What is the voltage across six 1.5-V batteries when they are connected (a) in series, (b) in parallel, (c) three in parallel with one another and this combination wired in series with the remaining three?

Answer:

maximum value of the magnetic field  B  = 1 ×[tex]10^{-8}[/tex] T

average intensity of the light = 0.011937 W/m²

power of source =  14.40 J

Explanation:

given data

maximum electric field E = 3.0 V/m

distance from a point source r = 9.8 m

solution

first we get here maximum value of the magnetic field  

maximum value of the magnetic field   = [tex]\frac{E}{C}[/tex]   .........1

maximum value of the magnetic field   = [tex]\frac{3}{3 \times 10^8}[/tex]

maximum value of the magnetic field  B  = 1 ×[tex]10^{-8}[/tex] T

and

now we get average intensity of the light that is

average intensity of the light =  [tex]\frac{EB}{2\mu _o}[/tex]   .........2

average intensity of the light = [tex]\frac{3 \times 1 \times 10^{-8}}{2 \times 4\pi \times 10^{-7}}[/tex]  

average intensity of the light = 0.011937 W/m²

and

now we get power of source that is express as

power of source = average intensity × 4×π×r²   ..........3

power of source = 0.011937 × 4×π×9.8²

power of source =  14.40 J

Answer:

I₀ = 0.2 mA

Explanation:

        [tex]V = I_{0}*R \\\\ I_{0} = \frac{V}{R} = \frac{1.50V}{7.71e3\Omega} = 0.2 mA[/tex]

0.19mA

Given;

Capacitor of capacitance = 4.07 μF

Resistor of resistance = 7.71 kΩ = 7.71 x 1000Ω  = 7710Ω

Voltage = 1.50V

Since the capacitor is initially uncharged, it behaves like a short circuit. Therefore, the only element drawing current at that instant is the resistor.  This means that the initial current in the circuit is the one due to (flowing through) the resistor.

And since there is negligible internal resistance, the emf of the battery is equal to the voltage supplied by the battery and is used to supply current to the resistor. Therefore, according to Ohm's law;

V = I x R           ---------------(i)

Where;

V = voltage supplied or the emf

I = current through the resistor

R = resistance of the resistor

Substitute the values of V and R into equation (i) as follows;

1.50 = I x 7710

I = [tex]\frac{1.5}{7710}[/tex]

I = 0.00019A

Multiply the result by 1000 to convert it to milliamperes as follows;

0.00019 x 1000 mA = 0.19mA

Therefore, the initial current in the circuit, expressed in milliamperes is 0.19

Answer:

The final speed of the rod is 0.86 m/s.

Explanation:

Given that,

Mass of rod = 1.3 kg

Distance between rail= 0.42 m

Current = 2.6 A

Magnetic field = 0.35 T

Distance = 1.25 m

We need to calculate the acceleration

Using formula of magnetic force

[tex]F= Bil[/tex]

[tex]ma=Bil[/tex]

[tex]a=\dfrac{Bil}{m}[/tex]

Put the value into the formula

[tex]a=\dfrac{0.35\times2.6\times0.42}{1.3}[/tex]

[tex]a=0.294\ m/s^2[/tex]

We need to calculate the final speed of the rod

Using equation of motion

[tex]v^2-u^2=2as[/tex]

Put the value in the equation

[tex]v^2=2\times0.294\times1.25[/tex]

[tex]v=0.86\ m/s[/tex]

Hence, The final speed of the rod is 0.86 m/s.

The final speed of the rod is 0.86 m/s.

This refers to the rate of change of the position of an object in a specified direction.

To calculate the acceleration

We use the formula of magnetic force

a=Bil/m

a= (0.35 x 2.6 x 0.42)/1.3

a= 0.294m/s^2

Then the final speed of the rod

We use the equation of motion

v^2 - u^2= 2as

=> v^2= 0.86m/s

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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:

Current I=6.34A

Explanation:

Given data

L=L₁=L₂=4.42 m

I₁=7.33A

Angle α=69.4°

B=0.547T

Force F=2.24N

Required

Current I

Solution

The length of each wire ,the magnetic field B,and the angle are same for both wires.

