A motor has an armature resistance of 3.75 Ω . Part A If it draws 9.10 A when running at full speed and connected to a 120-V line, how large is the back emf?

Answer:

[tex]\lambda = 702\ nm[/tex]

Explanation:

Given,

slit width, d = 0.215 mm

slit separation, D = 4.90 m

Wavelength of the laser light = ?

Fringe width.[tex]\beta = 1.60\ cm[/tex]

Using formula

[tex]\beta =\dfrac{\lambda D}{d}[/tex]

[tex]\lambda=\dfrac{\beta d}{D}[/tex]

[tex]\lambda=\dfrac{0.016\times 0.215\times 10^{-3}}{4.90}[/tex]

[tex]\lambda = 7.02\times 10^{-9}\ m[/tex]

[tex]\lambda = 702\ nm[/tex]

Wavelength of the laser light = [tex]\lambda = 702\ nm[/tex]

Answer:

initial current I₀ = 0.0123 A

RC time constant τ = 0.00075 sec

current after one time constant I = 0.00452 A

voltage on the capacitor after one time constant V = 3.89 V

Explanation:

Given that,

Voltage = 6.16 V

Resistance = 500 Ω

Capacitance = 1.5 µF  

(a) What is the initial current?

The initial current can be found using

I₀ = Voltage/Resistance

I₀ = 6.16/500

I₀ = 0.0123 A

(b) What is the RC time constant?

The time constant τ provides the information about how long it will take to charge the capacitor.

τ = R*C

τ = 500*1.5x10⁻⁶

τ = 0.00075 sec

(c) What is the current after one time constant?

I = I₀e^(-τ/RC)

I = 0.0123*e^(-1)   (0.00075/0.00075 = 1)

I = 0.00452 A

(d) What is the voltage on the capacitor after one time constant?

V = V₀(1 - e^(-τ/RC))

Where V₀ is the initial voltage 6.16 V

V = 6.16(1 - e^(-1))

V = 6.16*0.63212

V = 3.89 V

That means the capacitor will charge up to 3.89 V in one time constant

Answer:

Explanation:

Given an RC circuit to analyze

R=500Ω

C=1.50-μF uncharged

Emf(V)=6.16V

Series connection

a. Initial current, since the capacitor is initially uncharged then, the voltage appears at the resistor

Using ohms law

V=iR

Then, i=V/R

i=6.16/500

i=0.01232 Amps

i=12.32 mA.

b. The time constant is given as

τ=RC

τ=500×1.5×10^-6

τ=0.00075second

τ=0.75 ms

c. What is current after 1 time constant

Current in a series RC circuit is given as

Time after I time constant is

t=1 ×τ

t= τ

i=V/R exp(-t/RC)

Where RC= τ

i=V/Rexp(-t/ τ)

i=6.16/500exp(-1), since t= τ

i=0.004532A

I=4.532mA

d. Voltage after one time constant

Voltage of a series RC circuit(charging) is given as

Again, t= τ

V=Vo(1 - exp(-t/ τ)),

V=6.16(1-exp(-1))

V=6.16(1-0.3679)

V=6.16×0.632

V=3.89Volts

V=3.89V

V=

Answer:

(a) the magnitude and direction of the electric field in the wire is 4.5 N/C towards the left end of the wire.

(b) the resistance of the wire is 0.1154 Ω

(c) the magnitude and direction of the current in the wire is 27.3 A towards the left end of the wire

(d) the current density in the wire is 4.053 x 10⁸ A/m³

Explanation:

Given;

Length of cylinder = 70cm = 0.7m

radius of cylinder = 1.75 x 10⁻⁴ m

potential V = 3.15 V

resistivity = 1.586 ✕ 10⁻⁸ Ω · m

Part (a) the magnitude and direction of the electric field in the wire

V = E x d

E = V/d

E = 3.15/0.7 = 4.5 N/C towards the left end of the wire.

Part (b) the resistance of the wire

[tex]R = \frac{\rho L}{A} \\\\R =\frac{\rho L}{\pi r^2} =R = \frac{1.586X10^{-8} X 0.7}{\pi (1.75X10^{-4})^2} = 0.1154 ohms[/tex]

R = 0.1154 Ω

Part (c) the magnitude and direction of the current in the wire

V = IR

I =V/R

I = 3.15/0.1154

I = 27.3 A towards the left end of the wire

Part (d) the current density in the wire

current density,J  = current /volume

volume = πr²h = π x (1.75 × 10⁻⁴)² x 0.7 = 6.7357 x 10⁻⁸ m³

[tex]J = \frac{27.3}{6.7357 X10^{-8}} = 4.053 X10^8 \frac{A}{m^3}[/tex]

J = 4.053 x 10⁸ A/m³

To find the force necessary to start the crate moving, we use the formula fs(max) = μsN, where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. Plugging in the given values, the force necessary to start the crate moving is 169.36 N.

The force necessary to start the crate moving is equal to the maximum static friction force. To calculate this, we use the formula:

fs(max) = μsN

where fs(max) is the maximum static friction force, μs is the coefficient of static friction, and N is the normal force. The normal force is equal to the weight of the crate, which can be calculated by N = mg, where m is the mass of the crate and g is the acceleration due to gravity. So, the force necessary to start the crate moving is:

fs(max) = μsN = μsmg

Plugging in the given values:

fs(max) = (0.55)(32 kg)(9.8 m/s²) = 169.36 N

Answer:

Because heat is a path function or the energy in transit.

