

x_cm = (m1x1 + m2x2) / (m1 + m2),

where x_cm is the x-coordinate of the center of mass, m1 and m2 are the masses, and x1 and x2 are the positions along the x-axis.

Given:

m1 = 8 kg,

x1 = 3 m,

m2 = 10 kg,

y_cm = 4.5 m.

To solve for x2, we need to find the x-coordinate of the center of mass (x_cm) by using the y-coordinate:

y_cm = (m1y1 + m2y2) / (m1 + m2),

where y1 and y2 are the positions along the y-axis.

Rearranging the equation and substituting the given values:

4.5 = (83 + 10y2) / (8 + 10).

Simplifying the equation:

4.5 = (24 + 10*y2) / 18.

Multiplying both sides by 18:

81 = 24 + 10*y2.

Rearranging the equation:

10*y2 = 81 - 24,

10*y2 = 57.

Dividing both sides by 10:

y2 = 5.7.

Therefore, the y-coordinate of the 10 kg mass should be 5.7 m.

To find the x-coordinate of the 10 kg mass, we can use the equation for the center of mass:

x_cm = (m1x1 + m2x2) / (m1 + m2).

Substituting the given values:

x_cm = (83 + 10x2) / (8 + 10).

Since the center of mass is at x_cm = 0 (the origin), we can solve for x2:

0 = (83 + 10x2) / (8 + 10).

Rearranging the equation:

83 + 10x2 = 0.

24 + 10*x2 = 0.

10*x2 = -24.

Dividing both sides by 10:

x2 = -2.4.

Answer:

a

Explanation:

Answer:

Approximately \(3.967\; {\rm kg\cdot m\cdot s^{-1}}\).

Explanation:

Convert velocity to the standard units (meters per second):

\(\begin{aligned}v &= 246.2 \; {\rm km \cdot h^{-1}} \\ &= 246.2 \; {\rm km \cdot h^{-1}}\times \frac{1\; {\rm h}}{3600\; {\rm s}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \\ &\approx 68.389\; {\rm m\cdot s^{-1}}\end{aligned}\).

Convert mass to standard units (kilograms):

\(\begin{aligned} m &= 58\; {\rm g} \\ &= 58\; {\rm g} \times\frac{1\; {\rm kg}}{1000\; {\rm g}}\\ &= 0.058\; {\rm kg}\end{aligned}\).

When an object of mass \(m\) travels at a velocity of \(v\), momentum of that object would be \(p = m\, v\). In standard units, the momentum of this tennis ball would be:

\(\begin{aligned}p &= m\, v \\ &\approx (0.058\; {\rm kg})\, (68.389\; {\rm m\cdot s^{-1}}) \\ &\approx 3.967\; {\rm kg \cdot m\cdot s^{-1}}\end{aligned}\).

Answer:

bc it was a universal explosion and It started the future

Explanation:

FACTS

Answer:

i wouldn't believe so.

Explanation:

because there was no room or air for the sound to move through. this is because of immense heat and the amount of hyperactive neutrons, electrons and protons clouding everywhere. This would mean that even if there was sound it would a. not travel far or b. go in a completely different direction than expected.

Answer:

All planets including the outer larger planets were formed at the same time somewhere around 4.5 Billion years ago.

Explanation:

the young sun drove away most of the gas from the inner solar system, leaving behind the rocky cores also known as the terrestrial planets.

The molecules of O2 that are  present in 3.90 L flask at  a temperature of 273 K and a pressure of 1.00 atm is 1.047 x 10^23 molecules  of O2

Step  1:  used the ideal gas equation to calculate the moles of O2

that is Pv=n RT  where;

P(pressure)= 1.00 atm

V(volume) =3.90 L

n(number of moles)=?

