1. Draw four illustrations of a globe and paper that are positioned to yield equatorial, transverse, oblique, and polar aspect projections. Label the equator in each. Use your textbook or lecture material if you need a reference.2. On any map, why is there distortion at areas that do not fall on lines of tangency or secancy?

Answer:

the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

Explanation:

Given the data in the question;

centripetal acceleration a[tex]_c[/tex]₁ = 241 m/s²

radius r₁ = 0.0189 m

radius r₂ = 0.0897 m

centripetal acceleration a[tex]_c[/tex]₂ = ? m/s²

since the rotational period will be the same for the two disk,

we use the centripetal acceleration formula a[tex]_c[/tex] = (4π²r/T²) to find the rotational period for the first disk.

a[tex]_c[/tex]₁ = (4π²r₁/T²)

make T² subject of formula

T² = 4π²r₁ / a[tex]_c[/tex]₁

we substitute

T² = ( 4 × π² × 0.0189 )  / 241  

T² = 0.00309602528 s²

Now we use the same formula to find a[tex]_c[/tex]₂

a[tex]_c[/tex]₂ = ( 4π²r₂ / T² )

we substitute

a[tex]_c[/tex]₂ = ( 4 × π² × 0.0897 )  / 0.00309602528

a[tex]_c[/tex]₂ = 1143.8 m/s²

Therefore, the centripetal acceleration at a point that is 0.0897 m from the center of the disc is 1143.8 m/s²

Answer:

Explanation:

First last of thermodynamics, just discusses the changes that a system is undergoing and the processes involved in it. It explains conservation of energy for a system undergoing changes or processes.

Second law of thermodynamics helps in defining the process and also the direction of the processes. It tells about the possibility of a process or the restriction of a process. It states that the entropy of a system always increases.

For this to occur the energy contained by a body has to diminish without converting to work or internal energy. So imagine a machine which works with less than efficiency, this means there are losses but they don’t show up anywhere. But the energy is obtained from a higher energy source to lower.

The easy way to do this is with an imaginary device that extracts zero-point energy to heat a quantity of gas. Energy is being created, so the first law is violated, and the entropy of the system is increasing as the gas heats up.

First law is violated since the energy conversion don't apply but the direction of work is applied so second law is satisfied.

Answer:

135J

Explanation:

So we know ΔKinetic Energy= ΔWork

Kinetic energy=1/2mv²

So Kf-Ki=ΔK

ΔK=1/2*0.45(25²-5²)=135J

135J=ΔWork

Complete Question

Four small masses of 0.2 kg each are connected by light rods 0.4m long to form a square. What is the moment of inertia of this object for an axis through the middle of the square and parallel to two sides.

Answer:

[tex]I=0.032kgm^2[/tex]

Explanation:

From the question we are told that:

Mass[tex]m=0.2kg[/tex]

Length [tex]l=0.4m[/tex]

Generally the equation for Inertia is mathematically given by

 [tex]I=md^2[/tex]

 [tex]I=0.8*0.20(\frac{0.40}{2})^2[/tex]

 [tex]I=0.032kgm^2[/tex]

(a) As it gets compressed by a distance x, the spring does

W = - 1/2 (52.1 N/m) x ²

of work on the object (negative because the restoring force exerted by the spring points in the opposite direction to the object's displacement). By the work-energy theorem, this work is equal to the change in the object's kinetic energy. At maximum compression x, the object's kinetic energy is zero, so

W = ∆K

- 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.118 m

(b) Taking friction into account, the only difference is that more work is done on the object.

By Newton's second law, the net vertical force on the object is

∑ F = n - mg = 0

where n is the magnitude of the normal force of the track pushing up on the object. Solving for n gives

n = mg = 2.45 N

and from this we get the magnitude of kinetic friction,

f = µn = 0.120 (2.45 N) = 0.294 N

Now as the spring gets compressed, the frictional force points in the same direction as the restoring force, so it also does negative work on the object:

W (friction) = - (0.294 N) x

W (spring) = - 1/2 (52.1 N/m) x ²

==>   W (total) = W (friction) + W (spring)

Solve for x :

- (0.294 N) x - 1/2 (52.1 N/m) x ² = 0 - 1/2 (0.250 kg) (1.70 m/s)²

==>   x ≈ 0.112 m

For the 0.250 kg object moving along a horizontal track and collides with and compresses a light spring, with a spring constant of 52.1 N/m, we have:

a) The spring's maximum compression when the track is frictionless is 0.118 m.

