A locomotive creates a
59,400 N force, which creates
an acceleration of 0.145 m/s.
What is the mass of the
locomotive?
Unit=kg

Answer:

Explanation:

Let the velocity of projection be V .

component of velocity in vertically downward direction = v cos 44.1

component of velocity in horizontal direction = v sin 44.1

s = ut + 1/2 gt²

611 = v cos 44.1 x 5.4 + 1/2 x 9.8 x 5.4²

= 611 = 3.88 v + 142.88

v = 120.65 m /s

b )

horizontal component of velocity of projectile = v sin 44 .1 = 120.65 sin44.1

= 83.96 m /s

horizontal displacement = 83.96 x 5.4 = 453.38 m

c ) horizontal component will remain unchanged so horizontal component

=  v sin 44 .1 = 120.65 sin44.1

= 83.96 m /s

d ) velocity of projectile just before striking the ground

v = u + gt

= v cos 44.1  + 9.8 x 5.4

=  120.65 xcos 44.1  + 9.8 x 5.4

= 86.79 + 52.92

= 139.71 m /s

If the radius of a circle of area A and circumference C is doubled, the new circumference of the circle, in terms of C, will be equal to 2C.

The area (A) of a circle is given by:

\( A = \pi r^{2} \)   (1)

And a circumference (C) is:

\( C = 2\pi r \)   (2)

Where:

r: is the radius

If the radius of a circle and circumference is doubled, the new circumference of the circle is:

\( C_{2} = 2 (2 \pi r) = 2C \)

Therefore, the new circumference of the circle in terms of C is equal to 2C.

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Answer:

The organisms in a population may be distributed in a uniform, random, or clumped pattern. Uniform means that the population is evenly spaced, random indicates random spacing, and clumped means that the population is distributed in clusters.

Explanation:

Answer:

\(3.34 * 10^{-2}\)

Explanation:

Scientific notation is writing a number between 1 and 10, and multiplying the number by a power of 10.

Our number is 0.0334

Let's move the decimal point of the number to make it so that the number is at least 1 and less than 10.

We get: 3.34

We moved the decimal point 2 places to the right, so the power of 10 would be -2

Putting what we got together, the scientific notation for the number is:

\(3.34 * 10^{-2}\)

Answer:

4 joules

Explanation:

Answer:

Given: V=12 V

            I=2.5 mA

Let the resistance be R

By Ohm's Law, 

V=IR

12=2.5×10−3R

R=4.8×103 Ω

Answer:

What is the first velocity of the car with three washers

at the 0.25 meter mark? 0.19

What is the second velocity of the car with three

washers at the 0.50 meter mark? 0.45

Explanation:

Look at the question carefully.

The person above me got the answer wrong.

That answer is for a very similar question, but not this one.

(1) The first velocity of the car with two washers at the 0.25 meter mark is 0.125 m/s.

(2) The second velocity of the car with two washers at the 0.5 meter mark is 0.25 m/s.

Velocity is the rate of change of displacement with time. The velocity of the  cars depends on displacement and time of motion.

First velocity of the car at the 0.25 meter mark

The first velocity is calculated as follows;

v1 = 0.25 m / t1

let t₁ = 2 s

v₁ = 0.25/2

v₁ = 0.125 m/s

Second velocity of the car at the 0.5 meter mark

v2 = 0.25 m / (t2 – t1)

let t₂ = 3 s

v₂ = 0.25(3 - 2)

v₂ = 0.25 m/s

The ratio of distance covered by an object in a specific direction and the time taken to cover the distance is known as the velocity of the object. Mathematically, the expression for the velocity is,

v = d/t

Here, d is the distance covered.

And t is the average time taken to cover the distance.

Then the first velocity of the car at 0.25 m is,

v1 = d1/t1

v1 = 0.25 / t1

Here, t1 is the average time for the first distance.

And the second velocity of the car with four washers at the 0. 50 m mark is,

v2 = d2/t2

v2 = 0.50 /t2

Therefore, here, t2 is the average time for the second distance.

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Answer:

Force

Explanation:

Only a force can cause a stationary object to move or a moving object to change its speed or direction.

The time taken is approximately 2.08 seconds for the bag of sand to reach the ground from the moment of its release.