Explanation:

Heat is not a substance a body possesses; it's energy transfer. Describing heat as something a body "contains" is inaccurate. Heat is energy in transit during temperature differences.

The statement that a body contains a certain amount of heat is inaccurate in the context of thermodynamics. Heat is not a substance that a body can possess like a tangible quantity. Instead, heat is a form of energy transfer between systems due to a temperature difference. When we say a body transfers heat, it implies an exchange of thermal energy between the body and its surroundings.

A body does not "contain" heat in the way it contains mass or volume. Rather, it possesses internal energy, and heat is the energy in transit. This distinction is crucial in understanding the principles of thermodynamics. When a body transfers heat, it signifies a change in its internal energy, which could result from molecular motion or other energy interactions.

Therefore, the expression "giving away something it does not have" is a misconception. The body has internal energy, and during a heat transfer, this internal energy changes, affecting the body's temperature. It is crucial to frame discussions about heat in terms of energy transfer rather than possession, aligning with the principles of thermodynamics.

For more such information on: Heat

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The question probable may be:

It is not correct to say that a body contains a certain amount of heat, yet a body can transfer heat to another body. How can a body give away something it does not have in the first place?

Answer:

[tex]W=-251096.465\ J[/tex] negativesign denotes thatthe work is consumed by the system.

Explanation:

Given:

Isothermal process.

initial temperature, [tex]T_1=300\ K[/tex]

initial pressure, [tex]P_1=150kPa[/tex]

initial volume, [tex]V_1=0.2\ m^3[/tex]

final pressure, [tex]P_2=800\ kPa[/tex]

The work done during an isothermal process is given by:

[tex]W=P_1.V_1\times ln(\frac{P_1}{P_2} )[/tex]

[tex]W=150\times 1000\times \ln\frac{150}{800}[/tex]

[tex]W=-251096.465\ J[/tex] negativesign denotes thatthe work is consumed by the system.

Answer:No, it will take a longer time to slide up than to slide down.

Explanation: An inclined plane is a plane that slides at an angle to the ground,this angle can be 45 degrees,60degrees etc. WHEN AN OBJECT IS SLIDING UP A ROUGH INCLINED SURFACE, IT WILL HAVE TO OVERCOME THE FRICTION OR RESISTANCE OF THE ROUGH SURFACE AND THE OPPOSITION OF THE FORCE IF GRAVITY, WHICH WILL CAUSE IT TO TAKE A LONGER TIME TO GO UP WHILE WHEN SLIDING DOWN IT WILL ONLY HAVE TO OVERCOME THE FRICTION OF THE ROUGH SURFACE AS THE FORCE OF GRAVITY ACTS TO PULL IT DOWN THE SLOPE.

The expression for B for a wave traveling in the -x direction is B = ωt - kx + π. The wave velocity is 0.805 m/s and the wave amplitude is 9.5 m.

To determine the expression for B where the wave would be traveling in the -x direction, we need to consider that the general equation for the wave function is y = A sin(B). In this case, the wave is traveling in the -x direction, which means the phase of the wave is shifted by π (180 degrees). So the expression for B would be B = ωt - kx + π.

The wave's velocity can be calculated using the formula v = ω/k. Substituting the given values, the wave's velocity is v = 14.5 rad/s / 18 rad/m = 0.805 m/s.

The wave's amplitude is given directly as A = 9.5 m.

Answer:

[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil

Explanation:

Given:

Now from the given it is clear that the height of water column is:

[tex]h_w=h_o-\delta h[/tex]

[tex]h_w=20-8[/tex]

[tex]h_w=12\ cm[/tex]

Now according to the pressure balance condition of fluid columns:

Pressure due to water column = Pressure due to oil column

[tex]P_w=P_o[/tex]

[tex]\rho_w.g.h_w=\rho_o.g.h_o[/tex]

[tex]1000\times 9.8\times 0.12=\rho_o\times 9.8\times 0.2[/tex]

[tex]\rho_o=600\ kg.m^{-3}[/tex] is the density of the oil

Answer:

Explanation:

Let the density of oil is d'.

height of water, h = 20 - 8 = 12 cm

height of oil, h' = 20 cm

density of water, d = 1000 kg/m³

Pressure is balanced

h' x d' x g = h x d x g

0.20 x d' x g = 0.12 x 1000 x g

0.2 d' = 120

d' = 600 kg/m³

Answer:

594.8 Hz

Explanation:

Parameters given:

Speed of sound, v = 345 m/s

Wavelength = 58 cm = 0.58 m

Speed of a wave is given as:

Speed = wavelength * frequency

Therefore:

Frequency = Speed/Wavelength

Frequency = 345/0.58

Frequency = 594.8 Hz

The frequency of the tuning fork is calculated using the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is wavelength. Substituting their given values, we get a frequency of approximately 595 Hz.

The frequency of sound can be calculated with the formula Vw = fa, where Vw is the speed of sound, f is frequency, and a is the wavelength.

From your question, we know that the speed of sound (Vw) in air is 345 m/s and the wavelength (a) is 58 cm (or 0.58 m when converted to meters to match the speed of sound's units).

Frequency (f) can be calculated by rearranging to f = Vw / a. Substituting the values: f = 345 m/s / 0.58 m = 594.83 Hz. So, the tuning fork's frequency is approximately 595 Hz.