b) The spring's maximum compression when the track is not frictionless, with a coefficient of kinetic friction of 0.120 is 0.112 m.

a) We can calculate the spring's compression when the object collides with it by energy conservation because the track is frictionless:

[tex] E_{i} = E_{f} [/tex]

[tex] \frac{1}{2}m_{o}v_{o}^{2} = \frac{1}{2}kx^{2} [/tex]  (1)

Where:

[tex]m_{o}[/tex]: is the mass of the object = 0.250 kg

[tex]v_{o}[/tex]: is the velocity of the object = 1.70 m/s

k: is the spring constant = 52.1 N/m

x: is the distance of compression

After solving equation (1) for x, we have:

[tex] x = \sqrt{\frac{m_{o}v_{o}^{2}}{k}} = \sqrt{\frac{0.250 kg*(1.70 m/s)^{2}}{52.1 N/m}} = 0.118 m [/tex]

Hence, the spring's maximum compression is 0.118 m.

b) When the track is not frictionless, we can calculate the spring's compression by work definition:

[tex] W = \Delta E = E_{f} - E_{i} [/tex]

[tex] W = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} [/tex]   (2)

Work is also equal to:

[tex] W = F*d = F*x [/tex]     (3)

Where:  

F: is the force

d: is the displacement = x (distance of spring's compression)  

The force acting on the object is given by the friction force:

[tex] F = -\mu N = -\mu m_{o}g [/tex]   (4)

Where:

N: is the normal force = m₀g

μ: is the coefficient of kinetic friction = 0.120

g: is the acceleration due to gravity = 9.81 m/s²

The minus sign is because the friction force is in the opposite direction of motion.

After entering equations (3) and (4) into (2), we have:

[tex]-\mu m_{o}gx = \frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2}[/tex]

[tex]\frac{1}{2}kx^{2} - \frac{1}{2}m_{o}v_{o}^{2} + \mu m_{o}gx = 0[/tex]

[tex] \frac{1}{2}52.1 N/m*x^{2} - \frac{1}{2}0.250 kg*(1.70)^{2} + 0.120*0.250 kg*9.81 m/s^{2}*x = 0 [/tex]        

Solving the above quadratic equation for x

[tex] x = 0.112 m [/tex]  

Therefore, the spring's compression is 0.112 m when the track is not frictionless.

Read more here:

I hope it helps you!  

Answer:

you can measure by scale beacause we dont no sorry i cant help u but u can ask me some other Q

Answer:

Because gas contains free molecules but not liquid.

Please mark as brainliast

Answer:

The charged particle will follow a circular path.

Explanation:

Formula for the magnetic force is;

F = qvb sin θ

Where;

where;

q = the charge

v = the velocity

B = the magnetic field

θ = the angle between the velocity and magnetic field

We are told that velocity vector is perpendicular to the magnetic field lines. Thus, angle is 90.

So sin θ = sin 90 = 1

Thus,

F = qvB

Now, since the velocity vector is perpendicular to the magnetic field line,it also means from flemmings right hand rule, that the magnetic force is as well perpendicular to both of them.

Therefore, we have:

- a force that is always perpendicular to the velocity and as well constant in magnitude since magnitude of velocity or magnetic field does not change.

What this statement implies is that the force is acting as a centripetal force, and therefore, the charged particle will be kept in a uniform circular motion.

Answer:

The maximum wavelength of light that could liberate electrons from the aluminum metal is 303.7 nm

Explanation:

Given;

wavelength of the UV light, λ = 248 nm = 248 x 10⁻⁹ m

maximum kinetic energy of the ejected electron, K.E = 0.92 eV

let the work function of the aluminum metal = Ф

Apply photoelectric equation:

E = K.E + Ф

Where;

Ф is the minimum energy needed to eject electron the aluminum metal

E is the energy of the incident light

The energy of the incident light is calculated as follows;

[tex]E = hf = h\frac{c}{\lambda} \\\\where;\\\\h \ is \ Planck's \ constant = 6.626 \times 10^{-34} \ Js\\\\c \ is \ speed \ of \ light = 3 \times 10^{8} \ m/s\\\\E = \frac{(6.626\times 10^{-34})\times (3\times 10^8)}{248\times 10^{-9}} \\\\E = 8.02 \times 10^{-19} \ J[/tex]

