A diverging lens has a focal length of magnitude 10 cm. At what object distance will the magnification be +0.20? Express your answer with the appropriate units. H Value Units Submit Request Answer

Wavelength and frequency are related to energy in the same way that they are to light. Shorter wavelengths and higher frequencies are associated with more energy.

Therefore, lower energy is produced by longer wavelengths and lower frequencies.

Energy decreases with increasing wavelength length and vice versa. Visible light is more energetic than infrared radiation or radio waves and less energetic than, for example, ultraviolet light or X-rays.

The amount of energy they carry is dependent on both their frequency and amplitude. The amount of energy increases with frequency and amplitude, respectively.

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Answer:

\(\boxed {\boxed {\sf 400 \ J}}\)

Explanation:

Work is a force that causes the displacement of an object. It is the product of force and displacement.

\(W=Fd\)

The force is 40 Newtons and the displacement is 10 meters.

Substitute the values into the formula.

\(W= 40 \ N * 10 \ m\)

Multiply.

\(W= 400 \ N*m\)

Convert the units. 1 Newton meter is equal to 1 Joule, so our answer is equal to 400 Joules.

\(W= 400 \ J\)

400 Joules of work was done to move the block.

Answer:

400 J

Explanation:

This is the correct answer for k12

Answer:

If a coin and a feather are dropped from the same height at the same time on the moon’s surface, they fall together. Ans: It is because Gravity of the moon is six times less than that of the earth, therefore coin and a feather are fall together.

Answer:

The term “INTERmolecular forces” is used to describe the forces of attraction. BETWEEN atoms, molecules, and ions when they are placed close to each other. •

Hope it helps

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Answer: are the forces which mediate interaction between molecules, including forces of attraction or repulsion which act between atoms and other types of neighboring particles, e.g. atoms or ions.

Explanation:

The moment of inertia about the rotation axis for the given playground toy, with two children sitting opposite each other, is approximately 145.35 kg·m².

To determine the moment of inertia about the rotation axis for the given playground toy, we need to consider the contributions from the seats and the two children.

Given:

Mass of each seat = 6.4 kg

Length of the rods (r) = 1.5 m

Mass of the first child (m₁)= 16 kg

Mass of the second child (m₂) = 23 kg

The moment of inertia of each seat can be calculated using the formula for the moment of inertia of a point mass about an axis:

\(I_{seat} = m_{seat times} r^2\)

For each seat, the moment of inertia is:

\(I_{seat} = 6.4 kg times (1.5 m)^2= 14.4 kg\cdot m^2\)

Now, to calculate the moment of inertia contributed by the children, we need to consider that the children are located opposite each other. Assuming the axis of rotation passes through the center of mass of the children-seats system, the moment of inertia for each child is:

\(I_{child} = m_{child times} r^2\)

For the first child (m₁):

\(I_1 = 16 kg times (1.5 m)^2 = 36 kgm^2\)

For the second child (m₂):

\(I_2 = 23 kg times (1.5 m)^2 = 51.75 kgm^2\)

Finally, we can calculate the total moment of inertia by summing the contributions from the seats and the children:

Total moment of inertia =\(4 times I_{seat} + I_1 + I_2\)

= \(4 times (14.4 kgm^2) + 36 kgm^2 + 51.75 kgm^2\)

= \(57.6 kgm^2 + 36 kgm^2 + 51.75 kgm^2\)

= \(145.35 kgm^2\)

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Mechanical energy is of two types,

1. Kinetic energy

2. Potential energy

When the girl is at the top of her jump, the speed of the girl becomes zero.

Thus, the kinetic energy of the girl at the top of her jump is,

where m is the mass of the girl and v is her speed,

Substituting the known values,

Thus, the kinetic energy of the girl is zero at the top of her jump.

The potential energy of the girl at the top of her jump is,

where g is the acceleration due to gravity and h is the height of the girl,

Thus, the potential energy of the girl at the maximum height is maximum.

Hence, the mechanical energy of the girl at the top is in potential energy form because her kinetic energy is zero at the top of her jump.

Answer:

Explanation:

Begin by remembering that at the stone's max height, the final velocity there is 0. Being in possession of that information along with the fact that the pull of gravity due to acceleration is -9.8 m/s/s, we can use the equation

\(v^2=v_0^2+2a\)Δx where v is the final velocity, v₀ is the initial velocity, a is the pull of gravity, and Δx is the displacement. Filling in:

\(0^2=v_0^2+2(-9.8)(5.0)\)

0 = v₀² - 98 so

98 = v₀² and

v = 9.9 m/s Now we can put that into the KE equation.