To determine the time it takes for the bag of sand to reach the ground, we can use the kinematic equation for vertical motion. The equation is given as:

h = ut + (1/2)gt^2

Where:

h = height (37.8 m)

u = initial velocity (0 m/s)

g = acceleration due to gravity (-9.81 m/s^2, considering downward motion)

t = time

Since the balloon is ascending with a constant upward velocity of 2.95 m/s, the initial velocity of the bag is also 2.95 m/s in the upward direction. Therefore, we need to consider the initial velocity as negative.

Substituting the known values into the equation, we have:

37.8 = (-2.95)t + (1/2)(-9.81)t^2

Simplifying the equation further, we get:

-4.905t^2 - 2.95t + 37.8 = 0

Solving this quadratic equation for time, we find two solutions: t = 2.08 s and t = -3.61 s. Since time cannot be negative, we discard the negative value.

Therefore, it takes approximately 2.08 seconds for the bag of sand to reach the ground from the moment of its release.

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Answer:

57 mm

Explanation:

57 mm is equivalent to 5.7 cm

The answer is  energy 3 times as much. option A.

To calculate the amount of energy released when a 5 kg object is converted to pure energy, we can use Einstein's famous equation: E = mc². In this equation, E represents energy, m represents mass, and c represents the speed of light.

Given that the mass of the object is 5 kg, we can calculate the energy using the equation:

E = (5 kg) * (c²)

Now, to compare this energy with the amount of energy the Earth receives from the Sun every second, we need to determine the Earth's solar energy input.

The solar constant is the amount of solar radiation received per unit area at the Earth's distance from the Sun. Its average value is approximately 1361 Watts per square meter (W/m²). Multiplying this value by the surface area of the Earth (approximately 510 million square kilometers), we can estimate the total energy received by the Earth from the Sun every second.

Energy from the Sun = (1361 W/m²) * (510,000,000,000 m²)

To compare the energy released from converting a 5 kg object to energy with the energy received from the Sun, we divide the former by the latter:

Energy conversion / Energy from the Sun = [(5 kg) * (c²)] / [(1361 W/m²) * (510,000,000,000 m²)]

Simplifying the equation, we find:

Energy conversion / Energy from the Sun = (5 kg * c²) / (1361 W/m² * 510,000,000,000 m²)

The value of c² is approximately (3x10^8 m/s)² = 9x10^16 m²/s².

Plugging in the values, we get:

Energy conversion / Energy from the Sun = (5 kg * 9x10^16 m²/s²) / (1361 W/m² * 510,000,000,000 m²)

Simplifying further:

Energy conversion / Energy from the Sun ≈ 3.52

Therefore, the amount of energy released when a 5 kg object is converted to pure energy is approximately 3.52 times larger than the amount of energy the Earth receives from the Sun every second.

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Answer:

\(d=5\ g/cm^3\)

Explanation:

Given that,

Mass of the object, m = 100 grams

Volume of the object, V = 20 cm³

We need to find the density of the object. We know that, density is equal to mass per unit volume. So,

\(d=\dfrac{m}{V}\\\\d=\dfrac{100\ g}{20\ cm^3}\\\\d=5\ g/cm^3\)

So, the density of the object is equal to \(5\ g/cm^3\).

Answer: A flower pot falling

Explanation:

The car on the curve (its direction is changing) and the falling flower pot (its speed is changing) are both undergoing acceleration.

Given that,

The earth's moon has a gravitational field strength of about 1.6 n/kg

Mass of Moon, \(M=7.35\times 10^{22}\ kg\)

To find,

The radius of the Moon.

Solution,

The formula for the acceleration due to gravity is given by :

\(g=\dfrac{GM}{r^2}\)

r is radius of the Moon

\(r=\sqrt{\dfrac{GM}{g}} \\\\r=\sqrt{\dfrac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.6}} \\\\r=1750437.44\ m\\\\r=1.75\times 10^6\ m\)

So, the radius of the Moon is \(1.75\times 10^6\ m\).

Hi there!

A.

Since the can was launched from ground level, we know that its trajectory forms a symmetrical, parabolic shape. In other words, the time taken for the can to reach the top is the same as the time it takes to fall down.

Thus, the time to its highest point:
\(T_h = \frac{T}{2}\)

Now, we can determine the velocity at which the can was launched at using the following equation:
\(v_f = v_i + at\)

In this instance, we are going to look at the VERTICAL component of the velocity, since at the top of the trajectory, the vertical velocity = 0 m/s.

