Determine the two coefficients of static friction at B and at C so that when the magnitude of the applied force is increased to 360 N , the post slips at both B and C simultaneously.

Answer:

Poynting vector oscillate at a frequency of 2omega

Explanation:

This is because The poynting vector is proportional to the cross product of electric and magnetic field vectors. So Because both fields oscillate sinusoidally with frequency w, trigonometric identities show that their product is a sinusoidal function of frequency of 2w.

Answer:

The maximum  emf generated in the coil is 60527.49 V

Explanation:

Given;

area of coil, A = 0.320 m²

angular frequency, f = 100 rev/s

magnetic field, B = 0.43 T

number of turns, N = 700 turns

The maximum emf generated in the coil is calculated as,

E = NBAω

where;

ω is the angular speed = 2πf

E = NBA(2πf)

Substitute in the given values and solve for E

E = 700 x 0.43 x 0.32 x 2π x 100

E = 60527.49 V

Therefore, the maximum  emf generated in the coil is 60527.49 V

Answer:

Explanation:

The volume of the spherical balloon is expressed as V = 4/3πr³ where r is the radius of the spherical balloon. If the spherical balloon is inflated with gas at the rate of 500 cubic centimetres per minute then dV/dt = 500cm³.

Using chain rule to express dV/dt;

dV/dt = dV/dr*dr/dt

dr/dt is the rate at which the radius of the gallon is increasing.

From the formula, dV/dr = 3(4/3πr^3-1))

dV/dr = 4πr²

dV/dt = 4πr² *dr/dt

500 =  4πr² *dr/dt

If radius r = 40;

500 = 4π(40)² *dr/dt

500 = 6400π*dr/dt

dr/dt = 500/6400π

dr/dt = 5/64π

dr/dt  = 0.245cm/min

Hence, the radius of the balloon is increasing at the rate of 0.245cm/min

Answer:

  F_net = 264, 26 pound

Explanation:

For this exercise we use Newton's second law

               F_net = F_int - F_outside

where the force can be found from the definition of pressure

              P = F / A

              F = P A

we substitute

               F_net = P_inside  A - P_outside A

               F_net = A (P_inside - P_outside)

indicate that the pressure on the outside is 25% less than the pressure on the inside

               P_outside = 0.25 P_inside

The area is

               A = L W

we substitute

              F_net = L W P_inside (1-0.25)

let's calculate

suppose the pressure inside is atmospheric pressure

              P_inside= P_atmospheric = 1,013 10⁵ Pa = 14.7 PSI

              F_net = 8.1 3 14.5 0.75

              F_net = 264, 26 pound

Answer:

Explanation:

Work required = q x V

where q is charge on electron and V is potential difference

= 1.6 x 10⁻¹⁹  x 12

= 19.2 x 10⁻¹⁹ J

Answer:

Option D is the correct option.

Explanation:

Given,

Mass ( m ) = 3 kg

Acceleration due to gravity ( g ) = 9.8 m/s²

Weight ( w ) = ?

Now, let's find the weight :

[tex]w \: = \: m \times g[/tex]

plug the values

[tex] = 3 \times 9.8[/tex]

Multiply the numbers

[tex] = 29.4 \: [/tex] Newton

Hope this helps!!

best regards!!

Answer:

Car 3

Explanation:

Answer:

7.91 m/s

Explanation:

Speed: This can be defined as the ratio of this to time. Speed can also be the magnitude of a velocity or speed is velocity without direction.

The S.I unit of speed is m/s.

From the question,

K.E = 1/2(mv²)................ Equation 1

Where K.E = Kinetic Energy, m = mass of the runner, v = velocity of the runner.

make  v the subject of the equation

v = √(2K.E/m).................Equation 2

Given: K.E = 1.09×10³ J, m = 34.8 kg.

Substitute into equation 2

v = √(2× 1.09×10³/34.8)

v = √(62.64)

v = 7.91 m/s.

Hence the speed of the runner = 7.91 m/s

Answer:

A. plot an H-R diagram for the stars in the cluster.

Explanation:

A star cluster can be defined as a constellation of stars, due to gravitational force, which has the same origin.

