A 180 g model airplane charged to 18 mC and traveling at 2.2 m/s passes within 8.6 cm of a wire, nearly parallel to its path, carrying a 30 A current. What acceleration (in g's) does this interaction give the airplane?

Answers

Answer 1

Answer:

[tex]a=0.2*10^{-5}g[/tex]

Explanation:

From the question we are told that:

Mass [tex]M=180=>0.18kg[/tex]

Charge [tex]Q=18mC=18*10^-^3C[/tex]

Velocity [tex]v=2.2m/s[/tex]

Length of Wire [tex]L=8.6cm=>0.086[/tex]

Current [tex]I=30A[/tex]

Generally the equation for Magnetic Field of Wire B is mathematically given by

 [tex]B=\frac{\mu_0*I}{2\pi*l}[/tex]

 [tex]B=\frac{4*3.14*10^-^7*I}{2*3.14*8.6}[/tex]

 [tex]B=6.978*10^{-5}T[/tex]

Generally the equation for Force on the plane F is mathematically given by

 [tex]F=qvB[/tex]

Therefore

 [tex]ma=qvB[/tex]

 [tex]a=\frac{qvB}{m}[/tex]

 [tex]a=\frac{18*10^{-5}83.4*6.978*10^{-5}}{0.18kg}[/tex]

 [tex]a=2.37*10^{-5}[/tex]

Therefore in Terms of g's

 [tex]a=\frac{2.37*10^{-5}}{9.8}[/tex]

 [tex]a=0.2*10^{-5}g[/tex]


Related Questions

How can magnetic levitation be improved?

Answers

https://www.researchgate.net/publication/335238337_A_New_Strategy_for_Improving_the_Tracking_Performance_of_Magnetic_Levitation_System_in_Maglev_Train/fulltext/5d5a958d299bf1b97cf546ba/A-New-Strategy-for-Improving-the-Tracking-Performance-of-Magnetic-Levitation-System-in-Maglev-Train.pdf?origin=publication_detail


The higher the phase margin the more stable is the system and for these tuned parameters, the phase margin is around
. Some researcher given their theory on the phase margin that there are changes of getting sluggish response for larger phase margin but using TLBO algorithm the settling time and as well as peak overshoot of the system shows better response as compared to conventional techniques.

In a certain region of space near earth's surface, a uniform horizontal magnetic field of magnitude B exists above a level defined to be y = 0. Below y = 0 , the field abruptly becomes zero (seethe figure). A vertical square wire loop has resistivity rho mass density rhom, diameter d, and side length l. It is initially at rest with its lower horizontal side at y = 0 and is then allowed to fall under gravity, with its plane perpendicular to the direction of the magnetic field.
a) While the loop is still partially immersed in the magnetic field (as it fallsinto the zero-field region), determine the magnetic "drag" forcethat acts on it at the moment when its speed is v.
b) Assume that the loop achieves a terminal velocity vt before its upper horizontal side exits the field. Determine a formulafor vt
c) If the loop is made of copper and B = 0.80 T find vt

Answers

Answer:

a) F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]  

b) attached below

c) 0.037 m/s

Explanation:

a) Determine the magnetic "drag" force acting  at the moment

speed = v

first step: determine current in the loop

I = [tex]\frac{\pi d^2}{16pl} B lv[/tex]   ----- ( 1 )

given that the current will induce force on the three sides of the loop found in the magnetic field

forces on vertical sides = + opposite

we will cancel out

hence equation 1 becomes

F = [tex]\frac{\pi d^2B^2lv}{16p}[/tex]   ( according to Lenz law we can say that the direction of force is upwards and this force will slow down the decrease in flux )

b) Determine the formula for Vt

attached below

c) Find Vt

given :

B = 0.80 T

density of copper = 8.9 * 10^3 kg/m^3

resistivity of copper = 1.68 * 10^-8 Ωm

∴ Vt = 16 ( 8.9 * 10^3 kg/m^3 ) ( 1.68 * 10^-8 Ωm ) ( 9.8 m/s^2 ) / ( 0.08 T)^2

       = 0.037 m/s

Calculate the Combined resistance of the Circuit voltage across each resistor Current Passing through each resistor of 6,8,12ohms​

Answers

Answer:

Sorry I don't know the answer

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Answers

The question is incomplete. The complete question is :

Assume that the energy lost was entirely due to friction and that the total length of the PVC pipe is 1 meter. Use this length to compute the average force of friction (for this calculation, you may neglect uncertainties).