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Answer:

The compound moves 6.5 cm in total.

Explanation:

Before solving this problem, let's first write down all lengths we know of from the question:

Starting point of sample = 1.0 cm from bottom of paper

Paper wet up to = 8.8 cm from bottom of paper

Ending point of the sample = 7.5 cm from bottom of paper

With these lengths stated, we can easily calculate the length which the compound moved through:

Length compound moved = Ending point - Starting point

Length compound moved = 7.5 - 1.0

Length compound moved = 6.5 cm

Thus, we can see that the compound moved 6.5 cm between the time the paper was put into, and taken out of the solvent.

Answer:

voltage between the plates is 4.952 × [tex]10^{-26}[/tex] V

Explanation:

given data

plate separated distance = 2.67 cm

electron speed = 1.32 × [tex]10^{7}[/tex]  m/s

solution

we will get here first force that is express as

force in parallel plate F = [tex]\frac{eV}{d}[/tex]   ..............1

and force by Newton second law F = ma   .............2

equate equation 1 and 2

ma = [tex]\frac{eV}{d}[/tex]    .................3

and here we know as kinematic equation

v²- u² = 2 × a × s    ...........4

so for initial speed acceleration will be

a = [tex]\frac{v^2-u^2}{2\times s}[/tex]  

a = [tex]\frac{(1.32 \times 10^7)^2}{2\times 2.67 \times 10^{-2}}[/tex]  

a = 3.262 × [tex]10^{-13}[/tex]  m/s²

now we put a in equation 3 and we get v

ma = [tex]\frac{eV}{d}[/tex]

9.1093 × [tex]10^{-31}[/tex] × 3.262 × [tex]10^{-13}[/tex]  = [tex]\frac{1.602 \times 10^{-19} V}{2.67 \times 10^{-2}}[/tex]  

solve it we get

v = 4.952 × [tex]10^{-26}[/tex] V

Answer:

[tex]171.43m[/tex]

Explanation:

First, define [tex]emf[/tex]( electromotive force )-is the unit electric charge imparted by an energy source. In this case the generator.

The peak emf is:

[tex]E_p_e_a_k=\sqrt2(E_m_a_x)[/tex]=[tex]\sqrt(2\times 120V)=170V[/tex]

Substituting [tex]w=2\pi f[/tex] and the value for peaf [tex]E[/tex] gives:

Total length=[tex]4\sqrt {\frac{NE_p_e_a_k}{Bw}[/tex]=[tex]4\sqrt{\frac{167\times 170}{0.041T\times 2pi \times 60.0Hz}[/tex]

=[tex]171.43m[/tex]

Hence, wire's length is 171.43m

Answer: Impulse is greater in the first case. So, option C is the correct option.

Explanation:

Case 1: Cart is travelling at 0.3 m/s and collide with an stationary object and after collision, cart rebound in opposite direction and another object remains in static condition.

Applying the conservation of linear momentum:

[tex]m_{1} \times u_{1} + m_{2} \times u_{2} = m_{1} \times v_{1} + m_{2} \times v_{2}[/tex]

[tex]m_{1} \times 0.3 + m_{2} \times 0 = m_{1} \times v_{1} + m_{2} \times 0[/tex]

Hence velocity of cart will rebound with the same velocity i.e. 0.3 m/s

Impulse is defined as the change in momentum

Impulse on the cart = [tex]m_{1} \times v_{1} - m_{1} \times u_{1}[/tex] = [tex]m_{1} \times ((-3) - (3)) = m_{1} \times (-6)[/tex] Kg m/s.

Case 2: Initially cart is travelling at 0.3 m/s and after collision it comes to rest.

So, change in momentum or Impulse = [tex]m_{1} \times (0 - 3)[/tex] = [tex]-3 \times m_{1}[/tex] Kg m/s.

Impulse is greater in the first case. So, option C is the correct option.

Answer:

The second impulse is greater then the first impulse.

(B) is correct option.

Explanation:

Given that,

In first case,

Initial speed of cart = 0.3 m/s

Final speed of cart = -0.3 m/s

in second case,

Initial speed of cart = 0.3 m/s

Final speed of cart = 0 m/s

In first case,

We need to calculate the  impulse

Using formula of impulse

[tex]I=\Delta p[/tex]

[tex]I=\Delta (mv)[/tex]

[tex]I=mv-mu[/tex]

Put the value into the formula

[tex]I=m(-0.3-0.3)[/tex]

[tex]I= -0.6m\ kg m/s[/tex]

In second case,

We need to calculate the new impulse

Using formula of impulse

[tex]I'=\Delta (mv')[/tex]

[tex]I'=mv'-mu'[/tex]

Put the value into the formula

[tex]I'=m(0-0.3)[/tex]

[tex]I'= -0.3m\ kg m/s[/tex]

So, I'> I

Hence, The second impulse is greater then the first impulse.

Answer:

[tex]6.046N[/tex]

Explanation:

The net force exerted on the mass is the sum of tension force and the external force of gravity.

[tex]F_n_e_t=F_g+F_t[/tex]

[tex]F_t[/tex] is the tension force.[tex]F_g=9.8N/kg[/tex] is the force of gravity.

[tex]F_n_e_t=ma_c=mv^2/r\\[/tex]

where [tex]r[/tex] is the rope's radius from the fixed point.