The astronomy student would have to plot an H-R diagram for the stars in the cluster and determine the age of the cluster by observing the turn-off point. The turn-off is majorly as a result of gradual depletion of the source of energy of the star. Thus, it projects off the constellation.

Answer:

The  pressure is  [tex]P = 1652 \ Pa[/tex]

Explanation:

From the question we are told that

    The  volume of the container is  [tex]V = 1.83 \ L = 1.83 *10^{-3 } \ m^3[/tex]

     The mass of  [tex]N_2[/tex] is  [tex]m_n = 0.246 \ g = 0.246 *10^{-3} \ kg[/tex]

     The root-mean-square velocity is  [tex]v = 192 \ m/s[/tex]

The  root -mean square velocity is mathematically represented as

      [tex]v = \sqrt{ \frac{3 RT}{M_n } }[/tex]

Now the ideal gas law is mathematically represented as

      [tex]PV = nRT[/tex]

=>   [tex]RT = \frac{PV}{n }[/tex]

Where n is the number of moles which is mathematically represented as

         [tex]n = \frac{ m_n }{M }[/tex]

Where  M  is the molar mass of  [tex]N_2[/tex]

So  

        [tex]RT = \frac{PVM_n }{m _n }[/tex]

=>    [tex]v = \sqrt{ \frac{3 \frac{P* V * M_n }{m_n } }{M_n } }[/tex]

=>    [tex]v = \sqrt{ \frac{ 3 * P* V }{m_n } } }[/tex]

=>   [tex]P = \frac{v^2 * m_n}{3 * V }[/tex]

substituting values

    =>    [tex]P = \frac{( 192)^2 * 0.246 *10^{-3}}{3 * 1.83 *10^{-3} }[/tex]

=>         [tex]P = 1652 \ Pa[/tex]

Answer:

x-component = 7.1

Explanation:

x-component = 9cos38 = 7.09 =7.1

Answer:

7.1

Explanation:

use cos

Answer:

Explanation:

The rod will act as pendulum for small oscillation .

Time period of oscillation

[tex]T=2\pi\sqrt{\frac{l}{g} }[/tex]

angular frequency ω = 2π / T

= [tex]\omega=\sqrt{\frac{g}{l} }[/tex]

b )

ω = 20( given )

velocity = ω r = ω l

Let the maximum angular displacement in terms of degree be θ .

1/2 m v ² = mgl ( 1 - cosθ ) ,

[ l-lcosθ is loss of height . we have applied law of conservation of mechanical energy .]

.5 (  ω l )² = gl( 1 - cos θ )

.5 ω² l = g ( 1 - cosθ )

1 - cosθ  = .5 ω² l /g

cosθ = 1 - .5 ω² l /g

θ can be calculated , if value of l is given .

Answer:

1248m

The time that wave moves from the wave source to the ocean floor is half the total travel time: t = 2.5/2 = 1.25s

The speed of sound in seawater is 1560 m/s

Therefore, s = vt = (1560 m/s)(1.25s) =1248 m = 1.2km

..........................................................

The radiation pressure due to a bulb that emits 25 W of EM radiation at a distance of 6.5 cm from the center of the bulb and the light is completely absorbed is 1.5707x10⁻⁶ N/m².

Radiation pressure was defined as the mechanical pressure exerted upon any surface due to the exchange of momentum between the object and the electromagnetic field.

Radiation pressure always includes the Momentum of light or electromagnetic radiation of any wavelength that can be absorbed, reflected, or otherwise emitted by matter on any scale.

E.g: Black-body radiation

Given the values are,

Wattage of bulb = W = 25 W

distance = d = 6.5 cm = 0.065 m

To know the Radiation Pressure,

It can be given by

P = I/c

Where, c = 299792458 m/s is the speed of light,

I is the intensity of radiation and given by

I = W/4πd²

Where W is the Wattage of bulb and d is the distance

I = 25/4π*0.065²

I = 470.872 w/m²

so, the radiation pressure becomes

P = 470.872/299792458

P = 1.5707x10⁻⁶ N/m²

Therefore, the radiation pressure due to a 25 W bulb at a distance of 6.5 cm is 1.5707x10⁻⁶ N/m²

To know more about the radiation pressure,

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Answer:

w = √ 1 / CL

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

Explanation:

This problem refers to electrical circuits, the circuits where this phenomenon occurs are series RLC circuits, where the resistor, the capacitor and the inductance are placed in series.