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

Solution :

Given that :

The predicted range is 0.3503 m

Time of the fall is :

[tex]$t=\sqrt{\frac{2H}{g}}$[/tex]

[tex]v_1t= 0.35[/tex]  ...........(i)

[tex]v_0t= 1.09[/tex]  ...........(ii)

Dividing the equation (ii) by (i)

[tex]$\frac{v_0t}{v_1t}=\frac{1.09}{035} = 3.11$[/tex]

∴ [tex]v_0=3.11 \ v_1[/tex]

Now loss of energy  = change in the kinetic energy

[tex]$W=\frac{1}{2} m [v_0^2-v_1^2]$[/tex]

[tex]$W=\frac{1}{2} \times (16.3 \times 10^{-3}) \times [v_0^2-\left(\frac{v_0}{3.11}\right)^2]$[/tex]

[tex]$W=7.307\times 10^{-3} \ v_0^2$[/tex]

If f is average friction force, then

(f)(L) = W

(f) (1) = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

(f)  = [tex]$7.307\times 10^{-3} \ v_0^2$[/tex]

The Average force of friction is ( F )  = 7.307 * 10⁻³ v₀²

Given data:

Predicted range ( v₁t ) = 0.3503 m

Actual range ( v₀t ) = 1.09 m

mass = 16.3 g

First step : Determine the value of  V₀

[tex]t = \sqrt{\frac{2H}{g} }[/tex]    ,    v₁t  =  0.3503 ,    ( v₀t ) = 1.09 m

To obtain the value of  V₀  

Divide ( v₀t ) by ( v₁t )  =  1.09 / 0.3503 = 3.11 v₁

V₀ = 3.11 v₁

Next step : Determine the average force of friction ( f )

given that loss of energy results in a change in kinetic energy

W = [tex]\frac{1}{2} m ( vo^{2} - v1^{2} )[/tex]

    = 1/2 * 16.3 * 10⁻³ * [ v₀² - [tex](\frac{v_{0} }{3.11} )^{2}[/tex] ]

W = 7.307 * 10⁻³ v₀²

Average force of friction = W / Actual length

                                         = 7.307 * 10⁻³ v₀² / 1  

∴ Average force of friction ( F )  = 7.307 * 10⁻³ v₀²

Hence we can conclude that the average force of friction is 7.307 * 10⁻³ v₀²

Learn more about average force of friction : https://brainly.com/question/16207943

Your question has some missing data below are the missing data related to your question

Mass of the ball :  16.3 g

Predicted range :  0.3503 m

Actual range : 1.09 m

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. How much of the sample is left after 1.98 x 10^4 seconds?

a) 8.03 x 10^16 nuclei

b) 4.01 x 10^16 nuclei

c) 2.02 x 10^16 nuclei

d) 1.61 x 10^17 nuclei

Answers

Answer:

c) 2.02 x 10^16 nuclei

Explanation:

The isotope decay of an atom follows the equation:

ln[A] = -kt + ln[A]₀

Where [A] is the amount of the isotope after time t, k is decay constant, [A]₀ is the initial amount of the isotope

[A] = Our incognite

k is constant decay:

k = ln 2 / Half-life

k = ln 2 / 4.96 x 10^3 s

k = 1.40x10⁻⁴s⁻¹

t is time = 1.98 x 10^4 s

[A]₀ = 3.21 x 10^17 nuclei

ln[A] = -1.40x10⁻⁴s⁻¹*1.98 x 10^4 s + ln[3.21 x 10^17 nuclei]

ln[A] = 37.538

[A] = 2.01x10¹⁶ nuclei remain ≈

c) 2.02 x 10^16 nuclei

FROM THE _____ WHOLE WATER CYCLE STARTS ALL OVER AGAIN

Answers

Clouds? I am not sure of your options!

From the water whole water cycle starts again.

Most possibly water should be the answer.

