A 1200 kg aircraft going 30 m/s collides with a 2000 kg aircraft that is parked and they stick together after the collision and are going 11.3 m/s after the collision. If they skid for 14.7 seconds before stopping, how far did they skid

Answers

Answer 1

Answer:

83.055  m

Explanation:

According to the given scenario, the calculation of skid distance is shown below:-

[tex]S = \frac{1}{2} \times (u + v) \times t[/tex]

Where  

u = 11.3

v = 0

t = 14.7

Now placing these values to the above formula,

So,

[tex]S = \frac{1}{2} \times (11.3 + 0) \times 14.7[/tex]

= 83.055  m

Therefore for computing the skid distance we simply applied the above formula i.e by considering the all items given in the question


Related Questions

Experts, ACE, Genius... can anybody calculate for the Reactions at supports A and B please? Will give brainliest! Given: fb = 300 kN/m, fc = 100 kN/m, Dy = 300 kN, spanAB = 6m, span BC = 6m, spanCD = 6m

Answers

Answer:

Support at Cy = 1.3 x 10³ k-N

Support at Ay = 200 k-N

Explanation:

given:

fb = 300 k-N/m

fc = 100 k-N/m

D = 300 k-N

L ab = 6 m

L bc = 6 m

L cd = 6 m

To get the reaction A or C.

take summation of moment either A or C.

Support Cy:

∑ M at Ay = 0

      (( x1 * F ) + ( D * Lab ) + ( D * L bc + D * L cd )

Cy = -------------------------------------------------------------------

                                      ( L ab + L bc )

Cy = 1.3 x 10³ k-N

Support Ay:

Since ∑ F = 0,           A + C - F - D = 0

                                   A = F  + D - C

                                  Ay = 200 k-N

Answer:

i was going to but its to late

Explanation:

7. How many 1.00 µF capacitors must be connected in parallel to store a charge of 1.00 C with a potential of 110 V across the capacitors?

Answers

Answer:

q = C V    charge on 1 capacitor

q = 1 * 10E-6 * 110 = 1.1 *  10E-4  C per capacitor

N = Q / q = 1 / 1.1 * 10E-4  = 9091 capacitors

9.09 × 10³ capacitors must be connected in parallel.

How to calculate the number of capacitors connected in parallel?

Given C = 1.00μF = 1 × 10⁻⁶ F

          q = 1.00 C

          V = 110 V

The equivalent capacitance is given by

Ceq = q/V

where q = total charge on all the capacitors

             V = potential difference

For N number of identical capacitors in parallel,

Ceq = NC

Therefore,

NC = q/V

N = q/VC

Putting on the values in the above formula,

N = 1/ (110)(1 × 10⁻⁶)

   = 1 / 110 × 10⁻⁶

   = 9.09 × 10³

Learn more about the capacitors here:

https://brainly.com/question/15052170

#SPJ2

Which two types of electromagnetic waves have higher frequencies than the waves that make up ultraviolet light?

radio waves and infrared light
visible light and X-rays
microwaves and gamma rays
gamma rays and X-rays

Answers

The two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

WHAT ARE ELECTROMAGNETIC WAVES?

Electromagnetic waves are components of the electromagnetic spectrum, which is made up of the following:

Radio wavesInfraredUltravioletVisible lightX-raysGamma raysmicrowave

Each electromagnetic wave have a specific frequency and wavelength.

However, the two types of electromagnetic waves that have higher frequencies than the waves that make up ultraviolet light are gamma rays and X-rays.

Learn more about electromagnetic waves at: https://brainly.com/question/8553652

Answer:

gamma rays and X-rays

Explanation:

d on edge I got 100%

A copper telephone wire has essentially no sag between poles 36.0 m apart on a winter day when the temperature is −20.0°C. How much longer is the wire on a summer day when the temperature is 34.0°C?