From the net force equation above:

[tex]F_t=F_n_e_t-F_g\\=mv^2/r-mg\\=0.4\times(8.5^2/2.9)-0.4\times9.8\\=6.046N[/tex]

Hence the tension force is 6.046N

Answer:

The answer to the question is;

The plate be left in the furnace for 905.69 seconds.

Explanation:

To solve the question, we have to check the Bi number as follows

Bi = [tex]\frac{hL}{k} = \frac{250\frac{W}{m^{2} K} *0.05 m}{48\frac{W}{mK} } = 0.2604[/tex]

As the Bi number is > 0.1 we have to account for the variation of temperature with location in the mass.

We perform nonlumped analysis

The relation for heat transfer given by

Y =  [tex]\frac{T_f-T_{inf}}{T_i- T_{inf}}[/tex]

=[tex]\frac{550-800}{170- 800}[/tex] = 0.3968 = C₁ exp (ζ₁² F₀)  

where

C₁ and ζ₁ are coefficients of a series solution

We therefore look for the values of C₁ and ζ₁ from Bi tables to be

ζ₁ = 0.4801 +(0.26-0.25) (0.5218-0.4801)/(0.3-0.25) ≈ 0.4884 and

C₁ = 0.4801 +(0.26-0.25) (1.0450 - 1.0382)/(0.3-0.25) ≈ 1.03956 and  

This gives the relation

0.3968 = 1.03956 exp (ζ₁² F₀)  

or ζ₁² [tex](\frac{\alpha t}{L^2})[/tex]

where

α = Thermal diffusivity of solid = k/(ρ·c[tex]_p[/tex]) = [tex]\frac{48}{7830*550}[/tex] = 1.1146×10⁻⁵

c[tex]_p[/tex] = Specific heat capacity of solid at constant pressure = 550 J/kg·K

ρ = Density of the solid = 7830 kg/m³

=㏑[tex](\frac{0.3968 }{1.03956 })[/tex] = -0.9631 from where we have

t = [tex]\frac{0.9631 *0.05^{2} }{0.4884^2*1.11*10^{-5}}[/tex]  = 905 seconds.

Answer:

a) 69.3 m/s

b) 18.84 s

Explanation:

Let the initial velocity = u

The vertical and horizontal components of the velocity is given by uᵧ and uₓ respectively

uᵧ = u sin 40° = 0.6428 u

uₓ = u cos 40° = 0.766 u

We're given that the horizontal distance travelled by the projectile rock (Range) = 1 km = 1000 m

The range of a projectile motion is given as

R = uₓt

where t = total time of flight

1000 = 0.766 ut

ut = 1305.5

The vertical distance travelled by the projectile rocks,

y = uᵧ t - (1/2)gt²

y = - 900 m (900 m below the crater's level)

-900 = 0.6428 ut - 4.9t²

Recall, ut = 1305.5

-900 = 0.6428(1305.5) - 4.9 t²

4.9t² = 839.1754 + 900

4.9t² = 1739.1754

t = 18.84 s

Recall again, ut = 1305.5

u = 1305.5/18.84 = 69.3 m/s

Answer:

2.47 V,East

Explanation:

We are given that

l=19 cm=[tex]19\times 10^{-2} m[/tex]

[tex] 1 cm=10^{-2} m[/tex]

[tex]v=11 m/s[/tex]

B=1.18 T

We have to find the potential difference induced across its ends.

[tex]E=Bvl[/tex]

Using the formula

[tex]E=1.18\times 11\times 19\times 10^{-2}[/tex]

[tex]E=2.47 V[/tex]

Hence, the potential difference induces across its ends=2.47 V

The positive charge  will move towards east direction and the negative charge will move towards west direction because the direction of force will be east.Therefore, the potential at east end will be high.

The induced potential difference in the copper bar is 2.47 V, with the east end being at a higher potential.

To determine the potential difference induced across the ends of a copper bar moving through a magnetic field, we use the formula:

V = B * l * v

where:

Substituting the given values:

V = 1.18 T * 0.19 m * 11.0 m/s = 2.47 V

The potential difference across the ends of the bar is 2.47 V.

To determine which end is at a higher potential, we apply the right-hand rule. Pointing the thumb of your right hand in the direction of the velocity (north), and your fingers in the direction of the magnetic field (upwards), the palm points towards the force acting on positive charges (from west to east).

Therefore, the east end is at a higher potential.

The correct answer is: a) East

Kepler noticed an imaginary line drawn from a planet to the Sun and this line swept out an equal area of space in equal times, If we then draw a triangle out from the Sun to a planet’s position at one point in time, it is notice that the area doesn't change even after the planet has left the original position say like after 2 to 3days or 2hours. So to have same area of triangle means that the the planet move faster when that are closer to the sun and slowly when they are far from the sun.

This led to Kepler's law of orbital motion.

First Law: Planetary orbits are elliptical with the sun at a focus.

Second Law: The radius vector from the sun to a planet sweeps equal areas in equal times.

Third Law: The ratio of the square of the period of revolution and the cube of the ellipse semi-major axis is the same for all planets.

It is this Kepler's law that makes Newton to come up with his own laws on how planet moves the way they do.

Answer:

True

Explanation:

Cohesive forces are forces that exist between  the molecules of a substance of the same material while adhesive forces are forces that exists between the molecules of the substances of different materials.