R(gas constant) = 0.0821 L.atm/mol.K

T(temperature) = 273 k

by  making n the subject of the formula by  dividing  both side by RT

n= Pv/RT

n=[( 1.00 atm x 3.90 L)  /(0.0821 L.atm/mol.k  x273)]=0.174  moles

Step 2: use the Avogadro's  law constant  to calculate  the number of molecules

that  is  according to Avogadro's law

                          1  mole =  6.02 x10^23  molecules

                            0.174 moles=? molecules

by  cross  multiplication

the number of  molecules

= (0.174  moles x  6.02 x10^23  molecules)/ 1 mole  =1.047 x 10^23 molecules of O2

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We have that the total pipe drip of \(X=155 mL/min.\) expressed in gallons/day is

\(X=58.96gal/day\)

From the Question we are told that

Pipe Drip\(=155 mL/min.\)

Generally

We have that for mL to gallons conversion

\(1mL=0.000264172\)

And

A minute to day con version is

\(1min=0.000694444\)

Therefore

\(X=155 mL/min.\)

\(X=155(\frac{0.000264172}{0.000694444})\)

\(X=58.96gal/day\)

In conclusion

The total pipe drip of \(X=155 mL/min.\) expressed in gallons/day is \(X=58.96gal/day\)

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Answer:

the energy in the oscillations between the blocks is 3.025 J

Explanation:

Given the data in the question;

Force f = 138 N

stiffness of spring k = 605 kg/s²

mass of block = 202 g = 0.202 kg

pushing the block a distance 15 cm, the rightmost block has moved a distance 5 cm

i.e

x₁ = 15 cm

x₂ = 5cm

the energy in the oscillations between the blocks will be;

E\(_A\) = E\(_B\) = \(\frac{1}{2}\)k( Δx )²

we substitute

= \(\frac{1}{2}\) × k( 15 - 5 )² × 10⁻⁴

= \(\frac{1}{2}\) × 605 × ( 10 )² × 10⁻⁴

= \(\frac{1}{2}\) × 605 × 100 × 10⁻⁴

= 3.025 J

Therefore, the energy in the oscillations between the blocks is 3.025 J

The potential at the center of the square is 507 V. Option (a) is correct answer.

The electric potential at a distance r from a point charge q is given by:

V = kq/r

where k is the Coulomb constant.

In this case, the charges q₂ and q₄ are negative, so the potentials due to these charges will be negative. The charges q₁ and q₃ are positive, so the potentials due to these charges will be positive. The potential at the center of the square is given by:

\(V = \dfrac{kq_1}{a} + \dfrac{kq_2}{\sqrt{2}a} + \dfrac{kq_3}{a} + \dfrac{kq_4}{\sqrt{2}a}\)

Plugging in the values of the charges and the distance a, we get:

\(V = \dfrac{9 \times 10^9 \times 1 \times 10^{-8}}{1} + \dfrac{9 \times 10^9\times -2\times 10^{-8}}{\sqrt{2}\times 1} + \dfrac{9 \times 10^9\times 3 \times 10^{-8} }{1} + \dfrac{9 \times 10^9 \times 2 \times 10^{-8} }{\sqrt{2} \times 1}\)

Simplifying this expression,

V = 507 V

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--The complete question is, Four point charges q1 ,q2 ,q3 and q4 are placed at the corners of the square of side a, as shown in figure. The potential at the centre of the square is: (Given: q1 =1×10^−8 C , q2 =−2×10^−8 C, q3 =3×10^−8 C, q4 =2×10^−8C, a=1m).

a. 507 V

b. 607 V

c. 550 V

d. 650 V--

The frictional force of an object is the product of the normal force and coefficient of kinetic friction. Here the frictional force acting on the object  is 16.4 N.

Frictional force is a kind of force acting on a body to resist it from motion. Thus, the direction of the force will be in negative with the magnitude. Frictional force is the product of coefficient of friction and the normal force.

The normal  force acting on the object of mass 4.2 Kg is N = mg

N = 4.2 Kg × 9.8 m/s² = 41.16 N

Frictional force = ц N

                         = 0.40 × 41.16 N

                         = 16.4 N.

Therefore, the frictional force acting between the surface of the object and the floor is 16.4 N

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Your question is incomplete. But your complete question probably was:

The coefficient of kinetic friction between an object and the surface upon which it is sliding is 0.40 the weight of the object is 4.2 kg. What is the frictional force of the object?

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!