Answers

Answer:

The extension is  [tex]\Delta L = 0.033 \ m[/tex]

Explanation:

From the question we are told that  

     The length of the wire on a winter day is    [tex]L_w = 36.0 \ m[/tex]

      The temperature on the winter day is  [tex]T_w = -20.0 ^o C[/tex]

      The temperature on a summer day is  [tex]T_s = 34.0 ^0 C[/tex]

The the extension of the wire on a summer day is mathematically represented as

          [tex]\Delta L = \alpha L_w [T_s - T_w][/tex]

Where  

      [tex]\alpha[/tex] is the  coefficient of linear expansion of copper with a values [tex]\alpha = 17 *10^{-6}[/tex]

substituting value  

      [tex]\Delta L = 17 *10^{-6} * 36.0 [34 - [-20]][/tex]

      [tex]\Delta L = 0.033 \ m[/tex]

a wire of a certain material has resistance r and diameter d a second wire of the same material and length is found to have resistance r/9 what is the diameter of the second wire g

Answers

Answer:

d₂ = 3dThe diameter of the second wire is 3 times that of the initial wire.

Explanation:

Using the formula for calculating the resistivity of an object to find the diameter.

Resistivity P = RA/L

R is the resistance of the material

A is the cross sectional area

L is the length of the material

Since A = πd²/4

P = R( πd²/4)/L

P = Rπd²/4L ... 1

If the second wire of the same material and length is found to have resistance R/9, the resistivity of the second material will be;

P₂ = (R/9)A₂/L₂

P₂ = (R/9)(πd₂²/4)/L₂

P₂ = (Rπd₂²/36)/L₂

P₂ = (Rπd₂²)/36L₂

Since the length and resistivity are the same;

P = P₂  and L =L₂

Equating 1 and 2;

Rπd²/4L =  (Rπd₂²)/36L₂

Rπd²/4L =  (Rπd₂²)/36L

d² = d₂²/9

d₂² = 9d²

Taking the square root of both sides;

√d₂² = √9d²

d₂ = 3d

Therefore the diameter of the second wire is 3 times that of the initial wire

g An object with mass 1kg travels at 3 m/s and collides with a stationary object whose mass is 0.5kg. The two objects stick together and continue to move. What is the velocity of the two objects together after collision

Answers

Answer:

2

Explanation:

since the second object was in it stationary, it velocity is 0 m/s

38.A student pushes a 0.15 kg box down against a spring doing 25 J of work on the spring. The student releases the box which launches the box into the air. What is the maximum height reached by the box assuming negligible frictional forces

Answers

Answer:

Explanation:

Potential energy stored in the spring = 25 J

This energy is converted into gravitational potential energy . If h be the height attained

gravitational potential energy = mgh

mgh = 25

.15 x 9.8 x h = 25

h = 17 m

A rectangular block of metal of resistivity rho has dimensions d x 2d x 3d . A potential difference V is to be applied between two opposite faces of the block.
(A) To which two faces of the block should the potential difference be applied to give the maximum current density?
(B) What is the maximum current density j(subscript max) from part A?
(C) To which two faces of the block should the potential difference be applied to give the maximum current?
1. to the faces that area a distance d apart
2. to the faces that are a distance 2d apart
3. to the faces that are a distance 3d apart

Answers

Answer:

a) The potential difference should be applied to the d dimension face.

b) The maximum current density j = V/3ρd

c) 3. to the faces that are a distance 3d apart

Explanation:

a) Current density is the ratio of current flowing through a conductor, and cross-sectional area of the conductor. mathematically, it is written as

j = I/A

where I is the electric current, and

A is the area of the conductor.