Therefore:
\(0 = v_y + at\\\\0 = vsin\theta - g\frac{T}{2}\)

***vsinθ is the vertical component of the velocity.

Solve for 'v':
\(vsin(\alpha_0) = g\frac{T}{2}\\\\v = \frac{gT}{2sin(\alpha_0)}\)

Now, recall that:
\(W = \Delta KE = \frac{1}{2}m(\Delta v)^2\)

Plug in the expression for velocity:
\(W = \frac{1}{2}M (\frac{gT}{2sin(\alpha_0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{8sin^2(\alpha _0)}}\)

B.

We can use the same process as above, where T' = 2T and Th = T.

\(v = \frac{gT}{sin(\alpha _0)} }\\\\W = \frac{1}{2}M(\frac{gT}{sin(\alpha _0)})^2\\\\\boxed{W = \frac{Mg^2T^2}{2sin^2(\alpha _0)}}\)

C.

The work done in part B is 4 times greater than the work done in part A.

\(\boxed{\frac{W_B}{W_A} = \frac{4}{1} = 4.0}\)

A. To find the work done on the can by the launching device, we need to calculate the change in kinetic energy of the can. When the can is launched, it has an initial kinetic energy of 0 (since it starts from rest).

1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]

WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]

B. If the can stays in the air twice as long, the time of flight becomes 2T. We can repeat the above steps with the new value of T to find the initial velocity: Initial vertical velocity = g * (2T)

WB \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

C. To compare the results, we can calculate the ratio of WB to WA:

WB/WA \(= [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]] / [1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]]\)

Notice that the mass M and the angle α0 cancel out:

WB/WA \(= [((initial velocity)^2 * cos^2(α0) + (2gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)]\)

A.  At the highest point of its trajectory, the kinetic energy is again 0 (the vertical velocity component is 0). So the work done by the launching device equals the initial kinetic energy. The initial kinetic energy can be expressed as: Initial kinetic energy = \(1/2 * M * (initial velocity)^2\)

The initial velocity can be calculated using the launch angle and the time of flight T. At the highest point, the vertical component of the velocity is 0, so we only consider the horizontal component of the velocity. Horizontal component of velocity = initial velocity * cos(α0). The time T is the time taken to reach the highest point, so we can write:

T = time taken to reach the highest point = (initial vertical velocity) / g

Initial vertical velocity = g * T

Now, the initial velocity can be written as:

initial velocity = + (Initial vertical   \()√[(Horizontal component of velocity)^2\)                        

               =\(√[(initial velocity * cos(α0))^2 + (g * T)^2]\)

                =\(√[(initial velocity)^2 * cos^2(α0) + (g * T)^2]\)

Since the initial velocity is equal to the change in kinetic energy, we have: Initial kinetic energy :\(1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2] \\ WA = 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (g * T)^2]\)

B. Initial velocity:

                     \(√[(initial velocity * cos(α0))^2 + (g * (2T))^2]\\ = √[(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

Now, the initial kinetic energy for this case is: Initial kinetic energy (new) \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

WB \(= 1/2 * M * [(initial velocity)^2 * cos^2(α0) + (2gT)^2]\)

C. This shows that the mass and angle do not affect the work done on the can; only the time of flight and the acceleration due to gravity influence it. Since we know that T is doubled in part B (2T), we can write:

WB/WA  \(= [((initial velocity)^2 * cos^2(α0) + (2g * 2T)^2)] / [((initial velocity)^2 * cos^2(α0) + (g * T)^2)]\)

WB/WA \(= [((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)]\)

Now, we can see that WB is larger than WA by a factor of \([((initial velocity)^2 * cos^2(α0) + (4gT)^2)] / [((initial velocity)^2 * cos^2(α0) + (gT)^2)].\)

The value of this factor will depend on the specific values of the initial velocity, launch angle, and time of flight, but this ratio is greater than 1, indicating that more work is done on the can when it stays in the air for twice the time.

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therefore, the resistivity is 0.019210 miliOhms

Step 1

The resistance of a cylindrical segment of a conductor is equal to the resistivity of the material times the length divided by the area

so

Step 1

a)let

bI find the area

the area of a circle is given by:

so, replace

c) now, we can replace the values in the formula

therefore, the resistivity is 0.019210 miliOhms

I hope this helps yo u

No, the maximum range will not occur for θ = 45°. The maximum range occurs when the horizontal component of the initial velocity is equal to the initial vertical velocity at the instant the projectile hits the ground.

