A drug tagged with 9943Tc (half-life = 6. 05 h) is prepared for a patient. If the original activity of the sample was 1. 0 ✕ 104 Bq, what is its activity (R) after it has been on the shelf for 1. 8 h?

The frequency change, in accordance with the presented description, is 636 Hz.

The quantity of occurrences of a recurring event per unit of time is its frequency. It is distinct from angular frequency, and for clarification, it is sometimes referred to as sampling rate. One occurrence per second, or hertz (Hz), is the unit of frequency.

Speed (v1) = 36 m/s

Speed (v2) = 45 m/s

Frequency (f) = 500 Hz

Speed of sound (s) = 343 m/s

We are aware of the modification of frequency formula, which

Frequency = s+v1/ s+v2(f) -------(1)

Put a value here and apply equation (1).

We'll use v2 = -45 since it's the other way.

Frequency = s+v1/ s-v2(f)

Frequency = 343+36 / 343-45(500)

Frequency = 636

Thus, the frequency shift is 636 Hz.

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The complete question is-

A truck moving at 36 m/s passes a police car moving at 45 m/s in the opposite direction. If the frequency of the siren relative to the police car is 500 Hz, what is the frequency heard by an observer in the truck after the police car passes the truck? (The speed of sound in air is 343 m/s.)

a). 396 Hz

b). 636 Hz

c). 361 Hz

d). 393 Hz

e). 617 Hz

Answer:

attached below

Explanation:

The initial conditions :

x(0) = - 9 inches = -3/4 ft

x'(0) = - 2 ft/sec

spring stretched  3 inches = 1/4 ft

mass = w / g = 4 Ib / 32 ft/sec^2 = 1/8  slug

the spring constant ( k ) = w / l  = 4 / ( 1/4 ) = 16 Ib/ft

applying the second law of motion

m d^2x/dt^2 +  b dx/dt  + kx

= 1/8 d^2x/dt^2  + 2 dx/dt  + 16 x

= d^2x/dt^2  + 16 dx/dt  +  128 x  ------- ( 1 )

we will resolve the above equation to obtain the required equation of motion  x( t )

Attached below is the remaining part of the solution

When two tuning forks are sounded together, they can produce a phenomenon known as a beat frequency. This is caused by the interference of sound waves produced by the two forks. When two sound waves with slightly different frequencies meet, they will produce a periodic variation in sound intensity. This variation in sound intensity is perceived as a beat frequency.

In this case, the two tuning forks have frequencies of 342 Hz and 345 Hz, respectively. The difference between these two frequencies is:

345 Hz - 342 Hz = 3 Hz

Therefore, the beat frequency will be 3 Hz. This means that the sound intensity will vary at a rate of 3 times per second.

The beat frequency can be calculated by subtracting the frequency of one tuning fork from the frequency of the other. The resulting difference is the beat frequency.

The phenomenon of beat frequency has many practical applications. For example, musicians use it to tune their instruments. They can adjust the pitch of their instruments by listening to the beats produced by two tuning forks and matching the desired frequency.

Beat frequency is also important in the field of acoustics. It is used to measure the frequency of sound waves, which can be useful in a variety of applications such as sonar and acoustic imaging.

In conclusion, the beat frequency produced by two tuning forks with frequencies of 342 Hz and 345 Hz, respectively, will be 3 Hz. This phenomenon is caused by the interference of sound waves produced by the two forks and has many practical applications in the fields of music and acoustics.

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Time taken to complete one revolution for disk-shaped space station spins at uniform rate is 17.381 seconds

Given diameter of the disk(d) = 150 m

Radius = d/2 = 75 m

acceleration (a) = g

Ω = sqrt(g/r)

            = sqrt(9.8/75) rad/s^2

            = 0.361 rad/s^2

Time taken to complete one revolution = 2π/Ω

= 17.381 seconds

A circular motion is a body's movement along a circle-shaped route. Uniform Circular Motion is the term used to describe a body travelling at a consistent speed along a circular route. Here, the velocity varies while the speed is constant.

