The landing gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described determine the values of k and c that limit the amplitude of vibration of the airplane (x) to 0.1 m. assume

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Answer 1

Answer: The land of airplane gear of an airplane can be idealized as the spring-mass-damper system shown in fig. 3.52. if the runway surface is described

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A 1850 kg car traveling at 13.8 m/s collides with a 3100 kg car that is initally at rest at a stoplight. The cars stick together and move 1.91 m before friction causes them to stop. Determine the coefficient of kinetic friction between the cars and the road, assuming that the negative acceleration is constant and all wheels on both cars lock at the time of impact.

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To solve this problem, it is necessary to apply the concepts related to the conservation of momentum, the kinematic equations for the description of linear motion and the definition of friction force since Newton's second law.

The conservation of momentum can be expressed mathematically as

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

Where,

$m_(1,2)$ = Mass of each object

$v_(1,2)$ = Initial Velocity of each object

$v_f$ = Final velocity

Replacing we have that,

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

$1850*13.8+3100*0 = (1850+3100)v_f$

$v_f = 5.1575m/s$

With the final speed obtained we can determine the acceleration through the linear motion kinematic equations, that is to say

$v_f^2-v_i^2 = 2ax$

Since there is no initial speed, then

$v_f^2 = 2ax$

$5.1575^2 = 2a (1.91)$

$a = 6.9633m/s^2$

Finally with the acceleration found it is possible to find the friction force from the balance of Forces, like this:

$F_f = F_a \n\mu N = m*a \n\mu = (ma)/(N)\n\mu = (ma)/(mg)\n\mu = (a)/(g)\n\mu = (6.9633)/(9.8)\n\mu = 0.7105$

Therefore the Kinetic friction coefficient is 0.7105

A 55-liter tank is full and contains 40kg of fuel. Find using Sl units: • Density p. • Specific Weight y • Specific Gravity Answer tolerance = 1%. Be sure to include units. The sign of the answers will not be graded, use a positive value for your answer. Your answers: p= (Enter a positive value) y = (Enter a positive value) SG = (Enter a positive value)

Answers

Answer:

Density of the fuel is 727.3 kilograms per cubic meter.

Specific weight of the fuel is 7127.3 Newtons per cubic meter.

Specific gravity of the fuel is 0,727.

Explanation:

In order to use SI units, we have to convert liters to cubic meters. Knowing that a liter is a cubic decimeter and a cubic decimeter is $1*10^(-3)$ cubic meters, we know that the tank has 0,055 cubic meters of fuel (because it is full).

Now that we have things in SI units, we calculate density:

$p_(fuel)= (mass)/(volume) = (40 kg)/(0.055 m^(3) ) =727.3 (kg )/(m^(3) )$

Knowing the mass per unit of volume, we can calculate weight per unit of volume thanks to Newton's second law (mass times acceleration, g in this case, equals force (weight)), i.e. specific weight:

$y=p*g=727,3 (kg)/(m^(3))*9.8(m )/(s^(2))=7127,3 (N)/(m^(3))$

With density we can also calculate how dense the fuel is related to a reference (water), i.e. specific gravity. SG is a dimensionless number that tell us how much denser (SG>1) or lighter per unit of volume (SG<1) a substance is than water. We use water as a reference because it is one of the most used substances in our life, and it is a standard density (1000 kg per cubic meter at 4°C and 1 atm).

$SG=(p_(fuel) )/(p_(water) ) =(727.3 (kg )/(m^(3) ))/(1000 (kg )/(m^(3) )) =0,727$

Select the correct answer.If the coefficient of static friction is 0.35 and the normal force is 80 newtons, what is the maximum frictional force of the surface acting on the object?
O A.
B.
C.
OD.
OE
9.8 newtons
28 newtons
80 newtons
23 newtons
35 newtons

Answers

Final answer:

The maximum frictional force acting on the object is 28 newtons.

Explanation:

The maximum frictional force acting on an object can be calculated using the formula:

Frictional force = coefficient of static friction x normal force

Given that the coefficient of static friction is 0.35 and the normal force is 80 newtons, we can calculate the maximum frictional force as follows:

Maximum frictional force = 0.35 x 80 = 28 newtons

Learn more about Calculating maximum frictional force here:

brainly.com/question/20054484

Two long parallel wires are a center-to-center distance of 1.30 cm apart and carry equal anti-parallel currents of 2.40 A. Find the magnitude of the magnetic field at the point P which is equidistant from the wires. (R = 5.00 cm).