In these circuits the impedance is

             X = √ (R² +  ([tex]X_{C}[/tex] -[tex]X_{L}[/tex])² )

where Xc and XL is the capacitive and inductive impedance, respectively

            X_{C} = 1 / wC

           X_{L} = wL

From this expression we can see that for the resonance frequency

           X_{C} = X_{L}

the impedance of the circuit is minimal, therefore the current and voltage are maximum and an increase in signal intensity is observed.

This does not violate energy conservation because the voltage of the power source is equal to the voltage drop in the resistence

               V = IR

Since the contribution of the two other components is canceled, this occurs for

                X_{C} = X_{L}

                1 / wC = w L

                w = √ 1 / CL

Answer:

the frequency of revolution of the second particle is f

Explanation:

centripetal force is balanced by the magnetic force for object under magnetic field is given as

Mv²/r= qvB

But v= omega x r

Omega= 2pi x f

f= qB/2pi x M

So since frequency does not depend on the velocity.therefore the frequency of revolution of the second particle remains the same and its equal to f

Complete Question

Two parallel-plate capacitors C1 and C2 are connected in parallel to a 12.0 V battery. Both capacitors have the same plate area of 5.30 cm2 and plate separation of 2.65 mm. However, the first capacitor C1 is filled with air, while the second capacitor C2 is filled with a dielectric that has a dielectric constant of 2.10.

(a) What is the charge stored on each capacitor

 (b)  What is the total charge stored in the parallel combination?

Answer:

a

   i    [tex]Q_1 = 2.124 *10^{-11} \ C[/tex]

   ii    [tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]

b

  [tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]

Explanation:

From the question we are told that

   The  voltage of the battery is  [tex]V = 12.0 \ V[/tex]

    The  plate area of each capacitor is  [tex]A = 5.30 \ cm^2 = 5.30 *10^{-4} \ m^2[/tex]

    The  separation between the plates is  [tex]d = 2.65 \ mm = 2.65 *10^{-3} \ m[/tex]

     The permittivity of free space  has a value  [tex]\epsilon_o = 8.85 *10^{-12} \ F/m[/tex]

     The  dielectric constant of the other material is  [tex]z = 2.10[/tex]

The  capacitance of the  first capacitor is mathematically represented as

       [tex]C_1 = \frac{\epsilon * A }{d }[/tex]

substituting values

        [tex]C_1 = \frac{8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]

       [tex]C_1 = 1.77 *10^{-12} \ F[/tex]

The  charge stored in the first capacitor is  

       [tex]Q_1 = C_1 * V[/tex]

substituting values

        [tex]Q_1 = 1.77 *10^{-12} * 12[/tex]

       [tex]Q_1 = 2.124 *10^{-11} \ C[/tex]

The capacitance of the second  capacitor is mathematically represented as

       [tex]C_2 = \frac{ z * \epsilon * A }{d }[/tex]

substituting values

       [tex]C_1 = \frac{ 2.10 *8.85 *10^{-12 } * 5.30 *10^{-4} }{2.65 *10^{-3} }[/tex]

       [tex]C_1 = 3.717 *10^{-12} \ F[/tex]

The  charge stored in the second capacitor is  

      [tex]Q_2 = C_2 * V[/tex]

substituting values

     [tex]Q_2 = 3.717*10^{-12} * 12[/tex]

     [tex]Q_2 = 4.4604 *10^{-11} \ C[/tex]

Now  the total charge stored in the parallel combination is mathematically represented as

     [tex]Q_{eq} = Q_1 + Q_2[/tex]

substituting values

    [tex]Q_{eq} = 4.4604 *10^{-11} + 2.124*10^{-11}[/tex]