As we know that Force is:

[tex]F_{net}=I_{1}LBSin\alpha -I_{2}LBSin\alpha\\F_{net}=(I_{1}-I_{2})LBSin\alpha\\I_{2}=I_{1}-\frac{F_{net}}{LBSin\alpha}\\I_{2}=7.33A-\frac{2.24N}{(4.42m)(0.547T)Sin(69.4)} \\I_{2}=6.34A[/tex]    

Current I=6.34A

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:

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.

The tangential acceleration of a point on a rotating rigid body with constant angular acceleration depends on the change in the angular velocity. It's represented by the formula a_t = r * α, thus can change if the radius changes, even if the angular acceleration is constant.

In this case, a rigid body rotates about a fixed axis with constant angular acceleration. The tangential acceleration of any point on the body would depend on the change in the angular velocity, making the correct answer (b). This is because the tangential acceleration is directly proportional to the angular acceleration and the distance from the axis of rotation, as represented by the formula a_t = r * α, where a_t is the tangential acceleration, r is the radius, and α is the angular acceleration.

Therefore, if the angular acceleration is constant, the tangential acceleration can change if the radius changes. However, if the radius is also constant, then the tangential acceleration will be constant in magnitude, but its direction will change as the direction of the tangent to the motion changes.

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The tangential acceleration of a point on a rotating body with constant angular acceleration depends on the change in the angular velocity. This is due to the relationship defined by the equation for tangential acceleration at = α x r.

In the context of a rigid body rotating about a fixed axis with constant angular acceleration, the correct statement in relation to the tangential acceleration of any point on the body would be: b. The tangential acceleration depends on the change in the angular velocity.

This is because, in physics, tangential acceleration is a measure of how the tangential velocity of a point at a certain radius changes with time. It is directly proportional to the angular acceleration (α) and the radius (r), expressed by the equation at = α x r. Therefore, the tangential acceleration will change as the angular velocity changes, provided there is angular acceleration.

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Answer:

Explanation:

Check attachment for solution

Answer:

d = <23, 33, 0> m ,    F_W = <0, -9.8, 0> ,   W = -323.4 J

Explanation:

We can solve this exercise using projectile launch ratios, for the x-axis the displacement is

         x = vox t

Y Axis  

         y = [tex]v_{oy}[/tex] t - ½ g t²

It's displacement is

      d = x i ^ + y j ^ + z k ^

Substituting

      d = (23 i ^ + 33 j ^ + 0) m

Using your notation

   d = <23, 33, 0> m

The force of gravity is the weight of the body

         W = m g

        W = 1  9.8 = 9.8 N

In vector notation, in general the upward direction is positive

         W = (0 i ^  - 9.8 j ^ + 0K ^) N

         W = <0, -9.8, 0>

Work is defined

           W = F. dy

             W = F dy cos θ

In this case the force of gravity points downwards and the displacement points upwards, so the angle between the two is 180º

          Cos 180 = -1

           W = -F y

           W = - 9.8 (33-0)

           W = -323.4 J

Answer:

a. The energy of the air at the initial and the final states is 0kJ and 0.171kJ respectively

b. 0.171kJ

c. 0.143

Explanation:

a.

Because there are same conditions of the state of air at the surroundings and at the Initial stage, the energy of air at the Initial stage is 0kJ.

Calculating energy at the final state;

We start by calculating the specific volume of air in the environment and at the final state.

U2 = At the final state, it is given by

RT2/P2

U1= At the Initial state, it is given by

RT1/P1

Where R = The gas constant of air is 0.287 kPa.m3/kg

T2 = 150 + 273 = 423K

T1 = 25 + 273 = 298K

P2 = 600KPa

P1 = 100KPa

U2 = 0.287 * 423/600

U2 = 0.202335m³/kg

U1 = 0.287 * 298/100

U1 = 0.85526m³/kg

Then we Calculate the mass of air using ideal gas relation

PV = mRT

m = P1V/RT1 where V = 2*10^-3kg

m = 100 * 2 * 10^-3/(0.287 * 298)

m = 0.00234kg

Then we calculate the entropy difference, ∆s. Which is given by

cp2 * ln(T2/T1) - R * ln(P2/P1)

Where cp2 = cycle constant pressure = 1.005

∆s = 1.005 * ln (423/298) - 0.287 * ln(600/100)

∆s = -0.1622kJ/kg

Energy at the final state =

m(E2 - E1 + Po(U2 - U1) -T0 * ∆s)

E2 and E1 are gotten from energy table as 302.88 and 212.64 respectively

Energy at the final state

= 0.00234(302.88 - 212.64 + 100(0.202335 - 0.85526) - 298 * -0.1622)

Energy at the final state = 0.171kJ

b.