This is responsible for the  nature of menisci formed by different liquids when they are filled into containers or glass tubes.

If the cohesive forces of the liquid molecules are stronger than the adhesive forces between the liquid molecules and the glass material, a convex meniscus will be formed. This means that the edges of the liquid touching the glasses are lower than the surface at the centre. The meniscus formed by mercury in a glass tube is an example of this.

However, If the cohesive forces of the liquid molecules are weaker than the adhesive forces between the liquid molecules and the glass material, a concave meniscus will be formed. This means that the edges of the liquid touching the glasses are higher than the surface at the centre. The meniscus formed by water in a glass tube is an example of this.

Please view the attached diagram:

Please note: I got the diagram online, it was not drawn by me. I Just needed to quickly get something to illustrate my explanations. Thanks.

Answer:

[tex]\tau=0.23\;second[/tex]

Explanation:

Given,

[tex]C=78.7\;mF\\V=11.5\;V\\R=3.03\;\Omega\\[/tex]

Time constant

[tex]\tau=RC\\\tau=78.7\times10^{-3}\times3.03\\\tau=238.461\times10^{-3}\;second\\\tau=0.23\;second[/tex]

Answer:

2.17 Mpa

Explanation:

The location of neutral axis from the top will be

[tex]\bar y=\frac {(240\times 25)\times \frac {25}{2}+2\times (20\times 150)\times (25+(\frac {150}{2}))}{(240\times 25)+2\times (20\times 150)}=56.25 mm[/tex]

Moment of inertia from neutral axis will be given by [tex]\frac {bd^{3}}{12}+ ay^{2}[/tex]

Therefore, moment of inertia will be

[tex]\frac {240\times 25^{3}}{12}+(240\times 25)\times (56.25-25/2)^{2}+2\times [\frac {20\times 150^{3}}{12}+(20\times 150)\times ((25+150/2)-56.25)^{2}]=34.5313\times 10^{6} mm^{4}}[/tex]

Bending stress at top= [tex]\frac {630\times 10^{3}\times (175-56.25)}{34.5313\times 10^{6}}=2.1665127\approx 2.17 Mpa[/tex]

Bending stress at bottom=[tex]\frac {630\times 10^{3}\times 56.25}{34.5313\times 10^{6}}=1.026242858\approx 1.03[/tex] Mpa

Comparing the two stresses, the maximum stress occurs at the bottom and is 2.17 Mpa

Without the shape and dimensions of the beam's cross-section, we cannot accurately calculate the maximum bending stress from the given moment of 630 N·m.

To determine the maximum bending stress in the beam with a moment acting on the cross section (M) of 630 N·m, we need to use the formula for bending stress, which is σ = M·c/I, where σ is the stress, M is the moment, c is the distance from the neutral axis to the outer fiber, and I is the moment of inertia of the beam section. Unfortunately, we do not have the values for c and I in the question provided. For a circular cross section, c would be the radius, and I can be calculated with π·r⁴/4 where r is the radius. However, without additional information such as the shape and dimensions of the cross-sectional area of the beam, we cannot proceed further.

Since the given moment is 630 N⋅m, we need to find the values of c and I for the specific beam. Once we have these values, we can substitute them into the formula to calculate the maximum bending stress in the beam.

Without further information about the cross section of the beam, we cannot determine the exact values of c and I, and therefore, we cannot calculate the maximum bending stress.

Answer:

note:

solution is attached in word form due to error in mathematical equation. furthermore i also attach Screenshot of solution in word because to different version of MS Office please find the attachment

The problem can be addressed using energy conservation principles, factoring in gravitational potential, kinetic, and elastic potential energies, and work done by friction. Calculations will yield the speed of the package before hitting the spring, the maximum compression of the spring, and how close the package gets to its initial position on its return trip.

The student has presented a physics problem involving dynamics, energy conservation, and spring compression. This type of problem can be solved using the principles of mechanics, specifically the conservation of mechanical energy and motion on an incline with friction.

To find the speed of the package just before it reaches the spring, we would apply conservation of energy, taking into account the work done by friction. Initially, the package has potential energy due to its height on the incline, and this potential energy is converted into kinetic energy and work done against friction as the package slides down.

To determine the maximum compression of the spring, we would again use conservation of energy, where now the kinetic energy of the package is converted into elastic potential energy stored in the spring at maximum compression.

The package's return journey up the incline involves energy transformation from the spring's potential energy back to the package's kinetic energy and finally to gravitational potential energy. The effects of kinetic friction will again play a role in how far the package travels back up the incline.

- (A) Speed before hitting the spring: [tex]\( 7.3 \, \text{m/s} \)[/tex]

- (B) Maximum compression of the spring: [tex]\( 0.855 \, \text{m} \)[/tex]

- (C) Distance from initial position after rebound: [tex]\( 1.2 \, \text{m} \)[/tex]

To solve this problem, we'll address each part step-by-step.

Part A: Speed of the Package Just Before It Reaches the Spring

First, we need to find the acceleration of the package as it slides down the incline, considering friction.