A Michelson interferometer operating at a 400 nm wavelength has a 3.70-cm-long glass cell in one arm. To begin, the air is pumped out of the cell and mirror M2 is adjusted to produce a bright spot at the center of the interference pattern. Then a valve is opened and air is slowly admitted into the cell. The index of refraction of air at 1.00 atm pressure is 1.00028.
How many bright-dark-bright fringe shifts are observed as the cell fills with air?

Answers

Answer:

[tex]m=42\ fringes[/tex]

Explanation:

From the question we are told that:

Wavelength [tex]\lambda=400nm[/tex]

Length of cell arm [tex]h=3.70cm[/tex]

Refraction of air at  at 1.00 atm pressure [tex]n=1.00028.[/tex]

Generally the equation for Number of shifts is mathematically given by

[tex]m=N-N_o[/tex]

Since

[tex]N_0=\frac{2t}{\lambda_0}[/tex]

Therefore

[tex]m=\frac{2t}{\lambda_0/n}-\frac{2t}{\lambda_0}[/tex]

[tex]m=\frac{2t}{\lambda_0} n-1[/tex]

[tex]m=\frac{2(3.7*10^{-2})}{400*10^{-9}}*(1.00028-1)[/tex]

[tex]m=51.8[/tex]

[tex]m=42\ fringes[/tex]

Cual es l diferencia entre ruido y sonido

Answers

Answer:

E.l soni.do es un.a sensac.ión, en el órg.ano del oído, prod.ucida por el movimie/nto ondu>latorio de un m/edio elástico (normal/mente el aire), debi.do a ra.pidísimos ca/mbios de pre.sión, generado/s por el movimiento vibrat.orio d.e un cuerpo sonoro. ... /El ruido se consid/era a to/do sonid.o / o no de.seado.

Explanation:

An astronaut throws a wrench in interstellar space. How much force is required to keep the wrench moving continuously with constant velocity?
A.
a force equal to its weight on Earth
B.
a force equal to zero
C.
a force equal to half of its weight on Earth
D.
a force equal to double its weight on Earth

Answers

Answer:

0 N

Explanation:

This is a trick question, the mass of the wrench would be 0 due to it being in space and has no gravitational pull to weight it down. And since acceleration is defined as the rate and change of velocity with no respect of time and the wrench is moving at a constant velocity, that means the velocity is 0. and since F = m*a it would be F = 0 * 0 = 0 N

A car of mass 1000 kg is moving at 25 m/s. It collides with a car of mass 1200 kg moving at 30 m/s. When the cars collide, they stick together. What is the total momentum of the system after the collision? What is the total momentum of the system before the collision? What is the velocity of the cars after the collision?

Answers

Answer:

The total momentum of the cars before the collision is 61,000 kg.m/s

The total momentum of the cars after the collision is 61,000 kg.m/s

The velocity of the cars after the collision is 27.727 m/s

Explanation:

Given;

mass of the first car, m₁ = 1000 kg

initial velocity of the car, u₁ = 25 m/s

mass of the second car, m₂ = 1200 kg

initial velocity of the second car, u₂ = 30 m/s

The common velocity of the cars after collision = v

The total momentum of the cars before collision is calculated as;

P₁ = m₁u₁  +  m₂u₂

P₁ = (1000 x 25)  +  (1200 x 30)

P₁ = 61,000 kg.m/s

The total momentum of the cars after collision is calculated as;

P₂ = m₁v + m₂v

where;

v    is the common velocities of the cars after collision since they stick together.

P₂ = v(m₁ + m₂)

To determine "v" apply the principle of conservation of linear momentum for inelastic collision.

m₁u₁  +  m₂u₂  = v(m₁  + m₂)

(1000 x 25)  +  (1200 x 30) = v(1000 + 1200)

61,000 = 2,200v

v = 61,000/2,200

v = 27.727 m/s

The total momentum after collsion = v(m₁ + m₂)

                                                         = 27.727(1000 + 1200)

                                                          = 61,000 kg.m/s

Thus, momentum before and after collsion are equal.

In higher mass stars, repeating cycles of fusion will create heavier elements in layers
until which element is created at the center of the core?

hydrogen

iron

uranium

helium

Answers

the element is iron.

pls help! George pushes a wheelbarrow for a distance of 12 meters at a constant speed for 35 seconds by applying a force of 20 newtons. What is the
power applied to push this wheelbarrow?
A. 1.2 watts
B. 3.4 watts
C. 6.9 watts
D. 13 watts

Answers

Answer:

C. 6.9 watts

Explanation:

Power = work/time

if work = force×distance...