Minimum Work = ∆Energy

Minimum Work = Energy at the final state - Energy at the initial state

Minimum Work = 0.171 - 0

Minimum Work done = 0.171kJ

c. The second-law efficiency of this process is calculated by ratio of meaningful and useful work

= 0.171/1.2

= 0.143

Exergy measures the maximum work a system can produce. To calculate its change and consequently the minimum work supplied and second-law efficiency, additional data like specific heats are required.

This involves thermodynamics, a branch of physics that deals with energy transfer. Specifically, this question is about the concept of exergy, a measure of the maximum amount of work a system can produce with respect to its environment.

(a) The exergy (or available energy) of a system in a given state is the maximum theoretical work that can be obtained as the system communicates with an equilibrium state. In this case, the initial and final states of the system are given, but we need more data such as the specific heats, to compute the initial and final exergies.

(b) The minimum work that must be supplied is equivalent to the change in exergy from the initial to the final state, but again, it cannot be determined without knowing the specific heat values of air

(c) The second-law efficiency is defined as the ratio of the actual work to the work done in a reversible process. Here, it is the ratio of the useful work input (1.2 kJ) to the minimum work needed for the compression process. To find the exact efficiency, we need to compute the minimum required work, which would require the specific heat values.

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Answer:

The current in the primary is 0.026 A

Explanation:

Using the formula

I1 = (V1/V2)*I2

we have

I1 = (6.4/120)*0.500

I1 = 0.026 A

To find the current in the primary coil of a transformer, we use the power equivalence between the primary and secondary sides, and solve for the primary current using the given voltages and secondary current. This results in a primary current of 26.7 mA.

Calculating the Primary Current in a Transformer

To determine the primary current in a transformer, we need to use the relationship between the power in the primary and secondary circuits. The power delivered by an ideal transformer is the same on both sides, which means Powerprimary = Powersecondary. This can be written as:

Given parameters are:

We first calculate the power on the secondary side:

For an ideal transformer, this power must be equal to the power on the primary side:

Thus, we have:

Solving for [tex]Iprimary[/tex], we get:

Therefore, the current in the primary coil is 26.7 mA.

Answer:

2.33651226158 m

Explanation:

From the question the required data is as follows

f = Frequency of the initial note = 146.8 Hz

v = Velocity of sound in air = 343 m/s

The wavelength of a wave is given by

[tex]\lambda=\dfrac{v}{f}[/tex]

[tex]\Rightarrow \lambda=\dfrac{343}{146.8}[/tex]

[tex]\Rightarrow \lambda=2.33651226158\ m[/tex]

The wavelength of the initial note is 2.33651226158 m

Two circuits are considered equivalent when their output values are the same for all possible input combinations.

A circuit equivalent to another is one that meets the same conditions, (eg same current), under a different configuration.

Therefore, we can conclude that two circuits are considered equivalent when their output values are the same for all possible input combinations.

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Final answer:

Two circuits are considered equivalent if they produce the same output for all possible input combinations, as can be demonstrated using a truth table in digital logic. The correct option is C.

Explanation:

Two circuits are considered equivalent when their output values are the same for all possible input combinations. This means that regardless of the input values, both circuits will produce the same output across all scenarios. For instance, in digital logic, two circuits using different logic gates such as AND, OR, NAND, or XOR would be equivalent if they yield the same output for each combination of inputs, as represented in a truth table. An example of this is the use of De Morgan's laws where an AND gate followed by a NOT gate can be equivalent to a NAND gate, producing the same outputs. Understanding these principles is essential for fields like electronics, computer engineering, and programming. Hence, Option C is correct.