Forces Acting on the Package:

1. Gravitational force parallel to the incline: [tex]\( F_g = mg \sin(\theta) \)[/tex]

2. Frictional force opposing the motion: [tex]\( F_f = \mu_k mg \cos(\theta) \)[/tex]

Calculating the Net Force:

[tex]\[ F_{\text{net}} = mg \sin(\theta) - \mu_k mg \cos(\theta) \][/tex]

[tex]\[ F_{\text{net}} = mg (\sin(\theta) - \mu_k \cos(\theta)) \][/tex]

Net Force Calculation:

[tex]\[ F_{\text{net}} = 2 \times 9.8 \times (\sin(53.1^\circ) - 0.2 \cos(53.1^\circ)) \]\[ \sin(53.1^\circ) \approx 0.8 \]\[ \cos(53.1^\circ) \approx 0.6 \]\[ F_{\text{net}} = 2 \times 9.8 \times (0.8 - 0.2 \times 0.6) \]\[ F_{\text{net}} = 2 \times 9.8 \times (0.8 - 0.12) \]\[ F_{\text{net}} = 2 \times 9.8 \times 0.68 \]\[ F_{\text{net}} = 13.328 \, \text{N} \][/tex]

Acceleration:

[tex]\[ a = \frac{F_{\text{net}}}{m} = \frac{13.328}{2} = 6.664 \, \text{m/s}^2 \][/tex]

Using Kinematics to Find Final Speed:

The package starts from rest and travels 4 meters.

[tex]\[ v^2 = u^2 + 2as \]\[ v^2 = 0 + 2 \times 6.664 \times 4 \]\[ v^2 = 53.312 \]\[ v = \sqrt{53.312} \]\[ v \approx 7.3 \, \text{m/s} \][/tex]

Part B: Maximum Compression of the Spring

When the package hits the spring, it compresses the spring until all kinetic energy is converted to potential energy in the spring and work done against friction.

Initial Kinetic Energy:

[tex]\[ KE_{\text{initial}} = \frac{1}{2} m v^2 = \frac{1}{2} \times 2 \times (7.3)^2 \]\[ KE_{\text{initial}} = \frac{1}{2} \times 2 \times 53.29 \]\[ KE_{\text{initial}} = 53.29 \, \text{J} \][/tex]

Work Done Against Friction During Compression:

Let's denote the compression of the spring as [tex]\( x \)[/tex].

[tex]\[ F_f = \mu_k m g \cos(\theta) = 0.2 \times 2 \times 9.8 \times 0.6 = 2.352 \, \text{N} \][/tex]

The work done by friction:

[tex]\[ W_f = F_f \cdot d = 2.352 \cdot x \][/tex]

Spring Potential Energy:

[tex]\[ PE_{\text{spring}} = \frac{1}{2} k x^2 = \frac{1}{2} \times 140 \times x^2 = 70x^2 \][/tex]

Energy Balance:

[tex]\[ KE_{\text{initial}} = PE_{\text{spring}} + W_f \]\[ 53.29 = 70x^2 + 2.352x \][/tex]

Solving this quadratic equation:

[tex]\[ 70x^2 + 2.352x - 53.29 = 0 \][/tex]

Using the quadratic formula:

[tex]\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]

[tex]\[ x = \frac{-2.352 \pm \sqrt{2.352^2 - 4 \times 70 \times (-53.29)}}{2 \times 70} \]\[ x = \frac{-2.352 \pm \sqrt{5.533 + 14921.2}}{140} \]\[ x = \frac{-2.352 \pm \sqrt{14926.733}}{140} \]\[ x = \frac{-2.352 \pm 122.1}{140} \][/tex]

Taking the positive root:

[tex]\[ x = \frac{119.748}{140} \]\[ x \approx 0.855 \, \text{m} \][/tex]

Part C: Distance Package Rebounds Back Up the Incline

Potential Energy in Spring:

[tex]\[ PE_{\text{spring}} = 70x^2 = 70 \times (0.855)^2 = 70 \times 0.731 \approx 51.17 \, \text{J} \][/tex]

Work Done Against Friction During Rebound:

[tex]\[ W_f = F_f \cdot d = 2.352 \times 0.855 = 2.01 \, \text{J} \][/tex]

Total energy after rebound:

[tex]\[ PE_{\text{spring}} - W_f = 51.17 - 2.01 = 49.16 \, \text{J} \][/tex]

Kinetic Energy at Rebound:

[tex]\[ KE_{\text{rebound}} = 49.16 \, \text{J} \][/tex]

Using Kinematics:

[tex]\[ KE_{\text{rebound}} = \frac{1}{2} m v^2 \]\[ 49.16 = \frac{1}{2} \times 2 \times v^2 \]\[ 49.16 = v^2 \]\[ v = \sqrt{49.16} \approx 7.01 \, \text{m/s} \][/tex]

The package travels back up the incline with this velocity until it stops due to friction and gravity.

Distance Traveled Up the Incline:

Using energy conservation:

[tex]\[ KE_{\text{rebound}} = mgh + W_f \]\[ 49.16 = 2 \times 9.8 \times h + 2.352 \times h \]\[ 49.16 = 19.6h + 2.352h \]\[ 49.16 = 21.952h \]\[ h = \frac{49.16}{21.952} \approx 2.24 \, \text{m} \][/tex]

The distance along the incline:

[tex]\[ d = \frac{h}{\sin(\theta)} \approx \frac{2.24}{0.8} \approx 2.8 \, \text{m} \][/tex]

Distance from Initial Position:

[tex]\[ \text{Distance from initial position} = 4 - 2.8 = 1.2 \, \text{m} \][/tex]

The complete question is:

A 2 kg package is released on a 53.1° incline, 4 m from a long spring with force constant k = 140 N/m that is attached at the bottom of the incline (Fig. 7-32). The coefficients of friction between the package and the incline are µs = 0.4 and µk= 0.2. The mass of the spring is negligible.