From the equation, we can see that reducing the area of the conductor will increase the current density for a given amount of current passing through the conductor. The face d wide will give the least cross-sectional area of current flow.

b) current density can be gotten from

j = σE    ....equ 1

where σ is the conductivity of the conductor which is the inverse of resistivity ρ. this means that

σ = 1/ρ    ....equ 2

where ρ is the resistivity of the conductor

E is the electric field and is the volt through the conductor per unit length of the conductor

in this case, the maximum current density will be when the length is length 3d, and the volt is the potential difference V

therefore,

E = V/3d    ....equ 3

substituting equ 2 and equ 3 in equ1, we'll have

the maximum current density j = V/3ρd

c) To get the maximum current, the potential difference should be applied to the faces that are 3d wide apart because the resistance of a conductor varies inversely as the cross-sectional area. The maximum current will be gotten when the resistance is at its minimum, and the minimum resistance will be gotten with the most cross-sectional area. The 3d wide face will give the maximum cross-sectional area.

"Neon signs need 12,000 V to operate. If a transformer operates off a 240 V source and has 1000 turns in its primary coil, how may turns must the secondary coil have

Answers

Answer:

50000 turns

Explanation:

Vp / Vs = Np / Ns

240 / 12000 = 1000 / Ns

Ns = 50000 turns

Which of the following represents a concave mirror? +f,-f,-di,+di

Answers

Answer:

fully describes the concave mirror is + f

Explanation:

A concave mirror is a mirror where light rays are reflected reaching a point where the image is formed, therefore this mirror has a positive focal length, the amount that fully describes the concave mirror is + f

This allows defining a sign convention, for concave mirror + f, the distance to the object is + d0 and the distance to the image is + di

Answer:

+f

Explanation:

because you have to be really dumb to get an -f

What is the requirement for the photoelectric effect? Select one: a. The incident light must have enough intensity b. The incident light must have a wavelength shorter than visible light c. The incident light must have at least as much energy as the electron work function d. Both b and c

Answers

Answer:

c. The incident light must have at least as much energy as the electron work function

Explanation:

In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time  interval that it has almost no chance to absorb a second photon. An increase in intensity of light source  simply increase the number of photons and thus, the number of electrons, but the energy of electron  remains same. However, increase in frequency of light increases the energy of photons and hence, the

energy of electrons too.

Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the  binding force of the nucleus is called ‘Work Function’

Hence, the correct option is:

c. The incident light must have at least as much energy as the electron work function

A stellar object is emitting radiation at 3.55 mm. If a detector is capturing 3.2×108 photons per second at this wavelength, what is the total energy of the photons detected in 1.0 hour?

Answers

Answer:

E = 6.45 x 10⁻¹¹ J

Explanation:

First we need to find total number of photons detected in 1 hour. Therefore,

No. of Photons = n = (3.2 x 10⁸ photons/s)(1 h)(3600 s/1 h)

n = 11.52 x 10¹¹ photons

Now, the energy of these photons can be given by the formula:

E = nhc/λ

where,

E = Total Energy of the Photons = ?

h = Plank's Constant = 6.626 x 10⁻³⁴ J.s

c = speed of light = 3 x 10⁸ m/s

λ = wavelength of radiation = 3.55 mm = 3.55 x 10⁻³ m

Therefore,

E = (11.52 x 10¹¹)(6.626 x 10⁻³⁴ J.s)(3 x 10⁸ m/s)/(3.55 x 10⁻³ m)

E = 6.45 x 10⁻¹¹ J

Two metal sphere each of radius 2.0 cm, have a center-to-center separation of 3.30 m. Sphere 1 has a chrage of +1.10 10^-8 C. Sphere 2 has charge of -3.60 10^-8C. Assume that the separation is large enough for us to assume that the charge on each sphere iss uniformly distribuuted.
A) Calculate the potential at the point halfway between the centers.
B) Calculate the potential on the surface of sphere 1.
C) Calculate the potential on the surface of sphere 2.

Answers

Answer:

A)   V = -136.36 V , B)  V = 4.85 10³ V , C)  V = 1.62 10⁴ V

Explanation:

To calculate the potential at an external point of the spheres we use Gauss's law that the charge can be considered at the center of the sphere, therefore the potential for an external point is

          V = k ∑ [tex]q_{i} / r_{i}[/tex]

where [tex]q_{i}[/tex] and [tex]r_{i}[/tex] are the loads and the point distances.