     [tex]Q_{eq} = 6.5844 *10^{-11} \ C[/tex]

Answer:

The velocity is  [tex]v = 2.84 1 \ m/s[/tex]

Explanation:

The  diagram showing this set up is shown on the first uploaded image (reference Physics website )

From the question we are told that

    The mass is  m =  4 kg

    The  length of the string is [tex]L = 2.0 \ m[/tex]

    The constant angle is  [tex]\theta = 35.4 ^o[/tex]

Generally the vertical forces acting on the mass to keep it at equilibrium vertically is mathematically represented as

           [tex]Tcos (\theta ) - mg = 0[/tex]

=>        [tex]mg = Tcos (\theta )[/tex]

Now let the force acting on mass horizontally be k  so from SOHCAHTOA rule

         [tex]sin (\theta ) = \frac{k }{T}[/tex]

=>      [tex]k = T sin \theta[/tex]

Now this k is also equivalent to the centripetal force acting on the mass which is mathematically represented as

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

So

          [tex]k = F_v[/tex]

Which

=>       [tex]T sin \theta= \frac{ m v^2}{ r }[/tex]

So

        [tex]\frac{Tsin (\theta )}{Tcos (\theta )} = \frac{mg}{ \frac{mv^2}{r} }[/tex]

=>      [tex]Tan (\theta ) = \frac{v^2}{ r * g }[/tex]

=>      [tex]v = \sqrt{r * g * tan (\theta )}[/tex]

Now the radius is evaluated using SOHCAHTOA rule as

       [tex]sin (\theta) = \frac{ r}{L}[/tex]

=>    [tex]r = L sin (\theta)[/tex]

substituting values

       [tex]r = 2 sin ( 35.4 )[/tex]

       [tex]r = 1.1586 \ m[/tex]

So

       [tex]v = \sqrt{1.1586* 9.8 * tan (35.4 )}[/tex]

       [tex]v = 2.84 1 \ m/s[/tex]

Answer:

A. 2.97x 10^-6C

B. 1.48x10^ -6 C

Explanation:

Pls see attached file

Answer:

1) +9.4 x 10^-7 C

2) +4.72 x 10^-7 C  and  +1.9 x 10^-6 C

Explanation:

The two positive charges will repel each other

Repulsive force on charges = 0.200 N

distance apart = 20.0 cm = 0.2 m

charge on each sphere = ?

Electrical force on charged spheres at a distance is given as

F = [tex]\frac{kQq}{r^{2} }[/tex]

where F is the force on the spheres

k is the Coulomb's constant = 8.98 x 10^9 kg⋅m³⋅s⁻²⋅C⁻²

Q is the charge on of the spheres

q is the charge on the other sphere

r is their distance apart

since the charges are equal, i.e Q = q, the equation becomes

F = [tex]\frac{kQ^{2} }{r^{2} }[/tex]

making Q the subject of the formula

==> Q = [tex]\sqrt{\frac{Fr^{2} }{k} }[/tex]

imputing values into the equation, we have

Q = [tex]\sqrt{\frac{0.2*0.2^{2} }{8.98*10^{9} } }[/tex] = +9.4 x 10^-7 C

If one charge has four times the charge on the other, then

charge on one sphere = q

charge on the other sphere = 4q

product of both charges = [tex]4q^{2}[/tex]

we then have

F = [tex]\frac{4kq^{2} }{r^{2} }[/tex]

making q the subject of the formula

==> q =  [tex]\sqrt{\frac{Fr^{2} }{4k} }[/tex]

imputing values into the equation, we have

q = [tex]\sqrt{\frac{0.2*0.2^{2} }{4*8.98*10^{9} } }[/tex] = +4.72 x 10^-7 C

charge on other sphere = 4q = 4 x 4.72 x 10^-7 = +1.9 x 10^-6 C

Answer:

The glider’s speed after the skydiver lets go is 26 m/s

Explanation:

To calculate the glider’s speed just after the skydiver lets go, we will need to use the conservation of momentum