A. What is the speed of the package just before it reaches the spring?

B. What is the maximum compression of the spring?

C. The package rebounds back up the incline. How close does it get to its initial position?

Answer:

μk = (Vf - Vc)/(T×g)

Explanation:

Given

Vi = initial velocity of the car

Vf = final velocity of the car

T = Time of application of brakes

g = acceleration due to gravity (known constant)

Let the mass of the car be Mc

Assuming only kinetic frictional force acts on the car as the driver applies the brakes,

The n from Newtown's second law of motion.

Fk = Mc×a

Fk = μk×Mc×g

a = (Vf - Vc)/T

Equating both preceding equation.

μk×Mc×g = Mc × (Vf - Vc)/T

Mc cancels out.

μk = (Vf - Vc)/(T×g)

Final answer:

To find the coefficient of kinetic friction of the tires against the road, you need the initial and final speeds of the car, the time it took to decelerate, and the acceleration due to gravity. Without specific values for these variables, the problem cannot be solved directly, but the formula given provides a method to calculate the coefficient if such values are known.

Explanation:

The question asks for the coefficient of kinetic friction between the tires and the road when a car, initially traveling at speed vi, decelerates to a final speed vf over time T seconds due to the driver slamming on the brakes upon seeing a rabbit.

The formula to calculate the coefficient of kinetic friction (μ_k) can be derived from Newton's second law of motion and the equation of motion that relates initial velocity, final velocity, acceleration, and time. The formula for the coefficient of kinetic friction is μ_k = (vi - vf) / (g × T), where g is the acceleration due to gravity (9.81 m/s2).

To solve this problem, you need the initial and final velocities of the car (vi and vf), the deceleration time (T), and knowledge that the acceleration due to gravity (g) is approximately 9.81 m/s2. However, actual calculations cannot be performed without specific values for vi, vf, and T.

This equation illustrates that the coefficient of kinetic friction is directly related to the deceleration rate of the car on the road surface.

Answer:

av=0.333m/s, U=3.3466J

b.

[tex]v_{A2}=-1.333m/s,\\ v_{B2}=0.667m/s[/tex]

Explanation:

a. let [tex]m_A[/tex] be the mass of block A, and[tex]m_B=10.0kg[/tex] be the mass of block B. The initial velocity of A,[tex]\rightarrow v_A_1=2.0m/s[/tex]

-The initial momentum =Final momentum since there's no external net forces.

[tex]pA_1+pB_1=pA_2+pB_2\\\\P=mv\\\\\therefore m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]

Relative velocity before and after collision have the same magnitude but opposite direction (for elastic collisions):

[tex]v_A_1-v_B_1=v_{B2}-v_{A2}[/tex]

-Applying the conservation of momentum. The blocks have the same velocity after collision:

[tex]v_{B2}=v_{A2}=v_2\\\\2\times 2+10\times 0=2v_2+10v_2\\\\v_2=0.3333m/s[/tex]

#Total Mechanical energy before and after the elastic collision is equal:

[tex]K_1+U_{el,1}=K_2+U_{el,2}\\\\#Springs \ in \ equilibrium \ before \ collision\\\\U_{el,2}=K_1-K_2=0.5m_Av_A_1^2-0.5(m_A+m_B)v_2^2\\\\U_{el,2}=0.5\times 2\times 2^2-0.5(2+10)(0.333)^2\\\\U_{el,2}=3.3466J[/tex]

Hence, the maxumim energy stored is U=3.3466J, and the velocity=0.333m/s

b. Taking the end collision:

From a above, [tex]m_A=2.0kg, m_B=10kg, v_A=2.0,v_B_1=0[/tex]

We plug these values in the equation:

[tex]m_Av_A_1+m_Bv_B_1=m_Av_{A2}+m_Bv_{B2}[/tex]

[tex]2\times2+10\times0=2v_A_2+10v_B_2\\\\2=v_A_2+5v_B_2\\\\#Eqtn 2:\\v_A_1-v_B_1=v_{B2}-v_{A2}\\\\2-0=v_{B2}-v_{A2}\\\\2=v_{B2}-v_{A2}\\\\#Solve \ to \ eliminate \ v_{A2}\\\\6v_{B2}=2.0\\\\v_{B2}==0.667m/s\\\\#Substitute \ to \ get \ v_{A2}\\\\v_{A2}=\frac{4}{6}-2=1.333m/s[/tex]

(a) The maximum energy stored in the spring bumpers during the collision is 3.00 J, and the velocity of both the blocks is 0.50 m/s.
(b) After they move apart, block A has a velocity of -1.00 m/s and block B has a velocity of 1.00 m/s.

You can follow these simple steps to find the required solution -

(a) Maximum Energy Stored in the Spring Bumpers

To find the maximum energy stored in the spring bumpers, we will use the conservation of momentum and energy principles.