Answer:

μk = 0.408

Explanation:

Given:

m=0.50 Kg,

Let compressed distance x = 0.20 m, and

stretched distance after releasing y = 1.00 m

K = 100 N/M

Sol:

Law of conservation of energy

Energy dissipation due to friction = P.E stores in the spring

Ff * y = 1/2 K x ²   (Ff = μk Fn)  And (Fn = mg) so

μk mgy  =   1/2 K x ²

μk = 1/2 K x ² /mgy   Putting values

μk = (1/2 ) (100 N/M) (0.20 m)² / (0.50 Kg x 9.8 m/s² x 1 m)

μk = 0.408

0.4

(i) Since the mass is forced against the spring, an elastic energy ([tex]E_{E}[/tex]) due to the compression of the spring by the force is produced and is given according to Hooke's law by;

[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] k c²            --------------------------------(i)

Where;

k = spring's constant

c = compression caused on the spring.

From the question;

k = 100N/m

c = 0.20m

Substitute these values into equation (i) as follows;

[tex]E_{E}[/tex] = [tex]\frac{1}{2}[/tex] x 100 x 0.20²

[tex]E_{E}[/tex] = 2J

(ii) Now, when the mass is released, it causes the block to move some distance until it stops thereby doing some work within that distance. This means that the elastic energy is converted to workdone. i.e

[tex]E_{E}[/tex] = W          ------------(ii)

The work done (W) is given by the product of the net force(F) on the block and the distance covered(s). i.e

W = F x s           -----------------(iii)

But, the only force acting on the body as it moves is the frictional force ([tex]F_{R}[/tex]) acting to oppose its motion. i.e

F = [tex]F_{R}[/tex]

Where;

[tex]F_{R}[/tex] = μN      [μ = coefficient of kinetic friction, N = mg = normal reaction between the block and the tabletop, m = mass of the block, g = gravity]

[tex]F_{R}[/tex] = μ x mg

Substitute F = [tex]F_{R}[/tex] = μ x mg into equation (iii) as follows;

W = μ x mg x s             ----------------(iv)

Now substitute the value of W into equation (ii) as follows;

[tex]E_{E}[/tex] = μ x mg x s                ------------------(v)

Where;

[tex]E_{E}[/tex] = 2 J               [as calculated above]

m = 0.50 kg

s = distance moved by block = 1.00m

g = 10m/s²            [a known constant]

Substitute these values into equation (v) as follows;

2 = μ x 0.50 x 10 x 1

2 = 5μ

μ = 2 / 5

μ = 0.4

Therefore, the coefficient of kinetic friction between the block and the tabletop is 0.4

Explanation:

The mixing chamber will be well insulated when steady operating conditions exist such that there will be negligible heat loss to the surroundings. Therefore, changes in the kinetic and potential energies of the fluid streams will be negligible and there are constant fluid properties with no work interactions.

   [tex]T < T_{sat}[/tex] at 250 kPa = [tex]127.41^{o}C[/tex]

   [tex]h_{1}[/tex] approx equal to [tex]h_{f}[/tex] at [tex]80^{o}C[/tex]

              = 335.02 kJ/kg

    [tex]h_{2}[/tex] ≈ [tex]h_{f}[/tex] at [tex]20^{o}C[/tex]

                      = 83.915 kJ/kg

and,    [tex]h_{3}[/tex] ≈ [tex]h_{f}[/tex] at [tex]42^{o}C[/tex] = 175.90 kJ/kg

Therefore, mass balance will be calculated as follows.

   [tex]m^{o}_{in} - m^{o}_{out} = \Delta m^{o}_{system} \rightarrow m^{o}_{1} + m^{o}_{2} = m^{o}_{3}[/tex]

And, energy balance will be given as follows.

      [tex]E^{o}_{in} - E^{o}_{out} = \Delta E^{o}_{system}[/tex]

As we are stating steady conditions,

     [tex]\Delta m^{o}_{system}[/tex] and [tex]\Delta E^{o}_{system}[/tex] cancel out to zero.

So,    [tex]E^{o}_{in} = E^{o}_{out}[/tex]

     [tex]m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = m^{o}_{3}(h_{3})[/tex]

On combining the relations, we solve for [tex]m^{o}_{2}[/tex] as follows.