Then... power= (force×distance)/time

Power = (20×12)/35

= 6.9 watts

If a small child swallowed a safety pin, why
would an X-ray photograph clearly show the
location of the pin?

Answers

Answer:

yes

Explanation:

it is in the body system

Answer:

it would show clearly because it is a metal piece in the body.


What is Velocity in physics

Answers

Answer:

hii

Explanation:

i hope this helps you

Answer:

The velocity of an object is the rate of change of its position with respect to a frame of reference, and is a function of time. ... Velocity is a physical vector quantity; both magnitude and direction are needed to define it

Explanation:

hope it helps

pls maek me as brainliest thanks❤

A 45.00 kg person in a 43.00 kg cart is coasting with a speed of 19 m/s before it goes up a hill. there is no friction, what is the maximum vertical height the person in the cart can reach?

Answers

Answer:

the maximum vertical height the person in the cart can reach is 18.42 m

Explanation:

Given;

mass of the person in cart, m₁ = 45 kg

mass of the cart, m₂ = 43 kg

acceleration due to gravity, g = 9.8 m/s²

final speed of the cart before it goes up the hill, v = 19 m/s

Apply the principle of conservation of energy;

[tex]mgh_{max} = \frac{1}{2}mv^2_{max}\\\\ gh_{max} = \frac{1}{2}v^2_{max}\\\\h_{max} = \frac{v^2_{max}}{2g} \\\\h_{max} =\frac{(19)^2}{2\times 9.8} \\\\h_{max} = 18.42 \ m[/tex]

Therefore, the maximum vertical height the person in the cart can reach is 18.42 m

A magnetic field of 0.276 T exists in the region enclosed by a solenoid that has 517 turns and a diameter of 10.5 cm. Within what period of time must the field be reduced to zero if the average magnitude of the induced emf within the coil during this time interval is to be 12.6 kV

Answers

Answer:

The period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Explanation:

Given;

initial value of the magnetic field, B₁ = 0.276 T

number of turns of the solenoid, N = 517 turns

diameter of the solenoid, d = 10.5 cm = 0.105 m

induced emf, = 12.6 kV = 12,600 V

when the field becomes zero, then the final magnetic field value, B₂ = 0

The induced emf is given by Faraday's law;

[tex]emf = -\frac{NA\Delta B}{t} \\\\emf = -\frac{NA (B_2 -B_1)}{t} \\\\t = -\frac{NA (B_2 -B_1)}{emf}\\\\t = \frac{NA (B_1 -B_2)}{emf}\\\\where;\\\\t \ is \ the \ time \ when \ B = 0 \ \ (i.e\ B_2 = 0)\\\\A \ is \ the \ area \ of \ the \ coil\\\\A = \frac{\pi d^2}{4} = \frac{\pi (0.105)^2}{4} = 0.00866 \ m^2\\\\t= \frac{(517) \times (0.00866)\times (0.276 -0)}{12,600}\\\\t = 9.81 \times 10^{-5} \ s[/tex]

Therefore, the period the field must be reduced to zero is 9.81 x 10⁻⁵ s

Lighting is the movement of?

Answers

Explanation:

Movement:refers to the changing in the lights whether it be a change in intensity, color or direction of origin.

Space debris left from old satellites and their launchers is becoming a hazard to other satellites. (a) Calculate the speed in m/s of a satellite in an orbit 980 km above the Earth's surface.

Answers

Answer:

564

Explanation:

The viscid silk produced by the European garden spider (Araneus diadematus) has a resilience of 0.35. If 10.0 J of work are done on the silk to stretch it out, how many Joules of work are released as thermal energy as it relaxes?

Answers

Answer: The energy released as thermal energy is 6.5 J

Explanation:

Energy stored by the spider when it relaxes is given by:

[tex]E_o=\text{Resilience}\times \text{Work}[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]E_o=0.35\times 10\\\\E_o=3.5J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\text{Work done}-E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5J[/tex]

Hence, the energy released as thermal energy is 6.5 J

The energy released as thermal energy when 10 J of work is done to stretch silk will be 6.5 J

What is thermal energy?