When a projectile is launched at an angle θ from a height h1, the time it takes to reach the highest point of its trajectory is given by t = (V0 sin θ)/g, where V0 is the initial velocity and g is the acceleration due to gravity. At the highest point, the vertical component of the velocity is zero and the horizontal component of the velocity remains constant. The time of flight of the projectile is given by T = (2V0 sin θ)/g.

The range of the projectile is given by R = (V0 sin 2θ)/g, where sin 2θ = 2sin θ cos θ. When the projectile lands at a height h2 that is higher than h1, the maximum range will occur at an angle that depends on the initial velocity and the difference in height between the launch and landing points. In general, the maximum range will occur at an angle that is less than 45° when the landing height is greater than the launch height, and at an angle that is greater than 45° when the landing height is less than the launch height. The specific angle can be calculated using the equation for range and setting the derivative with respect to θ equal to zero.

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Let the stick be divided into two parts, A and B, by the point where the two ladybugs are sitting. Let the length of part A be x and the length of part B be (1-x). Then, we have:

x + (1-x) = 1

Let the distance of the midpoint from the end of part A be d. Then, we have:

d = x/2 - (1/7)

Let the mass of the stick be M and the mass of each ladybug be m. Then, we have:M = 7m

The gravitational force acting on the system produces a clockwise moment about the end of part A, which is given by:

Mg(x/2 - d) = Mg(x/2 - (x/2 - 1/7)) = Mg/7

Let the distance of the first ladybug from the end of part A be L1 and the distance of the second ladybug from the end of part A be L2. Then, we have:

L1 = x

L2 = 1 - (1-x) = x

The moments produced by the ladybugs are given by:

mgL1sinθ

mgL2sinθ

mgL1sinθ = mgL2sinθ = Mg/7

Substituting the given values and solving for θ, we get:

sinθ = M/14m = 1/14

θ = 3.87 degrees

Mg(x/2 - d)sinθ

2mgL1sinθ

Substituting the given values and solving for x, we get:

x = 0.315

Substituting this value into the equation for the moment of inertia, we get:

I = 1.08e-5.

Inertia is the property of an object to resist any change in its state of motion. It is a measure of an object's resistance to changes in its velocity, including changes in direction and speed. Objects with more mass have more inertia, and they require more force to be moved or to stop moving. Inertia is described by Newton's first law of motion, which states that an object at rest will remain at rest, and an object in motion will remain in motion with a constant velocity, unless acted upon by an external force.

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Answer:

1.) The reaction is at dynamic equilibrium.

A: Nitrogen and hydrogen combine at the same rate that ammonia breaks down.

2.) Which statement about the reaction is necessarily correct?

A: Both calcium carbonate and sodium carbonate are being produced.

3.) Both calcium carbonate and sodium carbonate are being produced.

A: The reaction is reversible.

4.) What is the fastest motion that can be measured in any frame of reference?

A: 300,000 km/s

5.) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north.

Why are we able to use these motions to describe the motion relative to the ground?

A: The people are moving much slower than the speed of light so the ground acts as a frame of reference.

Explanation:

these were mine i hope this helps ! :0

Answer:

(Question) A child is on a playground they start to slide down a large slide. At what point is the child in dynamic equilibrium with the slide?

(Answer) As the child is in motion as they are sliding down.

(Question) Which statement correctly defines dynamic equilibrium?

(Answer) Forces acting on a object are balanced and the object stays in motion.

(Question) While you push a box you begin to decrease the force you are exerting on the box. When will the box reach static equilibrium?

(Answer) When F^push = F^friction

(Question) What is the fastest motion that can be measured in any frame of reference?

(Answer) 300,000 km/s

(Question) Two people are on a train that is moving at 10 m/s north. They are walking 1 m/s south relative to the train. Relative to the ground, their motion is 9 m/s north;   Why are we able to use these motions to describe the motion relative to the ground?  

(Answer) The people are moving much slower than the speed of light so the ground acts as a frame of reference.