The work function of the aluminum metal is calculated as;

Ф = E - K.E

Ф = 8.02 x 10⁻¹⁹  -  (0.92 x 1.602 x 10⁻¹⁹)

Ф =  8.02 x 10⁻¹⁹ J   -  1.474 x 10⁻¹⁹ J

Ф = 6.546 x 10⁻¹⁹ J

The maximum wavelength of light that could liberate electrons from the aluminum metal is calculated as;

[tex]\phi = hf = \frac{hc}{\lambda_{max}} \\\\\lambda_{max} = \frac{hc}{\phi} \\\\\lambda_{max} = \frac{(6.626\times 10^{-34}) \times (3 \times 10^8) }{6.546 \times 10^{-19}} \\\\\lambda_{max} = 3.037 \times 10^{-7} m\\\\\lambda_{max} = 303.7 \ nm[/tex]

Answer:

(I)

[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 120 = (u \times 2) + \frac{1}{2} \times 0 \times {2}^{2} \\ 120 = 2u \\ { \tt{speed = 60 \: {ms}^{ - 1} }}[/tex]

(ii)

[tex]{ \bf{s = ut + \frac{1}{2}a {t}^{2} }} \\ 240 = (60t) \\ { \tt{time = 4 \: seconds}}[/tex]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

[tex]A_a(t) = 5m/s^2[/tex]

To get the velocity, we integrate over time:

[tex]V_a(t) = (5m/s^2)*t + V_0[/tex]

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

[tex]V_a(t) = (5m/s^2)*t[/tex]

To get the position equation we integrate again over time:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2 + P_0[/tex]

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

[tex]P_a(t) = 0.5*(5m/s^2)*t^2[/tex]

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

[tex]V_b(t) =20m/s[/tex]

To get the position equation, we can integrate:

[tex]P_b(t) = (20m/s)*t + P_0[/tex]

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

[tex]P_b(t) = (20m/s)*t + 100m[/tex]

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

[tex]P_a(t) = P_b(t)[/tex]

We can solve this for t.

[tex]0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0[/tex]

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

[tex]t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)} }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}[/tex]

We only care for the positive solution, which is:

[tex]t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s[/tex]

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

[tex]P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m[/tex]

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

[tex]V_a(t) = (5m/s^2)*11.48s = 57.4 m/s[/tex]

Answer:

Atmospheric pressure at Badwater is 1.01022 atm

Explanation:

Data given:

1 atmospheric pressure (Pi) = 1.01 * 10[tex]^{5}[/tex] Pa

Elevation (h) = 86m

gravity (g) = 9.8 m/s2

Density of air P = 1.225 kg/m3

Therefore pressure at bad water Pb = Pi + Pgh

Pb = (1.01 * 10[tex]^{5}[/tex]) + (1.225 * 9.8 * 86)

Pb = (1.01 * 10[tex]^{5}[/tex]) + 1032.43 = 102032 Pa

hence:

Pb = 102032 /1.01 * 10[tex]^{5}[/tex] = 1.01022 atm

Answer:

it is answer of u are question

Answer:

v = 87.46 m/s

Explanation:

The radial acceleration is the centripetal acceleration, whose formula is given as:

[tex]a_c = \frac{v^2}{r}[/tex]

where,

[tex]a_c[/tex] = centripetal acceleration = 17 m/s²

v = planes's speed = ?

r = radius of path = 450 m

Therefore,

[tex]17\ m/s^2 = \frac{v^2}{450\ m}\\\\v^2 = (17\ m/s^2)(450\ m)\\\\v = \sqrt{7650\ m^2/s^2}[/tex]

v = 87.46 m/s

Answer:

The work done on the block by the spring as it accelerates the block is 4kx².

Explanation:

Let initial distance is x.

It was compressed three times farther and then the block is released, new distance is 3x.

The work done in compressing the spring is given by :

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)[/tex]

[tex]W=\dfrac{1}{2}k(x_2^2-x_1^2)\\\\W=\dfrac{1}{2}k((3x)^2-x^2)\\\\W=\dfrac{1}{2}k((9x^2-x^2)\\\\W=\dfrac{1}{2}k\times 8x^2\\\\W=4kx^2[/tex]

So, the work done on the block by the spring as it accelerates the block is 4kx².