If a particle is travelling in a circle, it must be experiencing some acceleration that is pushing it in that direction and causing it to rotate around the centre. The motion is uniformly circular because the acceleration, which is perpendicular to the particle's velocity at every time, only modifies the direction of the velocity, not its quantity. The force pushing in the direction of the centre is known as the centripetal force, and we refer to this acceleration as centripetal acceleration (or radial acceleration).

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Answer:

square root of 25 belongs to rational number

The current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.

Ohm's Law states that the voltage across an electric conductor is directly proportional to the current(I) passing through it provided the resistant is constant.

So;

V ∝ I

V = IR  

where

The objective of this question want us to determine: How did the current change for each test provided that Avery uses a 1.5-volt battery, then she uses a 3-volt battery and lastly she uses a 9-volt battery, given that the resistance is constant through out the whole process.

In the first experiment;

In the second experiment;

In the third experiment;

Therefore, we can conclude that the current in each experiment increases with increase in the voltage. Similarly, the association between resistance and the current in a circuit shows that increase in the resistance shows a reduction in the current, vice versa.

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Answer:

A

Explanation:

because it is accelerating and as the vehicle accelerates the speed increas

Explanation:

If there are more reactants than products, it is a synthesis reaction. If oxygen is a part of the reactant, it is a combustion reaction. If one ion replaces another, it is a single replacement reaction. If there are two compounds in a reactant where cations switch, it is a double replacement reaction.

The end A of the rod react when the (re)charged ball approaches it after a great many previous contacts with end A is strongly repelled (option 1)

When a charged body comes into contact with a neutral body, electrons flow from the charged body to the neutral body. This is referred to as the generation of static charges.

According to the given problem,

Assuming the conducting rod is not grounded, a negative charge accumulates on both ends of the rod. Charges, on the other hand, cannot stay at the ends and must be distributed throughout the length.

The charged ball will now be strongly repelled by the rod as it approaches it after many previous contacts with end A.

As a result, we can conclude that the rod's end A is strongly repelled away from the charged ball as it approaches it.

The question is incomplete, it should be:

How does end A of the rod react when the charged ball approaches it after a great many previous contacts with end A? Assume that the phrase "a great many" means that the total charge on the rod dominates any charge movement induced by the near presence of the charged ball.

It is strongly repelled.

It is strongly attracted.

It is weakly attracted.

It is weakly repelled.

It is neither attracted nor repelled.

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1. The variation in temperature

2. The material's heat transfer coefficient

are two properties we have to look at.

By taking an example;

Use a circular rod made of a certain material (for example, steel) that is insulated all the way around.

One end of the rod is immersed in a huge reservoir of 100°C water, while the other is immersed in water at 40°C.

The cold water is kept in an insulated cylinder on both sides.

The temp of the chilly water is measured using a meter as a time - dependent.

Conclusion of experiment;

Heat is transferred from a hot location to a cooler region. Whenever heat is applied to a body, its thermal power rises, and its temperature rises.

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Cuando se sumerge una piedra en un recipiente con agua, la piedra experimenta dos fuerzas: su peso hacia abajo y una fuerza de empuje hacia arriba que equivale al peso del agua desalojada por la piedra.

Esto se debe al principio de Arquímedes, que establece que un cuerpo sumergido en un fluido experimenta una fuerza de empuje igual al peso del fluido desplazado.

Cuando se sumerge la piedra en el agua, la fuerza de empuje actúa en sentido contrario a la fuerza de gravedad, lo que hace que la piedra parezca "más liviana" en el agua. La magnitud de la fuerza de empuje es igual al peso del agua desplazada por la piedra, según el principio de Arquímedes.

El principio de Arquímedes establece que un cuerpo sumergido en un fluido experimenta una fuerza de empuje dirigida hacia arriba y de magnitud igual al peso del fluido desplazado por el cuerpo. Esto ocurre porque el cuerpo desplaza una cantidad de fluido equivalente a su propio volumen.

En el caso de la piedra sumergida en agua, el volumen del agua desplazada por la piedra es igual al volumen de la piedra. La fuerza de empuje actúa hacia arriba y contrarresta parcialmente la fuerza de gravedad, lo que hace que la piedra parezca "más liviana" en el agua.