Mathematically;

mv = mv + mv

so 680 * 26 = (680-60)v + 60 * 26

17680 = 620v + 1560

17680-1560 = 620v

16120 = 620v

v = 16120/620

v = 26 m/s

Answer:

  n = 2.0686

Explanation:

When an unpolarized ray of light is reflected on a surface, the reflected ray is partially polarized, complete polarization occurs when it is true that between the transmitted and reflected ray one has 90, the relationship is

        n = so tea

 let's calculate

        n = tan 64.2

        n = 2.0686

Answer:

a) h = 13,205.4 m

b)  r_f = 2.12 106 m

c)        e% = 0.68%

Explanation:

a) This is an exercise we are asked to use energy conservation,

Starting point. On the surface of Mimas

        Em₀ = K = ½ m v²

Final point. Where the ball stops

       [tex]Em_{f}[/tex] = U = m g h

        Em₀ = Em_{f}

        ½ m v² = m g h

         h = ½ v² / g

let's calculate

         h = ½ 41² / 0.0636

         h = 13,205.4 m

b) For this part we are asked to use the law of universal gravitation, write the energy

starting point. Satellite surface

           Em₀ = K + U = ½ m v² - GmM / r_o

final point. Where the ball stops

            [tex]Em_{f}[/tex]= U = - G mM / r_f

          Em₀ = Em_{f}

          ½ m v² - G m M / r_o = - G mM / r_f

In this case all distances are measured from the center of the satellite

         1 / rf = 1 / GM (-½ v² + G M / r_o)

let's calculate

         1 / rf = 1 / (6.67 10⁻¹¹ 3.75 10¹⁹) (- ½ 41 2 + 6.67 10⁻¹¹ 3.75 10¹⁹ / 1.98 105)

         1 / r_f = 3,998 10⁻¹¹(-840.5 + 12.63 10³)

          1 / r_f = 4,714 10⁻⁷

          r_f = 1 / 4,715 10⁻⁷

          r_f = 2.12 106 m

to measure this distance from the satellite surface

          r_f ’= r_f - r_o

          r_f ’= 2.12 106 - 1.98 105

         r_f ’= 1,922 106 m

c) the percentage difference is

          e% = 13 205.4 / 1,922 106 100

          e% = 0.68%

The estimate of part a is a little low

Answer:

a) The equation is [tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) Your velocity after collision is 2.64 m/s

c) The force you felt is 7392 N

d) you and your brother undergo an equal amount of acceleration

Explanation:

Your mass [tex]m_{y}[/tex] = 60 kg

your brother's mass [tex]m_{b}[/tex] = 30 kg

mass of the car [tex]m_{c}[/tex] = 80 kg

your initial speed [tex]u_{y}[/tex] = 0 m/s (since you've not started moving yet)

your brother's initial velocity [tex]u_{b}[/tex] = 3 m/s

your final speed [tex]v_{y}[/tex] after collision = ?

your brother's final speed [tex]v_{b}[/tex] after collision = ?

a) equations you need to use to figure out how fast you and your brother are moving after the collision is

[tex](m_{y}+m_{c} )u_{y} + (m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

but [tex]u_{y}[/tex] = 0 m/s

the equation reduces to

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

b) if your little brother reverses with velocity of 0.36 m/s it means

[tex]v_{b}[/tex] = -0.36 m/s (the reverse means it travels in the opposite direction)

then, imputing values into the equation, we'll have

[tex](m_{b}+m_{c} )u_{b} = (m_{y}+m_{c} )v_{y} + (m_{b}+m_{c} )v_{b}[/tex]

(30 + 80)3 = (60 + 80)[tex]v_{y}[/tex] + (30 + 80)(-0.36)

330 = 140[tex]v_{y}[/tex] - 39.6

369.6 = 140[tex]v_{y}[/tex]

[tex]v_{y}[/tex] = 369.6/140 = 2.64 m/s

This means you will also reverse with a velocity of 2.64 m/s

c) your initial momentum = 0  since you started from rest

your final momentum = (total mass) x (final velocity)