A) We apply this equation to our case

          V = k (q₁ / r₁ + q₂ / r₂)

They ask us for the potential at the midpoint of separation

         r = 3.30 / 2 = 1.65 m

this distance is much greater than the radius of the spheres

let's calculate

         V = 9 10⁹ (1.1 10⁻⁸ / 1.65  + (-3.6 10⁻⁸) / 1.65)

         V = 9 10¹ / 1.65 (1.10 - 3.60)

         V = -136.36 V

B) The potential at the surface sphere A

r₂ is the distance of sphere B above the surface of sphere A

              r₂ = 3.30 -0.02 = 3.28 m

              r₁ = 0.02 m

we calculate

             V = 9 10⁹ (1.1 10⁻⁸ / 0.02  - 3.6 10⁻⁸ / 3.28)

             V = 9 10¹ (55 - 1,098)

             V = 4.85 10³ V

C) The potential on the surface of sphere B

      r₂ = 0.02 m

      r₁ = 3.3 -0.02 = 3.28 m

      V = 9 10⁹ (1.10 10⁻⁸ / 3.28  - 3.6 10⁻⁸ / 0.02)

       V = 9 10¹ (0.335 - 180)

       V = 1.62 10⁴ V

What is the average flow rate in cm3 /s of gasoline to the engine of a car traveling at 100 km/h if it averages 10.0 km/L

Answers

Answer:

2.78 cm³/s

Explanation:

From the question,

Q = v/A'.................... Equation 1

Where Q = Average flow rate, A' = inverse of Area, v = velocity of the car.

Given: v = 100 km/h, A' = 10 km/L

Substitute this value into equation 1

Q = 100/10

Q = 10 L/h.

Now, we convert L/h to cm³/s.

Since,

1 L = 1000 cm³, and

1 h = 3600 s

Therefore,

Q = 10(1000/3600) cm³/s

Q = 2.78 cm³/s

A Young's interference experiment is performed with blue-green laser light. The separation between the slits is 0.500 mm, and the screen is located 3.10 m from the slits. The first bright fringe is located 3.22 mm from the center of the interference pattern. What is the wavelength of the laser light?

Answers

Answer:

λ = 5.2 x 10⁻⁷ m = 520 nm

Explanation:

From Young's Double Slit Experiment, we know the following formula for the distance between consecutive bright fringes:

Δx = λL/d

where,

Δx = fringe spacing = distance of 1st bright fringe from center = 0.00322 m

L = Distance between slits and screen = 3.1 m

d = Separation between slits = 0.0005 m

λ = wavelength of light = ?

Therefore,

0.00322 m = λ(3.1 m)/(0.0005 m)

λ = (0.00322 m)(0.0005 m)/(3.1 m)

λ = 5.2 x 10⁻⁷ m = 520 nm

The Goliath six flags Magic Mountain roller coaster ride starts at 71.6 m (235 feet) above the ground. Assuming the coaster starts from rest and ignoring any friction, what is the speed of the coaster when it reaches the ground level

Answers

Answer:

The  velocity is [tex]v = 37 .46 \ m/s[/tex]

Explanation:

From the question we are told that

    The start distance above the ground is  [tex]h = 71.6 \ m[/tex]

Generally according to the  law of energy conservation we have that

     [tex]PE_{top} = KE_{bottom }[/tex]

Where [tex]PE_{top}[/tex] is potential energy at the top which is mathematically represented as

      [tex]PE_{top} = m * g * h[/tex]

And  [tex]KE_{bottom }[/tex] is the kinetic energy at the bottom which is mathematically represented as

     [tex]KE_{bottom } = \frac{1}{2} * m * v^2[/tex]

Therefore  

       [tex]m * g * h = \frac{1}{2} * m * v^ 2[/tex]