Es importante tener en cuenta que la fuerza de empuje depende del volumen del cuerpo y de la densidad del fluido en el que se sumerge. En este caso, al conocer la densidad del agua, podemos determinar la magnitud de la fuerza de empuje como igual al peso del agua desplazada por la piedra.

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Según el principio de Arquímedes, una piedra u otro objeto sumergido en agua experimentará una fuerza de empuje hacia arriba igual al peso del agua que desplaza. Esto hace que el objeto parezca más ligero en el agua que en el aire.

En física, el fenómeno que describes se llama el principio de Arquímedes. Este principio establece que un objeto sumergido en un fluido experimenta una fuerza de empuje hacia arriba que es igual al peso del fluido que desplaza. En este caso, la piedra sumergida en el agua experimentará una disminución en su peso debido a esta fuerza de empuje. Supongamos que la piedra tiene una densidad mucho mayor que el agua, por lo que se hundirá. Sin embargo, sentirá menos peso que en el aire porque el agua empuja hacia arriba contra ella con una fuerza igual al peso del agua que ha desplazado. Este efecto es por el cual los objetos parecen más ligeros cuando están en el agua.

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The current in the circuit is  ≈ 2.07 amperes (A)

A current circuit is a closed path through which electric current can flow. It typically consists of a power source, such as a battery or generator, a load, such as a light bulb or motor, and conductive wires or other components that connect the power source and load to form a complete path for the current to flow.

The current in the circuit can be calculated using the formula:

current = charge / time

In this case, the charge moved around the circuit is 60 coulombs and the time taken is 29 seconds. Thus, the current in the circuit is:

current = 60 coulombs / 29 seconds

current ≈ 2.07 amperes (A)

Therefore, The current in the circuit is  ≈ 2.07 amperes (A)

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Answer:

\(\boxed {\boxed {\sf a=42.8 \ m/s^2}}\)

Explanation:

Acceleration can be found by dividing the change in velocity by the time.

\(a=\frac{v-u}{t}\) (v is the final velocity, u is the initial velocity, t is the time).

The velocity increased from 22 m/s to 236 m/s in 5 seconds. Therefore:

\(v=236 \ m/s\\u=22 \ m/s\\t= 5 \ s\)

Substitute the values into the formula.

\(a=\frac{236 \ m/s - 22 \ m/s}{5 \ s}\)

Subtract in the numerator.

\(a=\frac{214 \ m/s}{5 \ s}\)

Divide.

\(a=42.8 \ m/s^2\)

The acceleration of the object is 42.8 meters per square second.

Answer:

Initial velocity = 4 m/s

Explanation:

Acceleration, a = 1.2 m/s²

Final velocity, v = 16 m/s

Initial velocity, u = ?

Time, t = 10 seconds

So, we must use the formula:

v = u + at

Insert all the values into the formula.

16 = u + 1.2 × 10

16 = u + 12

Make 'u' the subject of the equation, and solve to find the value of u.

u = 16 - 12

u = 4 m/s

Answer:

35.14 m/s

Explanation:

The Law of Conservation of Momentum states that the momentum before and after a collision is the same.

Let's set the ball to have the subscript of 1 and the man to have the subscript of 2.

The initial and final mass of the ball is the same, so m₁ = 7.00 kg on both sides of the equation.

The initial velocity of the ball, v₁ on the left side of the equation, is the unknown variable we are trying to find.

The initial and final mass of the man is the same, so m₂ = 75.0 kg on both sides of the equation.

The man starts at rest, meaning that his initial velocity is v₂ = 0 m/s on the left side of the equation.

The final velocity of both the ball and the man is 3.00 m/s, so we can set v₁ and v₂ on the right side of the equation to equal 3.00 m/s.

Left side of the equation:

Right side of the equation:

Substitute these values into the Law of Conversation of Momentum formula.

Multiply and simplify.

Divide both sides of the equation by 7.

The ball needs to be moving at a speed of 35.14 m/s in order to send the man off at a speed of 3.00 m/s.