==>  (60 + 80) x 2.64 = 369.6 kg-m/s

If the collision lasted for 0.05 s,

then force exerted on you = (change in momentum) ÷ (time collision lasted)

force on you = ( 369.6 - 0) ÷ 0.05 = 7392 N

d) you changed velocity from 0 m/s to 2.64 m/s in 0.05 s

your acceleration is (2.64 - 0)/0.05 = 52.8 m/s^2

your brother changed velocity from 3 m/s to 0.36 m/s in 0.05 s

his deceleration is (3 - 0.36)/0.05 = 52.8 m/s

you and your brother undergo an equal amount of acceleration. This is because you gained the momentum your brother lost

Answer:

Explanation:

According to work energy theorem

change in kinetic energy of truck = work done against it

work done against it = force x displacement

= - 850 x 8 = 6800 J

change in kinetic energy of truck = - 6800 J .

energy will be reduced by 6800 J

Answer:-6800

Explanation:

Answer:

F = 213.75 N

Explanation:

First we need to calculate the angular acceleration of merry-go-round. For that purpose we use 1st equation of motion in angular form.

ωf = ωi + αt

where,

ωf=final angular velocity=(5.5 rev/min)(2π rad/1 rev)(1 min/60 s)=0.58 rad/s

ωi =initial angular velocity = 0 rad/s

t = time = 12 s

α = angular acceleration = ?

Therefore,

0.58 rad/s = 0 rad/s + α(12 s)

α = (0.58 rad/s)/(12 s)

α = 0.05 rad/s²

Now, we shall find the linear acceleration of the merry-go-round:

a = rα

where,

a = linear acceleration = ?

r = radius = 4.5 m

Therefore,

a = (4.5 m)(0.05 rad/s²)

a = 0.225 m/s²

Now, the force is given by Newton;s 2nd Law:

F = ma

where,

F = Force = ?

m = mass pf merry-go-round = 950 kg

Therefore,

F = (950 kg)(0.225 m/s²)

F = 213.75 N

Answer:

The current is changing at the rate of 0.20 A/s

Explanation:

Given;

inductance of the inductor, L = 5.0-H

current in the inductor, I = 3.0 A

Energy stored in the inductor at the given instant, E = 3.0 J/s

The energy stored in inductor is given as;

E = ¹/₂LI²

E = ¹/₂(5)(3)²

E = 22.5 J/s

This energy is increased by 3.0 J/s

E = 22.5 J/s + 3.0 J/s = 25.5 J/s

Determine the new current at this given energy;

25.5 = ¹/₂LI²

25.5 = ¹/₂(5)(I²)

25.5 = 2.5I²

I² = 25.5 / 2.5

I² = 10.2

I = √10.2

I = 3.194 A/s

The rate at which the current is changing is the difference between the final current and the initial current in the inductor.

= 3.194 A/s - 3.0 A/s

= 0.194 A/s

≅0.20 A/s

Therefore, the current is changing at the rate of 0.20 A/s.

The rate at which the current is changing is;

di/dt = 0.2 A/s

We are given;

Inductance; L = 5 H

Current; I = 3 A

Rate of Increase of energy; dE/dt = 3 J/s

Now, the formula for energy stored in inductor is given as;

E = ¹/₂LI²

Since we are looking for rate at which current is changing, then we differentiate both sides of the energy equation to get;

dE/dt = LI (di/dt)

Plugging in the relevant values gives;

3 = (5 × 3)(di/dt)

di/dt = 3/(5 × 3)

di/dt = 0.2 A/s

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Answer:

It becomes wider

Explanation:

Because The bigger the object the wave interacts with, the more spread there is in the interference pattern. Decreasing the size of the opening increases the spread in the pattern.

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.

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

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Answer:

the rate that the energy of a system is transformed

Explanation:

We can define energy as the capacity or ability to do work. Power is defined as the rate of doing work or the rate at which energy is transformed. It can also be regarded as the time rate of energy transfer. In older physics literature, power is sometimes referred to as activity.

Power is given by energy/time. Its unit is watt which is defined as joule per second. Another popular unit of power is horsepower. 1 horsepower = 746 watts.