Answer:

The correct answer will bs "2.41 m".

Explanation:

According to the question,

M = 50 g

or,

   = 0.050 kg

[tex]\Theta = 25^{\circ}[/tex]

k = 25.9 N/m

Δx = 0.200 m

Let the traveled distance be "x".

By using trigonometry, the height will be:

⇒ [tex]h = l Sin \Theta[/tex]

hence,

⇒ [tex]Potential \ energy \ at \ the \ top=Spring \ potential \ energy[/tex]

                                       [tex]Mgh=\frac{1}{2} k(\Delta x)^2[/tex]

By putting the values, we get

             [tex]0.050\times 9.8\times lSin 25^{\circ}=\frac{1}{2}\times 25.0\times (0.200)^2[/tex]

                                              [tex]l=2.41 \ m[/tex]      

Answer:

The maximum height is 0.33 m.

Explanation:

initial velocity, u = 8 m/s

final velocity, v = 0 m/s

10% of  kinetic energy is lost in friction.

The kinetic energy used to move up the top,

KE = 10 % of 0.5 mv^2

KE = 0.1 x 0.5 x m x 8 x 8 = 3.2 m

Let the maximum height is h.

Use conservation of energy

KE at the bottom = PE at the top

3.2 m = m x 9.8 x h

h = 0.33 m  

The height traveled vertically up the hill by the ball when it stops is 0.327 meter.

Given the following data:

Scientific data:

To determine how far (height) vertically up the hill the ball reaches when it stops:

By applying the law of conservation of energy, we have:

Kinetic energy lost at the bottom = Potential energy gained at the top.

Mathematically, the above expression is given by the formula:

[tex]0.1 \times \frac{1}{2} mv^2 = mgh\\\\0.1 \times \frac{1}{2} v^2 = gh\\\\h=\frac{0.1v^2}{2g}[/tex]

Substituting the given parameters into the formula, we have;

[tex]h=\frac{0.1 \times 8^2}{2\times 9.8} \\\\h=\frac{0.1 \times 64}{19.6} \\\\h=\frac{6.4}{19.6}[/tex]

Height, h = 0.327 meter.

Read more on kinetic energy here: https://brainly.com/question/17081653

Answer:

The angular acceleration is 4.5 rad/s^2.

Explanation:

Acceleration, a = 26.5 m/s2

length, L = 5.89 m

The angular acceleration is

[tex]\alpha =\frac{a}{L}\\\\\alpha = \frac{26.5}{5.89}=4.5 rad/s^2[/tex]

Answer:

The mass of human is 2898 times of the mass of cat.

Explanation:

A study finds that the metabolic rate of mammals is proportional to m^3/4 i.e.

[tex]M=\dfrac{km^3}{4}[/tex]

Where

k is constant

If m = 70 kg, the mass of human

[tex]M=\dfrac{70^3}{4}\\\\=85750[/tex]

If m = 4.91 kg, the mass of cat

[tex]M'=\dfrac{4.91^3}{4}\\\\=29.59[/tex]

So,

[tex]\dfrac{M}{M'}=\dfrac{85750}{29.59}\\\\=2897.93\approx 2898[/tex]

So, the mass of human is 2898 times of the mass of cat.

Answer:

Doing science could be defined as carrying out scientific processes, like the scientific method, to add to science's body of knowledge.

Answer: 11.36 m

Explanation:

Given

Mass of roller coaster is m=1200 kg

Initial speed of roller coaster is v=15 m/s

Energy at bottom and at the top is same i.e.

[tex]\Rightarrow \dfrac{1}{2}mv^2=mgh\\\\\Rightarrow \dfrac{1}{2}\times 1200\times 15^2=1200\times 9.8\times h\\\\\Rightarrow h=\dfrac{15^2}{2\times 9.8}\\\\\Rightarrow h=11.36\ m[/tex]

Thus, the highest point reach by the roller coaster is 11.36 m

Answer:

11.36m

Explanation:

Answer:

a)    v = 517.99 m / s,  b) θ = 296.3º

Explanation:

This is an exercise in kinematics, we are going to solve each axis independently

X axis

the acceleration is aₓ = 5.10 1 / S², they are on for t = 670 s and reaches a speed of vₓ=  3670 m / s, let's use the relation

           vₓ = v₀ₓ + aₓ t

           v₀ₓ = vₓ - aₓ t

           v₀ₓ = 3670 - 5.10 670

           v₀ₓ = 253 m / s

Y axis  

the acceleration is ay = 7.30 m / s², with a velocity of 4378 m / s after

t = 670 s

          v_y = v_{oy} + a_y t

          v_{oy} = v_y - a_y t

          v_oy} = 4378 - 7.30 670

          v_{oy}  = -513 m / s

to find the velocity modulus we use the Pythagorean theorem

          v = [tex]\sqrt{v_o_x^2 + v_o_y^2}[/tex]

          v = [tex]\sqrt{253^2 +513^2}[/tex]

          v = 517.99 m / s

to find the direction we use trigonometry

         tan θ ’= [tex]\frac{v_o_y}{v_o_x}[/tex]

         θ'= tan⁻¹  [tex]\frac{voy}{voy}[/tex]  

         θ'= tan⁻¹ (-513/253)

         tea '= -63.7

the negative sign indicates that it is below the ax axis, in the fourth quadrant

to give this angle from the positive side of the axis ax

          θ = 360 -   θ  

          θ = 360 - 63.7

          θ = 296.3º

P = F v

where P is power, F is the magnitude of force, and v is speed. So

50.0 hp = 37,280 W = F (24.6 m/s)

==>   F = (37,280 W) / (24.6 m/s) ≈ 1520 N

Answer:

Option "D" is the correct answer to the following question.

Explanation:

The gravitational potential energy of an item is determined by its mass, elevation, and gravitational acceleration. As a result, angular momentum and energy are preserved. The gravitational potential energy, on the other hand, varies with distance. When a consequence, kinetic energy varies during each orbit, resulting in a faster speed as a planet approaches the Sun.

Answer:

SPEED AND MASS

Explanation:

TOOK THE TEST

The magnetic field of the wire will be directed towards west. Using right thumb rule one can get the direction of field lines.

Answer:

the final velocity of the ball is 4.87 m/s

the final velocity of the ping ball is 9.87 m/s

Explanation:

Given;

mass of the ball, m₁ = 4.65 kg

mass of the ping ball, m₂ = 0.06 kg

initial velocity of the ping ball, u₂ = 0

initial velocity of the ball, u₁ = 5 m/s

let the final velocity of the ball = v₁

let the final velocity of the ping ball, = v₂

Apply the principle of conservation of linear momentum for elastic collision;

m₁u₁  +  m₂u₂  = m₁v₁   +  m₂v₂

4.65(5)  +   0.06(0)   =   4.65v₁   +   0.06v₂

23.25 + 0 = 4.65v₁  +  0.06v₂

23.25 = 4.65v₁  +  0.06v₂  ------ (1)

Apply one-directional velocity equation;

u₁ + v₁ = u₂  +  v₂

5 + v₁ = 0  +  v₂

5 + v₁ = v₂

v₁ = v₂ - 5  -------- (2)

substitute equation (2) into (1)

23.25 = 4.65(v₂ - 5)  +  0.06v₂

23.25 = 4.65v₂  -  23.25   +   0.06v₂

46.5 = 4.71 v₂

v₂ = 46.5/4.71

v₂ = 9.87 m/s

v₁ = v₂ - 5

v₁ = 9.87 - 5

v₁ = 4.87 m/s

Answer:

a.v=u+v/2

a.v=s/t

combining two equation we get,

u+v/2=s/t

(u+v)t/2=s

(u+v)t/2=s

{u+(u+at)}t/2=s

(u+u+at)t/2=s

(2u+at)t/2=s

2ut+at^2/2=s

2ut/2+at^2/2=s

UT +1/2at^2=s

proved

a=v-u/t

at=v-u

u+at=v

Answer:

a)     a = 27.44 m / s²,  b) a = 5.39 m / s², c)  a = 156.8 m / s², cabinet maximum acceleration does not change

Explanation:

a) In this exercise the wheels of the truck rotate to provide acceleration, but the contact point between the ground and the 2 wheels remains fixed, therefore the coefficient of friction for this point is static.