\(10=\frac{1}{2}m(9.8)\) and

\(m=\frac{2(10)}{98}\) so

m = .20 kg

The amount of information available, the length of time available to examine it, and consequently the complexity of weather occurrences limit one's ability to express their skill to predict the weather.

Models employ estimations and assumptions to forecast the weather because it is impossible to collect data from the future. These estimations from the models get less accurate as time goes on since the atmosphere is continually changing.

Probabilistic techniques accurately depict the uncertainty where deterministic plans provide exact average quantities with a known but unquantified uncertainty.

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Answer:

Saturn is almost as big as Jupiter despite its smaller mass because Jupiter is a more dense gas giant than Saturn.

Explanation:

Answer:

1.) B

2.) Polar Regions or high altitudes.

Explanation:

Star B appears brighter in the night sky. This is because star B is closer to us than star A, even though both stars A and B have equal luminosities.

Two main factors that determine how bright a star appears to be in the sky are the luminosity and the distance. Luminosity refers to a measure of radiant power of a star. The value of a star's luminosity is independent of an observer's distance from that star. If two stars have the same luminosity, that means they radiate the same amount of electromagnetic power. However, if one is closer to an observer, it will appear brighter to the observer than the other one will.

Star A and star B have equal luminosities as stated in the question, but Star A is twice as far as star B, which means star A is farther to us than star B. As a result, star B appears brighter in the night sky that star A does.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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The two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

In the given case, the figure shows an exaggerated view of a rifle that has been sighted in for a 91.4-meter target. Let the muzzle speed of the bullet be v0 = 427 m/s.

Now, we are required to find the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target.

It is known that the horizontal displacement of the bullet from the gun can be given by the equation: x = v0 t cosθ ..........(i)and the vertical displacement of the bullet from the gun can be given by the equation: y = v0 t sinθ - (1/2) g t^2..........(ii).

Here, t is the time of flight of the bullet and g is the acceleration due to gravity.

As the bullet hits the target, its final vertical displacement from the gun is equal to the height of the target, i.e.,y = 91.4m.Now, we can substitute equations (i) and (ii) in place of t and y in equation (ii) to get:x tanθ - (g/2v0^2) x^2 sec^2θ = 91.4 ..........(iii)This is a quadratic equation in tanθ.

On solving this equation using the quadratic formula, we get:tanθ = [-b ± √(b^2 - 4ac)]/2aWhere,a = -gx^2/(2v0^2) = -4.9x^2/v0^2, b = x, and c = -91.4.

Rearranging the terms, we get:2a tanθ^2 + b tanθ - 91.4 = 0On substituting the given values, we get:2(-4.9x^2/v0^2) tanθ^2 + x tanθ - 91.4 = 0θ1 and θ2 are the two possible angles which can be found by solving the above quadratic equation.

Using the trigonometric identity given in the hint, we can write: sin 2θ = 2 sinθ cos θ = 2 tanθ/ (1 + tan^2θ)Now, we can substitute tanθ = (-b ± √(b^2 - 4ac))/2a in the above equation to get: sin 2θ = (-4bx ± 2x√(b^2 - 4ac))/(b^2 + 4a^2)Now, we can substitute the given values to get: sin 2θ1 = -0.999sin 2θ2 = 0.998.

Thus, we get two values of sin 2θ, one is close to -1 and the other is close to 1. As sin 2θ = -1 when 2θ = -π/2 + nπ and sin 2θ = 1 when 2θ = π/2 + nπ, where n is an integer, we get two possible values of θ for each of these two cases.

Hence, the two possible angles θ1 and θ2 between the rifle barrel and the horizontal such that the bullet will hit the target are given by:θ1 = (-1/2) [arctan(2a/(-b - √(b^2 - 4ac))) + π/2 + nπ] andθ2 = (-1/2) [arctan(2a/(-b + √(b^2 - 4ac))) + π/2 + nπ]where n is an integer.

As one of these angles is so large that it is never used in target shooting, we only need to consider the other angle.

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Answer:

is the attraction or repulsion between charged particles

The electric force is the attraction or repulsion between charged particles. They are electrostatic force of attraction or electrostatic force of repulsion.