The dryness fraction is 0.96. Using steam tables the enthalpies at points 2, 3, and 4 can be calculated as 2936.4 kJ/kg, 2892.3 kJ/kg, and 1039.2 kJ/kg, respectively. The value of q is found to be 1438.3 kJ/kg.

a) 0.2, b) 2687 kJ/kg, c) 0.16 kJ/kg.K, d) 32%, e) 2549.52 kJ/kgPart (a): The steam is superheated at 300°C and 3 MPa, using steam tables it can be seen that the dryness fraction is 0.96.Part (b): This can be calculated using the formula shown below. The net work done by the turbine is given as follows: Net work = m (h1 - h2) + m (h3 - h4) Where m is the mass of the steam entering the turbine and h1, h2, h3, and h4 are the enthalpies at the different points in the cycle.h1 is given as 3478 kJ/kg, and using steam tables the enthalpies at points 2, 3, and 4 can be calculated as 2936.4 kJ/kg, 2892.3 kJ/kg, and 1039.2 kJ/kg, respectively.

Substituting the values in the formula gives the answer as 2687 kJ/kg.Part (c): Heat transfer per unit mass of steam to the boiler can be calculated using the formula shown below:q = h1 - h4 The value of q is found to be 1438.3 kJ/kg. Part (d): The thermal efficiency of the cycle can be calculated using the formula shown below: Efficiency = Net work output/ Heat inputHeat input can be calculated as follows: Heat input = m (h1 - h4) + m (h3 - h2) The value of heat input is calculated to be 4485.4 kJ/kg Substituting the values in the formula gives the answer as 32%.Part (e): The heat transfer to cooling water passing through the condenser is given as follows:q = m (h4 - h5)Where h4 is 1039.2 kJ/kg and h5 is 48.72 kJ/kg. The value of q is calculated to be 2549.52 kJ/kg.

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The trip takes a total of 1.5 hours and has an average speed of 40 mph.

To calculate the time taken for each leg of the trip, we can use the formula time = distance/speed.

For the first leg of the trip, the car travels 15 miles at a speed of 45 mph. Using the formula, we find that the time taken for this leg is 15/45 = 0.33 hours.

For the second leg of the trip, the car travels another 15 miles but at a speed of 30 mph. Using the formula, we find that the time taken for this leg is 15/30 = 0.5 hours.

To find the total time for the trip, we add the times for each leg: 0.33 hours + 0.5 hours = 0.83 hours.

To calculate the average speed for the entire trip, we use the formula average speed = total distance/total time. The total distance traveled is 15 miles + 15 miles = 30 miles. The total time taken is 0.83 hours. Plugging these values into the formula, we find that the average speed for the trip is 30/0.83 = 36.14 mph.

Therefore, the trip takes a total of 1.5 hours and has an average speed of 40 mph.

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In the peak of summer, it hasn't rained in a week, and the plants are looking rough, so watering the plants for an hour is a good idea.

However, the next day, you come back to the garden, and the plants look in worse shape than they did previously, as if none of that water made it to the plant. Plants absorb water through their roots. The root system of a plant is responsible for drawing water and nutrients from the soil. A plant's root system must be able to absorb water quickly in order for the plant to grow and thrive. When the soil around the root system is dry, the roots will stop growing and will not be able to absorb water.

It may even start to die. Watering plants during the peak of summer is important because it will help keep the soil moist and prevent the roots from drying out. However, watering a plant too much can be harmful. If a plant is overwatered, the water may not be able to penetrate the soil and reach the roots. Instead, it may just sit on top of the soil, causing the roots to rot and die. This can cause the plant to wilt and die.To summarize, if the soil around the plant is too dry, the roots may not be able to absorb the water you gave them, causing the plant to look worse than before. Conversely, overwatering can also be harmful because the water may not be able to penetrate the soil and reach the roots, causing the roots to rot and die.

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a. The third sphere must be placed between A and B, closer to A.

b.  the third sphere must be placed at 175 cm from the first sphere.

Since

Using Newton's law of universal gravitation, the net force on the first sphere on the third sphere is F = mAmC/x² where x = distance between sphere C and A

Also, the Force of attraction between there second sphere and the third sphere is F' = mBmC/(R - x)² where R = distance between mA and mB and x = distance between mA and mC.