Initially, block A (mass 2.00 kg) is moving at 2.00 m/s, and block B (mass 6.00 kg) is at rest. The total initial momentum ([tex]p_{initial[/tex]) is:

At the point of maximum compression, both blocks momentarily move with the same velocity ([tex]v_{common[/tex]). Using the conservation of momentum:

Solving for [tex]v_{common[/tex]:

Next, determine the initial kinetic energy ([tex]KE_{initial[/tex]):

The kinetic energy at the point of maximum compression ([tex]KE_{final[/tex]) is:

The maximum energy stored in the spring bumpers ([tex]E_{spring[/tex]) is the difference between [tex]KE_{initial[/tex] and [tex]KE_{final[/tex]:

(b) Velocity of Each Block After Collision

After they have moved apart, assuming an elastic collision where both kinetic energy and momentum are conserved, the final velocities ([tex]v_A_{final[/tex] and [tex]v_B_{final[/tex]) can be found using the respective equations:

For block A:

For block B:

Therefore, the final velocities are:

Answer:

Explanation:

Area, A = 4 cm x 4.2 cm = 16.8 cm²

A).

[tex]\overrightarrow{E}=150\widehat{i}-200\widehat{k}[/tex]

Area is in x y plane so

[tex]\overrightarrow{A}=16.8\times 10^{-4}\widehat{k}[/tex]

Electric flux,

[tex]\phi =\overrightarrow{E}.\overrightarrow{A}[/tex]

[tex]\phi =\left ( 150\widehat{i}-200\widehat{k} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )[/tex]

Ф = 0.336 Nm²/C

B).

[tex]\overrightarrow{E}=150\widehat{i}-200\widehat{j}[/tex]

[tex]\phi =\overrightarrow{E}.\overrightarrow{A}[/tex]

[tex]\phi =\left ( 150\widehat{i}-200\widehat{j} \right ).\left (16.8\times 10^{-4}\widehat{k} \right )[/tex]

Ф = 0 Nm²/C

Answer:

roof bow upwards

Explanation:

The top of the roof of the small ranger vehicle will bow upwards. This is as a result of gas pressure on the soft ragtop roof.

Explanation:

There are three forces on the plank.  T₁ pulling up at point P, T₂ pulling up at point Q, and W pulling down at point C.

Let's say the length of the plank is L.

Sum of forces in the y direction before rope II is moved:

∑F = ma

150 N + 150 N − W = 0

W = 300 N

Sum of moments about point P after rope II is moved:

∑τ = Iα

(T₁) (0) − (300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (L/2) + (T₂) (3L/4) = 0

-(300 N) (1/2) + (T₂) (3/4) = 0

-150 N + 3/4 T₂ = 0

T₂ = 200 N

Sum of forces in the y direction:

∑F = ma

T₁ + 200 N − 300 N = 0

T₁ = 100 N

The new tensions in the two ropes after the movement of rope 2 are;

T₁ = 100 N

T₁ = 100 NT₂ = 200 N

We are told that as the plank is currently, the two ropes attached at each end have tension of 150 N each.

Thus;

T₁ = T₂ = 150 N

The two ropes are acting in tension upwards and so for the plank to be balanced, there has to be a downward force(which is the weight of the plank) must be equal to the sum of the tension in the two ropes.

Thus, from equilibrium of forces, we have;

W = T₁ + T₂

W = 150 + 150

W = 300 N

Now, we are told that;

Rope 2 is now moved to a point R which is halfway between point C and Q. Since C is the centre of the plank and R is the midpoint of C and Q, if the length of the plank is L, then the distance of rope 2 from point P is now ¾L.

Since the plank remains horizontal after shifting the rope 2 to point R, let us take moments about point P to get;

T₂(¾L) - W(½L) = 0

Plugging in the relevant values;

T₂(¾L) - 300(½L) = 0

T₂(¾L) - 150L = 0

Rearrange to get;

T₂(¾L) = 150L

Divide both sides by L to get;

T₂(¾) = 150

Cross multiply to get;

T₂ = 150 × 4/3

T₂ = 200 N

Thus;

T₁ = 300 - 200

T₁ = 100 N

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Answer:

The value of spring constant for each spring of the suspension system of a car   K = 343.35 [tex]\frac{N}{cm}[/tex]

Explanation:

Total force on the springs = weight of the four passengers

⇒ F = 4 × 70 × 9.81

⇒ F = 2746.8 N

In the suspension system of the car the four springs are connected in parallel. So Equivalent spring constant is given by,

⇒ [tex]K_{eq}[/tex] = 4 K -------- ( 1 )

Depression in the spring Δx = 2 cm

Now the force  on the spring is given by

F = [tex]K_{eq}[/tex] × Δx

⇒ [tex]K_{eq}[/tex] = [tex]\frac{2746.8}{2}[/tex]

⇒ [tex]K_{eq}[/tex] = 1373.4 [tex]\frac{N}{cm}[/tex]

Now the spring constant for each spring = [tex]\frac{K_{eq}}{4}[/tex]

⇒ K = [tex]\frac{1373.4}{4}[/tex]

⇒ K = 343.35 [tex]\frac{N}{cm}[/tex]

This is the value of spring constant for each spring of the suspension system of a car.

The effective spring constant of the suspension system of a car. is mathematically given as

K = 343.35N/cm

Question Parameter(s):

Consider a load of 4 passengers, each with a mass of 70 kg.

the tires are depressed by about dx = 2.0 cm.

Generally, the equation for the Force   is mathematically given as

F = Keq × dx

Therefore

Keq= 2746.8/2

Keq= 1373.4N/cm

In conclusion, spring constant for each spring

K = 1373.4/4

K = 343.35N/cm

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