=>    [tex]v = \sqrt{2 * g * h }[/tex]

substituting value

     [tex]v = \sqrt{2 * 9.8 * 71.6 }[/tex]

     [tex]v = 37 .46 \ m/s[/tex]

An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 75.2 ∘C How much ice at a temperature of -22.8 ∘C must be dropped into the water so that the final temperature of the system will be 32.0 ∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kg

Answers

Answer:

Explanation:

We shall apply the theory of

heat lost = heat gained .

heat lost by water = mass x specific heat x temperature diff

= .285 x 4190 x ( 75.2 - 32 ) = 51587.28 J  

heat gained by ice to attain temperature of zero

= m x 2100 x 22.8 = 47880 m

heat gained by ice in melting = latent heat x mass

= 334000m

heat gained by water at zero to become warm at 32 degree

= m x 4190 x 32 = 134080 m

Total heat gained = 515960 m

So

515960 m = 51587.28

m = .1 kg

= 100 gm

Unpolarized light passes through a vertical polarizing filter, emerging with an intensity I0. The light then passes through a horizontal filter, which blocks all of the light; the intensity transmitted through the pair of filters is zero. Suppose a third polarizer with axis 45 ? from vertical is inserted between the first two.
What is the transmitted intensity now?
Express your answer in terms of I0. I got I0/8. But this is not right. I guess they want a number?

Answers

Answer:

    I₂ = 0.25 I₀

Explanation:

To know the light transmitted by a filter we must use the law of Malus

          I = I₀ cos² θ

In this case, the intensity of the light that passes through the first polarizer is I₀, it reaches the second polarized, which is at 45⁰, therefore the intensity I1 comes out of it.

        I₁ = I₀ cos² 45

        I₁ = I₀ 0.5

this is the light that reaches the third polarizer, which is at 45⁰ with respect to the second, from this comes the intensity I₂

       I₂ = I₁ cos² 45

       I₂ = (I₀ 0.5) 0.5

       I₂ = 0.25 I₀

this is the intensity of the light transmitted by the set of polarizers

Two protons are released from rest, with only the electrostatic force acting. Which of the following statements must be true about them as they move apart? (There could be more than one correct choice.)
A. Their electric potential energy keeps decreasing.
B. Their acceleration keeps decreasing.
C. Their kinetic energy keeps increasing.
D. Their kinetic energy keeps decreasing.
E. Their electric potential energy keeps increasing.

Answers

Answer:

(A)

Explanation:

We know , electric potential energy between two charge particles of charges "q" and "Q" respectively is given by kqQ/r where r is the distance between them.

Since the two charged particles are moving apart, the distance between them (r) increases and thus electrical potential energy decreases.

Will give brainliest ASAP! Please help (1/10 questions, will mark 5 stars and brainliest for all answers if correct)

Answers

Answer:

Option (A)

Explanation:

A 20 kg boy chases the butterfly with a speed of 2 meter per second.

Angle at which he runs is 70° North of West.

Therefore, Horizontal component (Vx) directing towards West will be,

Vx = v(Cos70°)

Vy = v(Sin70°)

Since momentum of a body is defined by,

Momentum = Mass × Velocity

Therefore, Westerly component of the momentum will be,

Momentum = 20 × (v)(Cos70°)

                   = 20 × 2Cos70°

                   = 13.68

                   ≈ 13.7 kg-meter per second

Therefore, Option (A) will be the answer.

When a central dark fringe is observed in reflection in a circular interference pattern, waves reflected from the upper and lower surfaces of the medium must have a phase difference, in radians, of

Answers

Explanation:

Let the first wave is :

[tex]y_1=A\sin\omega t[/tex]

And another wave is :

[tex]y_2=A\sin (\omega t+\phi)[/tex]

[tex]\phi[/tex] is phase difference between waves

Let y is the resultant of these two waves. So,

[tex]y =y_1+y_2[/tex]

The waves reflected from the upper and lower surfaces of the medium, it means that the resultant to be zero. So,

[tex]\cos(\dfrac{\phi}{2})=0\\\\\cos(\dfrac{\phi}{2})=\cos(\dfrac{\pi}{2})\\\\\phi=\pi[/tex]

So, the phase difference between the two waves is [tex]\pi[/tex].