Very large magnitude of power is measured in killowats and megawatts.

Answer:

a) The final speed of the 0.400-kg puck after the collision is 2.254 meters per second, b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards, c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.

Explanation:

a) Since collision is perfectly elastic and there are no external forces exerted on pucks system, the phenomenon must be modelled after the Principles of Momentum and Energy Conservation. Changes in gravitational potential energy can be neglected. That is:

Momentum

[tex]m_{1}\cdot v_{1,o} + m_{2}\cdot v_{2,o} = m_{1}\cdot v_{1,f} + m_{2}\cdot v_{2,f}[/tex]

Energy

[tex]\frac{1}{2}\cdot (m_{1}\cdot v_{1,o}^{2}+ m_{2}\cdot v_{2,o}^{2})=\frac{1}{2}\cdot (m_{1}\cdot v_{1,f}^{2}+ m_{2}\cdot v_{2,f}^{2})[/tex]

[tex]m_{1}\cdot v_{1,o}^{2} + m_{2}\cdot v_{2,o}^{2} = m_{1}\cdot v_{1,f}^{2} + m_{2}\cdot v_{2,f}^{2}[/tex]

Where:

[tex]m_{1}[/tex], [tex]m_{2}[/tex] - Masses of the 0.400-kg and 0.900-kg pucks, measured in kilograms.

[tex]v_{1,o}[/tex], [tex]v_{2,o}[/tex] - Initial speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

[tex]v_{1}[/tex], [tex]v_{2}[/tex] - Final speeds of the 0.400-kg and 0.900-kg pucks, measured in meters per second.

If [tex]m_{1} = 0.400\,kg[/tex], [tex]m_{2} = 0.900\,kg[/tex], [tex]v_{1,o} = +5.86\,\frac{m}{s}[/tex], [tex]v_{2,o} = 0\,\frac{m}{s}[/tex], the system of equation is simplified as follows:

[tex]2.344\,\frac{kg\cdot m}{s} = 0.4\cdot v_{1,f} + 0.9\cdot v_{2,f}[/tex]

[tex]13.736\,J = 0.4\cdot v_{1,f}^{2}+0.9\cdot v_{2,f}^{2}[/tex]

Let is clear [tex]v_{1,f}[/tex] in first equation:

[tex]0.4\cdot v_{1,f} = 2.344 - 0.9\cdot v_{2,f}[/tex]

[tex]v_{1,f} = 5.86-2.25\cdot v_{2,f}[/tex]

Now, the same variable is substituted in second equation and resulting expression is simplified and solved afterwards:

[tex]13.736 = 0.4\cdot (5.86-2.25\cdot v_{2,f})^{2}+0.9\cdot v_{2,f}^{2}[/tex]

[tex]13.736 = 0.4\cdot (34.340-26.37\cdot v_{2,f}+5.063\cdot v_{2,f}^{2})+0.9\cdot v_{2,f}^{2}[/tex]

[tex]13.736 = 13.736-10.548\cdot v_{2,f} +2.925\cdot v_{2,f}^{2}[/tex]

[tex]2.925\cdot v_{2,f}^{2}-10.548\cdot v_{2,f} = 0[/tex]

[tex]2.925\cdot v_{2,f}\cdot (v_{2,f}-3.606) = 0[/tex]

There are two solutions:

[tex]v_{2,f} = 0\,\frac{m}{s}[/tex] or [tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex]

The first root coincides with the conditions before collision and the second one represents a physically reasonable solution.

Now, the final speed of the 0.400-kg puck is: ([tex]v_{2,f} = 3.606\,\frac{m}{s}[/tex])

[tex]v_{1,f} = 5.86-2.25\cdot (3.606)[/tex]

[tex]v_{1,f} = -2.254\,\frac{m}{s}[/tex]

The final speed of the 0.400-kg puck after the collision is 2.254 meters per second.

b) The negative sign of the solution found in part a) indicates that 0.400-kg puck is moving westwards.

c) The speed of the 0.900-kg puck after the collision is 3.606 meters per second eastwards.