Thermal energy refers to the energy contained within a system that is responsible for its temperature. Heat is the flow of thermal energy.

Energy stored by the spider when it relaxes is given by:

[tex]\rm E_o=Resilience \ \times Work[/tex]

We are given:

Resilience = 0.35

Work done = 10.0 J

Putting values in above equation, we get:

[tex]\rm E_o=0.35\times 10[/tex]

[tex]E_o=3.5\ J[/tex]

Energy released at thermal energy is the difference between the work done and the energy it takes to relaxes, which is given by the equation:

[tex]E_T=\rm Work done -E_o[/tex]

Putting values in above equation, we get:

[tex]E_T=(10-3.5)=6.5\ J[/tex]

Hence, the energy released as thermal energy is 6.5 J

To know more about thermal energy follow

https://brainly.com/question/19666326

please help very easy 5th grade work giving brainliest

Answers

Answer:

the answer is option B because opposit sides of the magnets attract each other

make ansentance rkdloebebjekeoejbe​

Answers

Answer:

the man has returned from his trip

Answer:

just did by typing this lol

2. The given graph shows that the object is
(a) in non-uniform motion
(b) in uniform motion
(c) at rest
(d) in an oscillatory motion.
distance
time​

Answers

Answer:

(c) at rest

Explanation:

Given

See attachment for the distance time graph

Required

What does the graph illustrate?

From the graph, we can see that the line of distance is a horizontal line.

This suggests that a time increases, the distance remains unchanged

When distance remains unchanged over time, then it means the object is at rest.

Hence, (c) is correct

PLEASE HELP ME WITH THIS ONE QUESTION
The half-life of Barium-139 is 4.96 x 10^3 seconds. A sample contains 3.21 x 10^17 nuclei. What is the decay constant for this decay?

A) 1.67 x 10^-4 s^-1

B) 5.43 x 10^-4 s^-1

C) 1.40 x 10^-4 s^-1

D) 2.22 x 10^-4 s^-1

Answers

OPTION C is the correct answer.

according to the law of conservation of vhange , what must always be true in a nuclear reaction?​

Answers

Answer:

The Sum of mass and energy is always conserved in a nuclear reaction. Mass changes to energy, but the total amount of mass and energy combined remains the same

Explanation:

Every single radioactive decay, every single nuclear collision, every single nuclear reaction will conserve mass number and charge.

Equal masses of two different liquids are put into identical beakers.
Liquid 1 is heated for 100s and liquid 2 is heated for 200s by heaters of the same power.
The temperature of both liquids increases by the same amount.
Which statement is correct?
A Both liquids receive the same amount of energy.
B. Liquid 1 receives more energy than liquid 2.
C. Both liquids have equal thermal capacity.
D. The thermal capacity of liquid 1 is less than the thermal capacity of liquid 2.

Answers

Answer:

C

Explanation:

Because they both received the same temperature

From the top of the leaning tower of Pisa, a steel ball is thrown vertically downwards with a speed of 3.00 m/s. if the height of the tower is 200 m, how long will it take for the ball to hit the ground? Ignore air resistance.

Answers

Answer:

66,7 seconds

Explanation:

the formula for height/distance is : S=v.t

A ratio that compares the width and length of a garden is what type of model?

Answers

Answer:

physical

PLEASE MARK ME AS A BRAINLIEST

Answer: Mathematical

Explanation: I took the quiz

The bus travelled at velocity 15 ms-l for 5 minutes before it came to a stop. By using suitable linear equation, calculate the distance the bus has travelled.​

Answers

Answer:

ans: 2250 meters

Explanation:

initial velocity (U)= 15 m/s

final velocity (V) = 0m/s , since need to come in rest

total time taken (T) = 5 min= 300 seconds

total distance covered (S)= UT + 1/2 aT^2 ,

a= acceleration

S= 15 × 300 + 0.5 ×(0 - 15) × 300

since a = (V - U)/ T

S = 4500 - 2250

S= 2250 m

An unruly student with a spitwad (a lump of wet paper) of mass 20 g in his pocket finds himself in the school library where there is a ceiling fan overhead. He relieves his boredom by throwing the spitwad up at the ceiling fan where it collides with, and sticks to, the end of one of the blades of the stationary ceiling fan. Its horizontal velocity vector is perpendicular to the long axis of the blade. If the fan is free to rotate (no friction at all) and has moment of inertia I=1.4kgm2 , if the spitwad has horizontal velocity 4 m/s, and if the spitwad sticks to the fan blade at a distance of 0.6 m from the rotation axis of the fan, how much time will it take the fan to move through one complete revolution after the spitwad hits it (closest answer)?