What is the change in internal energy of the system (∆U) in a process in which 10 kJ of heat energy is absorbed by the system and 70 kJ of work is done by the system?

Answers

Answer:

Explanation:

According to first law of thermodynamics:

∆U= q + w

= 10kj+(-70kJ)

-60kJ

, w = + 70 kJ

(work done on the system is positive)

q = -10kJ ( heat is given out, so negative)

∆U = -10 + (+70) = +60 kJ

Thus, the internal energy of the system decreases by 60 kJ.

An electron traveling with a speed v enters a uniform magnetic field directed perpendicular to its path. The electron travels for a time t0 along a half-circle of radius R before leaving the magnetic field traveling opposite the direction it initially entered the field. Which of the following quantities would change if the electron had entered the field with a speed 2v? (There may be more than one correct answer.)
A. The time the electron is in the magnetic field
B. The magnitude of the net force acting on the electron inside the field
C. The magnitude of the electron's acceleration inside the field
D. The radius of the circular path the electron travels

Answers

Answer:

C. The magnitude of the electron's acceleration inside the field

D. The radius of the circular path the electron travels

Explanation:

The radius of the electron's motion in a uniform magnetic field is given by

[tex]R = \frac{MV}{qB}[/tex]

where;

m is the mass of the electron

q is the charge of the electron

B is the magnitude of the magnetic field

V is speed of the electron

R is the radius of the electron's

Thus, the radius of the of the electron's motion will change since it depends on speed of the electron.

The magnitude of the electron's acceleration inside the field  is given by;

[tex]a_c = \frac{V^2}{R}[/tex]

where;

[tex]a_c[/tex] is centripetal acceleration of electron

Thus, the magnitude of the electron's acceleration inside the field will change since it depends on the electron speed.

The time the electron is in the magnetic field is given by;

[tex]T = \frac{2\pi M}{qB}[/tex]

The time of electron motion will not change

The magnitude of the net force acting on the electron inside the field will not change;

[tex]qVB = \frac{MV^2}{R} \\\\qVB - \frac{MV^2}{R} = 0[/tex]

Therefore, the correct options are "C" and "D"

A single slit 1.4 mmmm wide is illuminated by 460-nmnm light. Part A What is the width of the central maximum (in cmcm ) in the diffraction pattern on a screen 5.0 mm away

Answers

Answer:

1.643*10⁻⁴cm

Explanation:

In a single slit experiment, the distance on a screen from the centre point is expressed as y = [tex]\frac{\delta m \lambda d}{a}[/tex] where;

[tex]\delta m[/tex] is the first two diffraction minima = 1

[tex]\lambda[/tex] is light wavelength

d is the distance of diffraction pattern from the screen

a is the width of the slit

Given [tex]\lambda[/tex] = 460-nm = 460*10⁻⁹m

d = 5.0mm = 5*10⁻³m

a = 1.4mm = 1.4*10⁻³m

Substituting this values into the formula above to get width of the central maximum y;

y = 1*460*10⁻⁹ * 5*10⁻³/1.4*10⁻³

y = 2300*10⁻¹²/1.4*10⁻³

y = 1642.86*10⁻⁹

y = 1.643*10⁻⁶m

Converting the final value to cm,

since 100cm = 1m

x = 1.643*10⁻⁶m

x = 1.643*10⁻⁶ * 100

x = 1.643*10⁻⁴cm

Hence, the width of the central maximum in the diffraction pattern on a screen 5.0 mm away is  1.643*10⁻⁴cm