Explanation:

just finished the quick check UwU

Answer:

a)   I = I₀/2, b)  I = I₀/2 cos² θ

Explanation:

To answer these questions, let's analyze a little the way of working of a polarized

* When a non-polarized light hits a polarizer, the electric field that is not in the direction of the polarizer is absorbed, so the transmitted light is

          i = I₀ / 2

and is polarized in the direction of the polarizer

* when a polarized light reaches the analyzer it must comply with Malus's law

          I = I₁ cos² θ

where the angle is between the polarized light and the analyzer.

With this, let's answer the questions

a) When a polarizer is placed in the non-polarized light path, half of it is absorbed and only the light that has polarization in the direction of the polarizer is transmitted with an intensity of

                  I = I₀/2

b) when a polarizer and an analyzer are fitted, the intensity of the light transmitted by the analyzer is

                I = I₀/2 cos² θ

where the final value depends on the angle between the polarizer and the analyzer.

Let's look at two extreme cases

θ = 0          I = Io / 2

θ = 90º      I = 0

Answer:

The new force becomes (1/9)th of the original force.

Explanation:

The gravitational force between two masses is given by :

\(F=G\dfrac{m_1m_2}{r^2}\)

Where

r is the distance between masses,

If the new distance is, r' = 3r

The new force is given by :

\(F'=G\dfrac{m_1m_2}{r'^2}\\\\F'=G\dfrac{m_1m_2}{(3r)^2}\\\\F'=\dfrac{1}{9}\times G\dfrac{m_1m_2}{r^2}\\\\F'=\dfrac{F}{9}\)

So, the new force becomes (1/9)th of the original force.

The point on the image that is the source of the sound / sound waves travelling would be B. 2.

When a sound is produced, such as a musical note or spoken word, it initiates a disturbance at the source, which sets the surrounding particles in motion. As these particles oscillate back and forth, they generate regions of compression and rarefaction, resulting in the formation of longitudinal waves.

The propagation of sound waves emanates from a central source, radiating outward in a spherical pattern, thereby creating an expanding wavefront. This fundamental principle of wave behavior stems from the point-source nature of sound waves, wherein they originate from a central locus and disperse in all directions.

This central source in the image is 2.

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Ice and snow in the polar regions play a crucial role in shaping the climate of these areas through various mechanisms.

Firstly, ice and snow have high albedo, meaning they reflect a significant portion of the incoming solar radiation back into space. This reflective property reduces the amount of solar energy absorbed by the Earth's surface, leading to cooler temperatures in the polar regions.

This phenomenon helps maintain the polar climate and contributes to the formation of ice caps and glaciers.

Secondly, the presence of ice and snow influences the movement of ocean currents and atmospheric circulation patterns. Melting ice and snow contribute to the formation of cold, dense water in polar regions, which sinks and initiates deep ocean currents.

These currents, such as the thermohaline circulation, play a vital role in redistributing heat globally and regulating climate patterns.

Furthermore, the vast ice sheets in polar regions act as a thermal insulator, preventing the transfer of heat between the atmosphere and the underlying ocean. This insulation effect helps maintain lower temperatures in the polar regions, influencing the overall climate and weather patterns.

Additionally, the freezing and melting of ice impact the salinity of the surrounding seawater. Changes in salinity affect the density of water, which can further influence ocean circulation and climate patterns.

In summary, the presence of ice and snow in the polar regions contributes to the climate by reflecting solar radiation, influencing ocean currents and atmospheric circulation, and acting as a thermal insulator. These factors play a significant role in shaping the unique climate conditions found in the polar regions.

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Answer:

6.7 seconds

Explanation:

d=(1/2)at^2

equation

1000=(1/2)45t^2.

substitute

2000=45t^2.

multiply by 2 for both sides

44.44=t^2.

divide both sides by 45

6.7=t

take the square root of both sides

Answer:

the required minimum thickness is 179.19 nm

Explanation:

Given the data in the question;

Refraction index of air n₁ = 1

Refraction index of sheet n₂ = 1.49

Refraction index of film n₃ = 1.62

wavelength of green light λ = 534 nm

In the data given, n₂ > n₃

so, this case has no pie-phase shift,

the condition for constructive interference will be;

mλ = 2Ln₂

L = mλ / 2n₂

so we substitute

L = m(534) / 2( 1.49 )

L = m( 534/ 2.98 )

L = m( 179.19 nm )

so for minimum value of L,

let m = 1

such that, L\(_{min\) will be;

L\(_{min\) = 1 × ( 179.19 nm )

L\(_{min\) = 179.19 nm

Therefore, the required minimum thickness is 179.19 nm

This question involves the concepts of the equation of motion and elastic collision.