Since the net force must be zero, then

F = F'

mAmC/x² = mBmC/(R - x)²

So, (R - x)²/x² = mB/mC/mAmC

(R - x)²/x² = mB/mA

Substituting the values of mA and mB into the equation, we have

(R - x)²/x² = 7/2

(R - x)²/x² = 3.5

(R - x)/x = √3.5

(R - x)/x = 1.87

R - x = 1.87x

R = x + 1.87x

R = 2.87x

x = R/2.87

x = 0.35R

Since x = 0.35R, the third sphere must be placed between A and B, closer to A.

Since the position of the third sphere is

So, x = 0.35 × 500 cm

x = 175 cm

So, the third sphere must be placed at 175 cm from the first sphere.

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Starting from rest, the disc completes \(\theta\) revolutions after \(t\) seconds according to

\(\theta=\dfrac\alpha2t^2\)

with angular acceleration \(\alpha\). It completes 10 rev in 2 s, which means

\(10\,\mathrm{rev}=\dfrac\alpha2(2\,\mathrm s)^2\implies\alpha=5\dfrac{\rm rev}{\mathrm s^2}\)

Find the time it takes to complete 20 rev with this acceleration:

\(20\,\mathrm{rev}=\dfrac12\left(5\dfrac{\rm rev}{\mathrm s^2}\right)t^2\implies t=\sqrt8\,\mathrm s\approx2.83\,\mathrm s\)

so it takes approximately 0.83 s to complete 10 more rev.

The pilot should look to the right side of the aircraft. In aviation, the term "3 o'clock" refers to a clock face analogy where the nose of the aircraft is at noon.

Therefore, when the ATC radar facility advises "TRAFFIC 3 O'CLOCK," it means that the traffic is located on the right side of the aircraft. Additionally, the advisory states that the traffic is "2 miles, westbound," indicating that the traffic is moving in a westward direction from the pilot's perspective. Therefore, the pilot should look to the right side of the aircraft and scan the airspace for traffic, keeping in mind the specified distance and direction. The clock face analogy is a common method used in aviation to describe the position of aircraft or objects relative to the nose of an aircraft. In this analogy, the nose of the aircraft is considered the 12 o'clock position, and the rest of the directions are determined accordingly.

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Hi there!

We can use the conservation of momentum for an INELASTIC collision:

\(m_1v_1 + m_2v_2 = v_f(m_1 + m_2)\)

Let:

m1 = mass of van (5000 kg)

m2 = mass of car (? kg)

v1 = initial velocity of van (10 m/s)

v2 = initial velocity of car (0 m/s)

vf = final velocity of BOTH objects (6 m/s)

We can rearrange and plug in values to solve for m2:

\(m_1v_1 + m_2(0) = v_f(m_1 + m_2)\\\\m_1v_1 = v_f(m_1 + m_2)\\\\(5000)(10) = 6(5000 + m_2)\)

Solve:

\(50000 = 30000 + 6m_2\\\\20000 = 6m_2\\\\m_2 = 20000/6 = \boxed{3333.33 kg}\)

The acceleration of the particle after 1 minute is \(18 units/min^2.\)

To find the acceleration, we need to differentiate the position function twice with respect to time.

Given:

\(s(t) = 2t^3 + 3t^2 + t\)

To find the velocity, differentiate s(t) with respect to t:

\(v(t) = d/dt [2t^3 + 3t^2 + t]\)

\(v(t) = 6t^2 + 6t + 1\)

To find the acceleration, differentiate v(t) with respect to t:

\(a(t) = d/dt [6t^2 + 6t + 1]\)

\(a(t) = 12t + 6\)

Now, substitute t = 1 to find the acceleration after 1 minute:

\(a(1) = 12(1) + 6\)

\(= 12 + 6\)

\(= 18\)

Therefore, the acceleration after 1 minute is \(18 units/min^2\)

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The simplified expression for Q using De Morgan's laws is Q = A . (A' . C')'.

a) Truth tables for the NAND gate and NOR gate with two inputs:

NAND gate:

| A | B | Q |

|---|---|---|

| 0 | 0 | 1 |

| 0 | 1 | 1 |

| 1 | 0 | 1 |

| 1 | 1 | 0 |

NOR gate:

| A | B | Q |

|---|---|---|

| 0 | 0 | 1 |

| 0 | 1 | 0 |

| 1 | 0 | 0 |

| 1 | 1 | 0 |

b) Truth table and logic expression for Z in terms of inputs A, B, and C:

| A | B | C | Z |

|---|---|---|---|

| 0 | 0 | 0 | 1 |

| 0 | 0 | 1 | 0 |

| 0 | 1 | 0 | 1 |

| 0 | 1 | 1 | 0 |

| 1 | 0 | 0 | 1 |

| 1 | 0 | 1 | 0 |

| 1 | 1 | 0 | 0 |

| 1 | 1 | 1 | 0 |

Logic expression for Z: Z = (A' AND B' AND C) OR (A' AND B AND C')

c) Simplification of the Boolean expression Q = (A. (A + C))' using De Morgan's laws:

Q = (A. (A + C))'

Apply De Morgan's law: (AB)' = A' + B'

Q = (A' + (A + C)')'

Apply De Morgan's law again: (A + B)' = A' . B'

Q = ((A')' . (A + C)')'

Simplifying the double negations: (A')' = A and (A + C)' = A' . C'

Q = (A . (A' . C'))'

Final simplified expression: Q = A . (A' . C')'

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The air density Difference = 0.829 kg/ \(m^{3}\)

Elaborating the question :

Ideal gas equation : PV= nRT

           R = 8.314J/mol/k

n =\(\frac{m}{MW}\) where , n = moles of gas

                           m = mass of gas

                           MW = molecular weight

PMW =pRT      

               p = air density ,  T = -19 + 273k = 254k

p = \(\frac{PMW}{RT}\)

p = 3.30 × \(10^{4}\) ×0.0188kg ÷ 8.314 ×254 k

p = 0.450kg/m³

Density of air under standard condition = 1.28kg/m³

difference = 1.28 - 0.45 = 0.829kg/m³.

What is air density ?

Air density is defined as the mass of air per unit volume, whereas the specific volume is defined in terms of unit mass of dry air.The density of air or atmospheric density, denoted ρ, is the mass per unit volume of Earth's atmosphere. Air density, like air pressure, decreases with increasing altitude. It also changes with variation in atmospheric pressure, temperature and humidity.

Question is incomplete , missing part is given below :

What is the air density atop of Mount Everest on a summer day (at a pressure of 3.30 x 104 N/m2 and a temperature of -190C)? How is it different from the density of air in standard conditions (at a pressure of 1.01 x 105 N/m2 and a temperature of 200 C the air density is1.28 kg/m3). The molar mass of air is 28.8g/mole.

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From what we know, we can confirm that water must gain or lose large amounts of energy when its temperature changes due to its high specific heat.

This is another term used to describe heat or thermal capacity. These terms describe the amount of energy needed to change the temperature of a substance. Due to this, water requires a high amount of energy to raise its temperature by one degree.

Therefore, we can confirm that water must gain or lose large amounts of energy when its temperature changes due to its high specific heat.

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The distance from the base of the cliff is 13.67 m.

The impact velocity of the block when it falls off the cliff is 15.3 m/s.

Initial velocity of the block, u = 15 m/s

Mass of the block, m = 5 kg

Coefficient of friction, μ = 0.3

Height of the cliff, h = 12 m

μmg = ma

a = μg

a = 0.3 x 9.8

a = 2.94 m/s²

At the initial surface,

Total energy of the block = 1/2mu² - mgh'

E = m(u²/2 - gh)

E = 5(15²/2 - 9.8 x 5)

E = 5 x 63.5

E = 317.5J

The impact velocity of the block when it falls off the cliff is given by,

1/2 mv² = mgh

v = √2gh

v = √(2 x 9.8 x 12)

v = √235.2

v = 15.3 m/s

The distance from the base of the cliff is given by,

R = v√(2h/g)

R = 15.3 x√(2 x 13/9.8)

R = 15.3 x 1.628

R = 13.67 m

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