Let's apply Newton's second law

we set a regency hiss where the x axis is in the direction of movement of the truck

Y axis y

        N- W = 0

        N = W = m g

X axis

       2fr = m a

the expression for the friction force is

      fr = μ N

      fr = μ m g

we substitute

      2 μ m g = m /2   a

     a = 4 μ g

      a = 4 0.7 9.8

      a = 27.44 m / s²

b) let's look for the maximum acceleration that can be applied to the cabinet

       fr = m a

       μ N = ma

       μ m g = m a

       a = μ g

       a = 0.55  9.8

       a = 5.39 m / s²

as the acceleration of the platform is greater than this acceleration the cabinet must slip

c) the friction force is in the four wheels as well

With when the truck had two-wheel Thracian the weight of distributed evenly between the wheels, in this case with 4-wheel Thracian the weight must be distributed among all

applying Newton's second law

         4 fr = (m/4) a

         16 mg = (m) a

         a = 16 g

         a = 16 9.8

         a = 156.8 m / s²

cabinet maximum acceleration does not change

Answer:

The car travels the distance of 225m before coming to rest.

Explanation:

v = 45m/s

t = 5s

Therefore,

d = v*t

= 45*5

= 225m

(a) Let x be the maximum elongation of the spring. At this point, the glider would have zero velocity and thus zero kinetic energy. The total work W done by the spring on the glider to get it from the given point (4.00 cm from equilibrium) to x is

W = - (1/2 kx ² - 1/2 k (0.0400 m)²)

(note that x > 4.00 cm, and the restoring force of the spring opposes its elongation, so the total work is negative)

By the work-energy theorem, the total work is equal to the change in the glider's kinetic energy as it moves from 4.00 cm from equilibrium to x, so

W = ∆K = 0 - 1/2 m (0.835 m/s)²

Solve for x :

- (1/2 (160 N/m) x ² - 1/2 (160 N/m) (0.0400 m)²) = -1/2 (0.190 kg) (0.835 m/s)²

==>   x ≈ 0.0493 m ≈ 4.93 cm

(b) The glider attains its maximum speed at the equilibrium point. The work done by the spring as it is stretched away from equilibrium to the 4.00 cm position is

W = - 1/2 k (0.0400 m)²

If v is the glider's maximum speed, then by the work-energy theorem,

W = ∆K = 1/2 m (0.835 m/s)² - 1/2 mv ²

Solve for v :

- 1/2 (160 N/m) (0.0400 m)² = 1/2 (0.190 kg) (0.835 m/s)² - 1/2 (0.190 kg) v ²

==>   v ≈ 1.43 m/s

(c) The angular frequency of the glider's oscillation is

√(k/m) = √((160 N/m) / (0.190 kg)) ≈ 29.0 Hz

The amplitude of the motion is 0.049 cm. The maximum speed of the glider is 1.429 m/s. The angular frequency of the oscillation is 29.02 rad/s

From the given information;

Using energy conservation E, the amplitude of the motion can be calculated by using the formula:

[tex]\mathbf{E = \dfrac{1}{2}mv^2 + \dfrac{1}{2}kx^2}[/tex]

[tex]\mathbf{E = \dfrac{1}{2}(0.19 \ kg )\times (0.835)^2 + \dfrac{1}{2}(160) (0.04)^2}[/tex]

[tex]\mathbf{E =0.194 \ J}[/tex]

Similarly, we know that:

[tex]\mathbf{E = \dfrac{1}{2}kA^2}[/tex]

Making amplitude A the subject, we have:

[tex]\mathbf{A = \sqrt{\dfrac{2E}{k}}}[/tex]

[tex]\mathbf{A = \sqrt{\dfrac{2(0.194)}{160}}}[/tex]

[tex]\mathbf{A =0.049 \ cm}[/tex]

Again, using the energy conservation, the maximum speed of the glider can be calculated by using the formula:

[tex]\mathbf{E =\dfrac{1}{2} mv^2 _{max}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2E}{m}}}[/tex]

[tex]\mathbf{v _{max} = \sqrt{\dfrac{2\times 0.194}{0.19}}}[/tex]

[tex]\mathbf{v _{max} = 1.429 \ m/s}[/tex]

The angular frequency of the oscillation can be computed by using the expression:

[tex]\mathbf{\omega = \sqrt{\dfrac{k}{m}}}[/tex]

[tex]\mathbf{\omega = \sqrt{\dfrac{160}{0.19}}}[/tex]

ω = 29.02 rad/s

Learn more about energy conservation here:

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