Answers

Image is missing, so i have attached it

Answer:

19.04 × 10⁻⁴ T in the +x direction

Explanation:

We are told that the point P which is equidistant from the wires. (R = 5.00 cm). Thus distance from each wire to O is R.

Hence, the magnetic field at P from each wire would be; B = μ₀I/(2πR)

We are given;

I = 2.4 A

R = 5 cm = 0.05 m

μ₀ is a constant = 4π × 10⁻⁷ H/m

B = (4π × 10⁻⁷ × 2.4)/(2π × 0.05)

B = 9.6 × 10⁻⁴ T

To get the direction of the field from each wire, we will use Flemings right hand rule.

From the diagram attached:

We can say the field at P from the top wire will point up/right

Also, the field at P from the bottom wire will point down/right

Thus, by symmetry, the y components will cancel out leaving the two equal x components to act to the right.

If the mid-point between the wires is M, the the angle this mid point line to P makes with either A or B should be same since P is equidistant from both wires.

Let the angle be θ

Thus;

sin(θ) = (1.3/2)/5

θ = sin⁻¹(0.13) = 7.47⁰

The x component of each field would be:

9.6 × 10⁻⁴cos(7.47) = 9.52 × 10⁻⁴ T

Thus, total field = 2 × 9.52 × 10⁻⁴ = 19.04 × 10⁻⁴ T in the +x direction

Final answer:

The magnetic field at point P, which is equidistant from two long parallel wires with equal anti-parallel currents, is calculated using Ampere's law. The net magnetic field is zero because the fields due to each wire cancel each other at that point.

Explanation:

The question concerns the calculation of the magnetic field at a point equidistant from two long parallel wires that carry equal anti-parallel currents. According to the right-hand rule and Ampere's law, each wire generates a magnetic field that circles the wire. For two wires carrying currents in opposite directions, the magnetic fields at the midpoint between the wires will point in opposite directions, thus they will subtract from each other when calculating the total magnetic field at point P.

To find the magnetic field at point P, we use the formula for the magnetic field due to a long straight current-carrying wire: B = (μ₀I)/(2πd), where B is the magnetic field, μ₀ is the permeability of free space (4π x 10-7 T·m/A), I is the current, and d is the distance to the point of interest from the wire. In this case, the distance d will be the radius R = 5.00 cm since point P is equidistant from both wires.

Substituting the values into the formula, the magnetic field due to each wire at point P can be calculated. However, since the currents are anti-parallel, the net magnetic field at P would be the difference between the two fields, which is zero.

If the balloon takes 0.19 s to cross the 1.6-m-high window, from what height above the top of the window was it dropped?

Answers

Answer:

$heigth=2.86m$

Explanation:

Given data

time=0.19 s

distance=1.6 m

To find

height

Solution

First we need to find average velocity

$V_(avg)=(distance)/(time)\nV_(avg)=(1.6m)/(0.19s)\nV_(avg)=8.42m/s$

Also we know that average velocity

$V_(avg)=(V_(i)+V_(f))/2\n$

Where

Vi is top of window speed

Vf is bottom of window speed

Also we now that

$V_(f)=V_(i)+gt\nV_(f)=V_(i)+(9.8)(0.19)\nV_(f)=V_(i)+1.862$

Substitute value of Vf in average velocity

$V_(avg)=(V_(i)+V_(f))/2\nwhere\nV_(f)=V_(i)+1.862\nand\nV_(avg)=8.42m/s\nSo\n8.42m/s=(V_(i)+V_(i)+1.862)/2\n2V_(i)+1.862=16.84\nV_(i)=(16.84-1.862)/2\nV_(i)=7.489m/s\n$

Vi is speed of balloon at top of the window

Now we need to find time

$V_(i)=gt\nt=V_(i)/g\nt=7.489/9.8\nt=0.764s$

So the distance can be found as

$distance=(1/2)gt^(2)\n distance=(1/2)(9.8)(0.764)^(2)\n distance=2.86m$

wo charged spheres are 1.5 m apart and are exerting an electrostatic force (Fo) on each other. If the charge on each sphere decreases by a factor of 9, determine (in terms of Fo) how much electrostatic force each sphere will exert on the other.