   [tex]m^{o}_{1}(h_{1}) + m^{o}_{2}(h_{2}) = (m^{o}_{1} + m^{o}_{2})(h_{3})[/tex]

   [tex]m^{o}_{2} = (\frac{(h_{1} - h_{3})}{(h_{3} - h_{2})}) \times m^{o}_{1}[/tex]

              = [tex]\frac{(335.02 - 175.90)}{(175.90 - 83.915)} \times 0.5[/tex]  

       [tex]m^{o}_{2}[/tex] = 0.865 kg/s

                       = 51.9 kg/min      (as 1 min = 60 sec)

Thus, we can conclude that the mass flow rate of cold stream is 51.9 kg/min.

Answer:

Photons Generated= ~1715 photons

Explanation:

The detailed explanation of answer is given in attached file.

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:A. Keep pushing the box forward at a steady speed.

Explanation: Frictional force is a resistant force which oppose the Movement of an object, frictional force can emanate from a rough surface.

When an object that is moving with a consistent force is opposed by the roughness of the surface through which it is moving,it will cause the object to continue to move with a reduced speed as it goes along.

WHEN YOU APPLY A CONSTANT FORCE ON A MOVING OBJECT THAT IS OPPOSED BY A ROUGH SURFACE IT WILL RESULT IN AN AVERAGE MINIMAL FORCE BEING APPLIED TO THE OBJECT.

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.

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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:

a. 0.21 rad/s2

b. 2.205 N

Explanation:

We convert from rpm to rad/s knowing that each revolution has 2π radians and each minute is 60 seconds

200 rpm = 200 * 2π / 60 = 21 rad/s

180 rpm = 180 * 2π / 60 = 18.85 rad/s

r = d/2 = 30cm / 2 = 15 cm = 0.15 m

a)So if the angular speed decreases steadily (at a constant rate) from 21 rad/s to 18.85 rad/s within 10s then the angular acceleration is

[tex]\alpha = \frac{\Delta \omega}{\Delta t} = \frac{21 - 18.85}{10} = 0.21 rad/s^2[/tex]

b) Assume the grind stone is a solid disk, its moment of inertia is

[tex]I = mR^2/2[/tex]

Where m = 28 kg is the disk mass and R = 0.15 m is the radius of the disk.

[tex] I = 28*0.15^2/2 = 0.315 kgm^2[/tex]

So the friction torque is

[tex]T_f = I\alpha = 0.315*0.21 = 0.06615 Nm[/tex]

The friction force is

[tex]F_f = T_f/R = 0.06615 / 0.15 = 0.441 N[/tex]

Since the friction coefficient is 0.2, we can calculate the normal force that is used to press the knife against the stone

[tex]N = F_f/\mu = 0.441/0.2 = 2.205 N[/tex]

Answer:

-0.3

Explanation:

F' = μmg ........... Equation 1

Where F' = Frictional force, μ = coefficient of kinetic friction, m = mass of the stone, g = acceleration due to gravity.

But,

F' = ma ............ Equation 2

Where a = acceleration of the stone.

Substitute equation 2 into equation 1

ma = μmg

dividing both side of the equation by m

a = μg

make μ the subject of the equation

μ = a/g............... Equation 3

From the equation of motion,

v² = u²+2as................. Equation 4

Where v and u are the final and the initial velocity respectively, s = distance.

Given: v = 0 m/s (to rest), u = 8.0 m/s, s = 11 m.

Substitute into equation 4

0² = 8² + 2×11×a

22a = -64

a = -64/22

a = -32/11 m/s² = -2.91 m/s²

substitute the values of a and g into equation 3

μ = -2.91/9.8

μ = -0.297

μ ≈ -0.3

Final answer:

The coefficient of kinetic friction is found using the work-energy principle by equating the initial kinetic energy of the stone to the work done by friction. Given that the stone travels 11 meters and comes to rest, the coefficient of kinetic friction is calculated as approximately 0.296.