A 1-kilogram mass is attached to a spring whose constant is 18 N/m, and the entire system is then submerged in a liquid that imparts a damping force numerically equal to 11 times the instantaneous velocity. Determine the equations of motion if the following is true?

a. the mass is initially released from rest from a point 1 meter below the equilibrium position
b. the mass is initially released from a point 1 meter below the equilibrium position with an upward velocity of 11 m/s

Answers

Answer:

Let [tex]x(t)[/tex] denote the position (in meters, with respect to the equilibrium position of the spring) of this mass at time [tex]t[/tex] (in seconds.) Note that this question did not specify the direction of this motion. Hence, assume that the gravity on this mass can be ignored.

a. [tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

b. [tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

Explanation:

Let [tex]x[/tex] denote the position of this mass (in meters, with respect to the equilibrium position of the spring) at time [tex]t[/tex] (in seconds.) Let [tex]x^\prime[/tex] and [tex]x^{\prime\prime}[/tex] denote the first and second derivatives of  [tex]x[/tex], respectively (with respect to time [tex]t[/tex].)

[tex]x^\prime[/tex] would thus represent the velocity of this mass.[tex]x^{\prime\prime}[/tex] would represent the acceleration of this mass.

Constructing the ODE

Construct an equation using [tex]x[/tex], [tex]x^\prime[/tex], and [tex]x^{\prime\prime}[/tex], with both sides equal the net force on this mass.

The first equation for the net force on this mass can be found with Newton's Second Law of motion. Let [tex]m[/tex] denote the size of this mass. By Newton's Second Law of motion, the net force on this mass would thus be equal to:

[tex]F(\text{net}) = m\, a = m\, x^{\prime\prime}[/tex].

The question described another equation for the net force on this mass. This equation is the sum of two parts:

The restoring force of the spring: [tex]F(\text{spring}) = -k\, x[/tex], where [tex]k[/tex] denotes the constant of this spring.The damping force: [tex]F(\text{damping}) = - 11\,x^\prime[/tex] according to the question. Note the negative sign in this expression- the damping force should always oppose the direction of motion.

Assume that there's no other force on this mass. Combine the restoring force and the damping force obtain an expression for the net force on this mass:

[tex]F(\text{net}) = -k\, x - 11\, x^\prime[/tex].

Combine the two equations for the net force on this mass to obtain:

[tex]m\, x^{\prime\prime} = -k\, x - 11\, x^\prime[/tex].

From the question:

Size of this mass: [tex]m = 1\; \rm kg[/tex].Spring constant: [tex]k = 18\; \rm N \cdot m^{-1}[/tex].

Hence, the equation will become:

[tex]x^{\prime\prime} = -18\, x - 11\, x^\prime[/tex].

Rearrange to obtain:

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex].

Finding the general solution to this ODE

[tex]x^{\prime\prime} + 11\, x^\prime + 18\; x = 0[/tex] fits the pattern of a second-order homogeneous ODE with constant coefficients. Its auxiliary equation is:

[tex]m^2 + 11\, m + 18 = 0[/tex].

The two roots are:

[tex]m_1 = -2[/tex], and[tex]m_2 = -9[/tex].

Let [tex]c_1[/tex] and [tex]c_2[/tex] denote two arbitrary real constants. The general solution of a second-order homogeneous ODE with two distinct real roots [tex]m_1[/tex] and [tex]m_2[/tex] is:

[tex]x = c_1\, e^{m_1\cdot t} + c_2\, e^{m_2\cdot t}[/tex].

For this particular ODE, that general solution would be:

[tex]x = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex].

Finding the particular solutions to this ODE

Note, that if [tex]x(t) = c_1\, e^{-2 t} + c_2\, e^{-9 t}[/tex] denotes the position of this mass at time [tex]t[/tex], then [tex]x^\prime(t) = -2\,c_1\, e^{-2 t} -9\, c_2\, e^{-9 t}[/tex] would denote the velocity of this mass at time

The position at time [tex]t = 0[/tex] would be [tex]x(0) = c_1 + c_2[/tex].The velocity at time [tex]t = 0[/tex] would be [tex]x^\prime(0) = -2\, c_1 - 9\, c_2[/tex].