The velocity of the ball as it rebounds from the floor will be "14 m/s".

The elastic collision is the one during which the momentum and kinetic energy of a body remain conserved. Hence, the velocity of the body also remains constant. Therefore, the rebound velocity of the ball will be the same as the striking velocity of the ball. This can be calculated using the third equation of motion.

\(2gh=v_f^2-v_i^2\)

where,

Therefore,

\((2)(9.81\ m/s^2)(10\ m)=v_f^2-(0\ m/s)^2\\v_f=\sqrt{196.2\ m^2/s^2}\\\\v_f=14\ m/s\)

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Answer:

The answer should be True

Explanation:

Answer:

False

Explanation:

Took the test and got it right

Although it is true that scientific models and hypotheses have been updated and improved over time, this is not sufficient justification to reject science as a whole or to doubt what scientists claim.

Models aid in our comprehension of systems and their characteristics. An atomic model, for instance, depicts what an atom's structure may like based on what is known about how atoms function. It may not accurately represent the precise makeup of an atom. Models are frequently condensed.

Over time, this atomic model has evolved. The model was used by scientists to make predictions. Their trials occasionally yielded unexpected outcomes that did not match the pre-existing model. The model was modified by scientists so that it could account for the fresh data.

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Answer: \(= (1.31i+2.42j)m/s\\\)

Explanation:

GIVEN DATA:

\(m1 = 2.03kg\\m2 = 3.02kg\\v1 = 1.90 i - 2.91 j) m/s\\v2 = (0.92 i -6.00 j) m/s\\\)

Solution:

the velocity for centre of the mass

\(vcm= \frac{m1v1+m2v2}{m1+m2}\)

input the values into the formula.

\(vcm=\frac{(2.03)(1.90 i - 2.91 j)+(3.02)(0.92 i -6.00 j) }{2.03+3.02}\)

open the bracket

\(= \frac{3.86i-5.91j +2.78i+18.12j}{5.05}\)

collect like terms

\(= \frac{6.64i+12.21j}{5.05}\)

\(= (1.31i+2.42j)m/s\)

total momentum

\(p= (m1+m2)vcm\\= (2.03+3.02) (1.31i+2.42j)\\= 5.05 (1.31i+2.42j)\\=(6.62i+12.22j)kg.m/s\)

Hi there!

We can begin by using Lenz's Law:
\(\epsilon = -N\frac{d\Phi _B}{dt}\)

N = Number of Loops

Ф = Magnetic Flux (Wb)
t = time (s)

Also, we can rewrite this as:
\(\epsilon = -NA\frac{dB}{dt}\)

A = Area (m²)

Since the area is constant, we can take it out of the derivative.

This is a single wire loop, so N = 1.

Now, we can develop an expression for the induced emf.

We can begin by solving for the area:

\(A = \pi r^2 \\\\d = r/2 r = 0.05cm \\\\A = \pi (0.05^2) = 0.007854 m^2\)

We can also express dB/dt as:
\(\frac{dB}{dt} = \frac{\Delta B}{t} = \frac{0-0.5}{t} = \frac{-0.5}{t}\)

Now, we can create an equation.

\(\epsilon = -(1)(0.007854)\frac{-0.5}{t} = \frac{0.003927}{t}\)

To solve the system, we must now develop an expression for current given an emf and resistance.

Begin by calculating the resistance of the copper wire:
\(R = \frac{\rho L}{A}\)

ρ = Resistivity of copper (1.72 * 10⁻⁸ Ωm)
L = Length of wire (0.01 m)

A = cross section area (m²)

Solve:
\(R = \frac{(1.72*10^{-8})(0.01)}{\pi (0.001^2)} = 5.475 * 10^{-5} \Omega m\)

Now, we can use the following relation (Ohm's Law):

\(\epsilon = iR\\\\\epsilon = \frac{Q}{t}R\)

*Since current is equivalent to Q/t.

Plug in the value of R and set the two equations equal to each other.

\(\frac{Q}{t}(5.475 * 10^{-5}) = \frac{0.003927}{t}\)

Cancel out 't'.

\(Q (5.475 * 10^{-5}) = 0.003927 \\\\Q = \frac{0.003927}{5.475*10^{-5}} = \boxed{71.73 C}\)