An electric force is generated from the attraction or repulsion between charged particles. There are two kinds of charges, positive charges and negative charges.

Negative charges are produced by electrons whereas, protons are carrying  positive charges. Two positive charges or two negative charges will repel whereas, a positive charge and a negative charge attracts together.

This force of attraction is the basic of chemical bonding. The nucleus of an atom attracts the bonded electrons and this force of attraction results in the combination of two atoms forming compounds.

Thus, the blank space indicates the electric force.

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(a) The potential at x = 0 is 18.6V, the potential at x = 3.00m is 5.1V

and the potential at x = 6.00m is -8.4V.

(b) The magnitude and direction of the electric field at x = 0 is 3.50 V/m directed towards the negative x-axis.

The magnitude and direction of the electric field at x = 3.00m is 3.50 V/m directed towards the positive x-axis.

The magnitude and direction of the electric field at x = 6.00m is 3.50 V/m directed towards the positive x-axis.

How to find the potential in a region between x = 0 and x = 6.00 m?

The potential in a region between x = 0 and x = 6.00 m is V = a + bx, where a = 18.6 V and b = -3.50 V/m.

To find the potential, substitute the given values of a, b and x into the given formula;

Potential at x = 0:V = a + bx = 18.6 + (-3.5 × 0) = 18.6 V

Therefore, the potential at x = 0 is 18.6V.

Potential at x = 3.00m:V = a + bx = 18.6 + (-3.5 × 3.00) = 5.1 V

Therefore, the potential at x = 3.00m is 5.1V.

Potential at x = 6.00m:V = a + bx = 18.6 + (-3.5 × 6.00) = -8.4 V

Therefore, the potential at x = 6.00m is -8.4V.

(b) Finding the magnitude and direction of the electric field.

To find the electric field, take the derivative of the potential with respect to x;

E = -dV/dx = -b

The negative sign indicates that the electric field is directed towards the negative x-axis.

The magnitude of the electric field is equal to the absolute value of the slope of the graph of the potential versus distance.

Electric field at x = 0:E = -dV/dx = -b = -(-3.50 V/m) = 3.50 V/m directed towards the negative x-axis

Therefore, the magnitude and direction of the electric field at x = 0 is 3.50 V/m directed towards the negative x-axis.

Electric field at x = 3.00m:E = -dV/dx = -b = -(-3.50 V/m) = 3.50 V/m directed towards the positive x-axis.

Therefore, the magnitude and direction of the electric field at x = 3.00m is 3.50 V/m directed towards the positive x-axis.

Electric field at x = 6.00m:E = -dV/dx = -b = -(-3.50 V/m) = 3.50 V/m directed towards the positive x-axis

Therefore, the magnitude and direction of the electric field at x = 6.00m is 3.50 V/m directed towards the positive x-axis.

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Answer:

Biologist >

Explanation:

A biologist is a scientist who examines organisms that are alive.

:)

Answer:

The force of gravitational attraction between the satellite and Earth is \(2.587\times 10^{10}\) newtons.

Explanation:

Statement is incorrect. Correct statement is:

A satellite (\(m = 4.44\times 10^{9}\,kg\)) travels in orbit around the Earth at a distance of \(1.9\times 10^{6}\,m\) above Earth's surface. What is the force of gravitational attraction between the satellite and Earth?

The gravitational force experimented by the satellite (\(F\)), in newtons, is calculated by Newton's Law of Gravitation, whose equation is defined by following formula:

\(F = \frac{G\cdot m \cdot M}{R^{2}}\) (1)

Where:

\(G\) - Gravitational constant, in cubic meters per kilogram-square second.

\(m\) - Mass of the satellite, in kilograms.

\(M\) - Mass of the Earth, in kilograms.

\(R\) - Distance of the satellite with respect to the center of the Earth, measured in meters.

If we know that \(G = 6.674\times 10^{-11}\,\frac{m^{3}}{kg\cdot s^{2}}\), \(m = 4.44\times 10^{9}\,kg\), \(M = 5.972\times 10^{24}\,kg\) and \(R = 8.271\times 10^{6}\,m\), then the force of gravitational attraction between the satellite and Earth is:

\(F = 2.587\times 10^{10}\,N\)

The force of gravitational attraction between the satellite and Earth is \(2.587\times 10^{10}\) newtons.