a. 1min
b. 2min
c. 3min
d. 4min
e. 5min
f. 6min

Answers

Answer:

T = 188.5 s, correct is  C

Explanation:

This problem must be worked on using conservation of angular momentum. We define the system as formed by the fan and the paper, as the system is isolated, the moment is conserved

         

initial instant. Before the crash

        L₀ = r m v₀ + I₀ w₀

the angular speed of the fan is zero w₀ = 0

final instant. After the crash

        L_f = I₀ w + m r v

        L₀ = L_f

        m r v₀ = I₀ w + m r v

angular and linear velocity are related

        v = r w

        w = v / r

        m r v₀ = I₀ v / r + m r v

         m r v₀ = (I₀ / r + mr) v

       v = [tex]\frac{m}{\frac{I_o}{r} +mr} \ r v_o[/tex]

let's calculate

       v = [tex]\frac{0.020}{\frac{1.4}{0.6 } + 0.020 \ 0.6 } \ 0.6 \ 4[/tex]

       v = [tex]\frac{0.020}{2.345} \ 2.4[/tex]

       v = 0.02 m / s

         

To calculate the time of a complete revolution we can use the kinematics relations of uniform motion

        v = x / T

         T = x / v

the distance of a circle with radius r = 0.6 m

         x = 2π r

we substitute

         T = 2π r / v

let's calculate

         T = 2π 0.6/0.02

         T = 188.5 s

reduce

         t = 188.5 s ( 1 min/60 s) = 3.13 min

correct is  C

A 3.25-gram bullet traveling at 345 ms-1 strikes and enters a 2.50-kg crate. The crate slides 0.75 m along a wood floor until it comes to rest.

Required:
a. What is the coefficient of dynamic friction between crate and the floor?
b. What is the average force applied by the crate on the bullet during collision if the bullet penetrates the 1.10cm into the crate?

Answers

Answer:

a)   μ = 0.0136, b)   F = 22.8 N

Explanation:

This exercise must be solved in parts. Let's start by using conservation of moment.

a) We define a system formed by the downward and the box, therefore the forces during the collision are internal and the momentum is conserved

initial instant. Before the crash

        p₀ = m v₀

final instant. After inelastic shock

        p_f = (m + M) v

the moment is preserved

        p₀ = p_f

        m v₀ = (m + M) v

        v = [tex]\frac{m}{m + M} \ v_o[/tex]

We look for the speed of the block with the bullet inside

        v = [tex]\frac{0.00325}{0.00325 + 2.50 } \ 345[/tex]

        v = 0.448 m / s

Now we use the relationship between work and kinetic energy for the block with the bullet

in this journey the force that acts is the friction

         W = ΔK

          W = ½ (m + M) [tex]v_f^2[/tex]  - ½ (m + M) v₀²

the final speed of the block is zero

the work between the friction force and the displacement is negative, because the friction always opposes the displacement

         W = - fr x

we substitute

           - fr x = 0 - ½ (m + M) vo²

           fr = ½ (m + M) v₀² / x

         

the friction force is

          fr = μ N

          μ = fr / N

equilibrium condition

          N - W = 0

          N = W

          N = (m + M) g

we substitute

         μ = ½ v₀² / x g

we calculate

          μ = ½ 0.448 ^ 2 / 0.75 9.8

          μ = 0.0136

b) Let's use the relationship between work and the variation of the kinetic energy of the block

          W = ΔK

initial block velocity is zero vo = 0

         F x₁ = ½ M v² - 0

         F = [tex]\frac{1}{2} M \frac{x}{y} \frac{v^2}{x1}[/tex]

         F = ½ 2.50 0.448² / 0.0110

         F = 22.8 N

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