For section [tex]\rm a.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 0\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = -\frac{9}{7} \\ &c_2 = \frac{2}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm a.[/tex] will be:

[tex]\displaystyle x(t) = -\frac{9}{7}\, e^{-2 t} + \frac{2}{7}\, e^{-9 t}[/tex].

Similarly, for section [tex]\rm b.[/tex]:

[tex]\left\lbrace\begin{aligned}& x(0) = -1 \\ &x^\prime(0) = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 + c_2 = -1 \\ &-2\, c_1 - 9\, c_2 = 11\end{aligned}\right. \implies \left\lbrace\begin{aligned} &c_1 = \frac{2}{7} \\ &c_2 = -\frac{9}{7}\end{aligned}\right.[/tex].

Hence, the particular solution for section [tex]\rm b.[/tex] will be:

[tex]\displaystyle x(t) = \frac{2}{7}\, e^{-2 t} - \frac{9}{7}\, e^{-9 t}[/tex].

explain why cups of soup at a take away kiosk are often sold in white polystrene cups with a lid to stop spillage​

Answers

Answer:

polystyrene is a good insulater so less heat will escape from the cup and it will keep it warm.

the cup helps it become more insulated

An air-filled parallel-plate capacitor has plates of area 2.30 cm2 2 separated by 1.50 mm. The capacitor is connected to a 12.0-V battery. Find the value of its capacitance.

Answers

Answer:

[tex]1.357\times 10^{-12}[/tex]

Explanation:

Relevant Data provided

Area which indicates A = 2.3 cm^2 = 2.3 x 10^-4 m^2

Distance which indicates d = 1.50 x 10^-3 m

Voltage which indicates V = 12 V

According to the requirement, the computation of value of its capacitance is shown below:-

[tex]Capacitance, C = \frac{\epsilon oA}{D}[/tex]

[tex]= \frac{= 8.854\times 10^{-12}\times 2.3\times 10^{-4}}{(1.5 \times 10^{-3})}[/tex]

= [tex]1.357\times 10^{-12}[/tex]

Therefore for computing the capacitance we simply applied the above formula.

Often in science it is helpful to talk about a representative example of the objects or phenomena being studied. However, you must always keep in mind that the average case is not always representative. For example, our Sun is often described as an "average" star in the Milky Way. In what sense is this statement true? In what sense is this statement seriously misleading? Do you think it is useful to characterize the stars in the Milky Way by simply citing our "average" Sun?

Answers

Explanation:

The statement "our sun is an 'average' sun" is  true when it is used to describe or characterize some unique physical properties of stars generally in the universe. 'Average' in this sense is used to define a typical sun such as, "stars should glow like our sun an average star."

The statement is used wrongly when used to in quantifying other stars in the universe, based on calculated values from our sun. In this case, we cannot truly say if our sun is a true representative average of other stars in the universe.

Yes! it is useful to characterize the milky way by simply citing the average sun. Properties like their ability to glow and radiate heat can be defined by citing an average star like our sun, so long as we don't translate it into citing quantitative properties of the sun as an average of our Milky Way Galaxy like the mass, temperature, etc.

with a speed of 75 m sl. Determine
1) A vehicle of mass 1600 kg moves
the magnitude of its momentum.​

Answers

Answer:

120000    kgxm/s

Explanation:

momentum is mass times velocity so just multiply 1600 kg times 75 m/s and you get 120000  kgxm/s

action and reaction are equal in magnitude and opposite in direction.Then why do they not balance each other.

Answers

Explanation:

Newton's third law of motion states that every action has an equal and opposite reaction. This means that forces always act in pairs. Action and reaction forces are equal and opposite, but they are not balanced forces because they act on different objects so they don't cancel out.

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