Answer:

0.25p.a

Explanation:

force=15N area=60m²

so pressure =force/area =15N/60m²=0.25p.a

Answer:

A, C, D

Explanation:

A, C and D are true.

B is false as energy can neither be created nor destroyed.

While measuring , there are a lot of errors which can happen due to which ""true distances"" be different for different groups even though they are measuring the same distance.

While measuring , there are a lot of errors which can happen , to find the most correct value in order to make the result precise and accurate number of measurements needed to be done in order to reduce the error

As nothing can be measured with full precision and accuracy, different measurements for different groups even though they are measuring the same distance.

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The formula for energy consumed to calculate the energy consumed by the motor in 23 seconds:

To calculate the energy consumed by the motor in operating the centrifuge for 23 seconds, we need to use the formula:

Energy Consumed = Power x Time

Since we are assuming that there is no loss of energy, the power consumed by the motor will be equal to the kinetic energy of the rotating parts of the centrifuge. We can calculate this using the formula:

Kinetic Energy = (1/2) x Moment of Inertia x Angular Velocity^2

We don't have information about the moment of inertia or the angular velocity, but we do have the data from a single measurement, which gives us the frequency of rotation. We can use this frequency to calculate the angular velocity:

Angular Velocity = 2π x Frequency

Substituting this value in the kinetic energy formula, we get:

Kinetic Energy = (1/2) x Moment of Inertia x (2π x Frequency)^2

So, to find the energy consumed by the motor for 23 seconds while operating the centrifuge without any sample, follow these steps and use the provided formula.

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Answer:

v₃ = 9.62[m/s]

Explanation:

To solve this type of problem we must use the principle of conservation of linear momentum, which tells us that the momentum is equal to the product of mass by velocity.

We must analyze the moment when the astronaut launches the toolkit, the before and after. In order to return to the ship, the astronaut must launch the toolkit in the opposite direction to the movement.

Let's take the leftward movement as negative, which is when the astronaut moves away from the ship, and rightward as positive, which is when he approaches the ship.

In this way, we can construct the following equation.

\(-(m_{1}+m_{2})*v_{1}=(m_{1}*v_{2})-(m_{2}*v_{3})\)

where:

m₁ = mass of the astronaut = 157 [kg]

m₂ = mass of the toolkit = 5 [kg]

v₁ = velocity combined of the astronaut and the toolkit before throwing the toolkit = 0.2 [m/s]

v₂ = velocity for returning back to the ship after throwing the toolkit [m/s]

v₃ = velocity at which the toolkit should be thrown [m/s]

Now replacing:

\(-(157+5)*0.2=(157*0.1)-(5*v_{3})\\(5*v_{3})= 15.7+32.4\\v_{3}=9.62[m/s]\)

The velocity with which the astronaut must throw the tool kit is 9.62 m/s.

The given parameters:

Apply the principle of conservation of linear momentum to determine the velocity of the tool kit;

\(m_1 u_1 + m_2u_2 = v(m_1 + m_2)\\\\-0.1(157) \ + 5u_2 = 0.2(157 + 5) \\\\-15.7 + 5u_2 = 32.4\\\\5u_2 = 32.4 + 15.7\\\\5u_2 = 48.1\\\\u_2 = \frac{48.1}{5} \\\\u_2 = 9.62 \ m/s\)

Thus, the velocity with which the astronaut must throw the tool kit is 9.62 m/s.

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To calculate the magnification of the Yerkes Telescope, we can use the following the formula:

Magnification = (-) Focal Length of Objective Lens / Focal Length of Eyepiece

Since the objective lens has a focal length of 19.4 m and the eyepiece has a focal length of 2.5 cm (or 0.025 m), we can plug these values into the formula:

Magnification = (-) 19.4 m / 0.025 m

Magnification = - 776

Therefore, the magnitude of the magnification of the Yerkes Telescope is 776, which means that it can magnify the objects 776 times their original size. Note that the negative sign indicates that the image is inverted.

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a. By the work-energy theorem, the total work required to stop the car is equal to the change in its kinetic energy,

W = 0 - 1/2 (750 kg) (25 m/s)² ≈ -230 kJ

b. The car covers a distance x as it stops such that

W = (-725 N) x   ==>   x ≈ 320 m

The x-component of the net force on q2 is -1.439 × 10^-10 N.

The given charges are Q1 = -4.60 * 10^-6 C, q2 = +3.75 * 10^-5 C, and q3 = -5.30 * 10^-6 C. We need to find the x-component of the net force on q2.

The formula for force is F = K(q1 * q2) / r^2, where K is Coulomb's constant = 8.99 × 10^9 Nm^2/C^2, q1 and q2 are the charges in Coulombs, and r is the separation distance in meters.

The force between q2 and Q1 on the x-axis can be calculated as:

F1 = K(Q1 * q2) / r^2 ... (i)

Here, K = 8.99 × 10^9 Nm^2/C^2, Q1 = -4.60 × 10^-6 C, q2 = 3.75 × 10^-5 C, and r = 0.05 m.

Calculating F1:

F1 = (8.99 × 10^9) * (-4.60 × 10^-6) * (3.75 × 10^-5) / (0.05)^2

F1 = -1.242 × 10^-10 N

Similarly, the force between q2 and q3 on the x-axis can be calculated as:

F2 = K(q2 * q3) / r^2 ... (ii)

Here, K = 8.99 × 10^9 Nm^2/C^2, q2 = 3.75 × 10^-5 C, q3 = -5.30 × 10^-6 C, and r = 0.06 m.

Calculating F2:

F2 = (8.99 × 10^9) * (3.75 × 10^-5) * (-5.30 × 10^-6) / (0.06)^2

F2 = -1.971 × 10^-11 N

The net force can be calculated by adding F1 and F2, considering their x-components:

Fnet = F1 + F2

Fnet = (-1.242 × 10^-10) + (-1.971 × 10^-11)

Fnet = -1.439 × 10^-10 N

As both forces act in opposite directions along the x-axis, the net force is in the negative x-direction.

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An instrument capable of measuring and displaying electrical impulses is the Electrocardiogram (EKG).  That instrument using electrodes placed on the skin.

In the term of science, Electrocardiography generally can be defined as the process of producing an electrocardiogram, a recording of the heart's electrical activity. Electrocardiography also can be defined as an electrogram of the heart which is a graph of voltage versus time of the electrical activity of the heart using electrodes placed on the skin. The main principle of  EKG or Electrocardiogram is by records impulses to show how fast the heart is beating, the rhythm of the heartbeat (steady or irregular), and the strength and timing of the electrical impulses as they move through different parts of the heart.

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Answer:

60miles/h=96560.6m/3600s=27m/s

sf=1/2at^2+v0t+s0

vf=at+v0

t=-v0/a

1/2(v0^2/a)-v0^2/a+s0=0

(v0^2/2a)-v0^2/a=-s0

(V0^2-2V0^2)/2a=-s0

a=-729/-200=4m/s^2

Answer:

Covalent bond.

Explanation:

There are 4 main types of bonds:

Covalent, ionic, metallic, and hydrogen.

Covalent bond: Involves the sharing of pairs of electrons, here the difference between the electronegativity of the atoms is not too large. Covalent bonds usually form an octet of electrons.

Ionic bond: This happens because the electrostatic attraction between the atoms whit very different electronegativities

Hydrogen bond: Electrostatic attractive force between an electronegative atom and a hydrogen atom covalently bonded to another electronegative atom.

Metallic bond: Type of bond that makes the metallic atoms to stay really tightly together. The atoms bond because of the electrostatic atractive force between conduction electrons and positively charged metal ions.

Now, in this case, we have the bond between Nitrogen (electronegativity = 2.0) and Hydrogen (electronegativity = 2.1)

So we can see that:

The elements are not metals, so we can discard metallic bond.

For a hydrogen bond, we need 3 atoms (one of which is hydrogen), here we have two, so we can discard this option.

Ionic bond needs different electronegativities, here the electronegativities are really close together, so the ionic bond can be discarded.

we can conclude that the bond will be a covalent bond.

Given :

\(m_1 = 82 \ kg\\\\v_1 = 257 m/s\\\\m_2 = 60 kg\)

Let, us assume that momentum is conserved during the process.

To Find :

The velocity \(v_2\) .

Solution :

Applying conservation of momentum :

Initial momentum = Final momentum

\(m_1v_1 = m_2v_2\\\\82\times 257=60\times v_2\\\\v_2 = \dfrac{82\times 257}{60}\ m/s\\\\v_2 = 351.23\ m/s\)

Hence, this is the required solution.