Explanation:

To find the coefficient of kinetic friction between the stone and the surface, we need to use the work-energy principle, which states that the work done by all the forces acting on an object is equal to the change in its kinetic energy. Since the stone comes to rest, all of its initial kinetic energy has been converted into work done against friction.

First, let's calculate the initial kinetic energy (KE) of the stone:

This energy is equal to the work done by friction (Wf):

Wf = Frictional force (f) x Distance (d)

Since the frictional force is equal to the kinetic friction coefficient (µk) multiplied by the normal force (N), and the normal force is equal to the weight of the stone (mg, where g is the acceleration due to gravity), we can express the work done by friction as:

Wf = µkmgd

Setting the work done by friction equal to the initial kinetic energy gives us:

32 J = µkmg(11 m)

Solving for the coefficient of kinetic friction:

µk = 32 J / (mg x 11 m)

Now, assuming the acceleration due to gravity (g) is 9.8 m/s²:

µk = 32 J / (m x 9.8 m/s² x 11 m)

Since the mass (m) cancels out, we don't need to know it:

µk = 32 J / (9.8 m/s² x 11 m)

µk = 0.296

Therefore, the coefficient of kinetic friction between the stone and the level surface is approximately 0.296.

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

Explanation:

It is known that relation between torque and angular acceleration is as follows.

                    [tex]\tau = I \times \alpha[/tex]

and,       I = [tex]\sum mr^{2}[/tex]

So,      [tex]I_{1} = 2 kg \times (1 m)^{2} + 2 kg \times (1 m)^{2}[/tex]

                       = 4 [tex]kg m^{2}[/tex]

      [tex]\tau_{1} = 4 kg m^{2} \times \alpha_{1}[/tex]

     [tex]\tau_{2} = I_{2} \alpha_{2}[/tex]

So,      [tex]I_{2} = 2 kg \times (0.5 m)^{2} + 2 kg \times (0.5 m)^{2}[/tex]

                     = 1 [tex]kg m^{2}[/tex]

 as [tex]\tau_{2} = I_{2} \alpha_{2}[/tex]

                   = [tex]1 kg m^{2} \times \alpha_{2}[/tex]        

Hence,     [tex]\tau_{1} = \tau_{2}[/tex]

                  [tex]4 \alpha_{1} = \alpha_{2}[/tex]

            [tex]\alpha_{1} = \frac{1}{4} \alpha_{2}[/tex]

Thus, we can conclude that the new rotation is [tex]\frac{1}{4}[/tex] times that of the first rotation rate.

Final answer:

The final angular velocity will be approximately 1 rev/s.

Explanation:

First, let's calculate the initial angular velocity using the formula:

Initial Angular Velocity = Initial Moment of Inertia * Initial Angular Velocity / Final Moment of Inertia

Given that the initial moment of inertia is 5 kg(m2), the initial angular velocity is 1 rev/s, and the final moment of inertia is approximately 5 kg(m2), we can calculate the initial angular velocity as follows:  (5 kg(m2) * 1 rev/s / 5 kg(m2)) = 1 rev/s. Therefore, the final angular velocity will also be approximately 1 rev/s.

Answer:

E = I(R + r)

Making I the subject of the formular by dividing both sides by R + r,

I = E/(R + r)

E = 4V, r = 0.7Ohm, R = 0

I = 4/(0 + 0.7) = 4/0.7

I = 5.174285714A

Explanation:

For a cell of emf E, internal resistance r, connected to an external resistance R, the current flowing through the circuit will be given as:

I = E/(R + r). I is measured in Amperes(A), emf in volts(V), R in Ohms and internal resistance r also in ohms

Explanation:

According to the free body diagram a block of mass m will have expression for force as follows.

                  N = ma

and,       [tex]f_{c} - mg[/tex] = 0  

             [tex]\mu_{s}N - mg[/tex] = 0

       [tex]\mu_{s} = \frac{mg}{N}[/tex] = [tex]\frac{mg}{ma}[/tex]

                  = [tex]\frac{g}{a}[/tex]

                  = [tex]\frac{9.8}{13}[/tex]

                  = 0.75

Therefore, we can conclude that the value of coefficient of static friction between the two blocks is 0.75.

The coefficient of static friction between the large block and smaller block is equal to 0.754.

Given the following data:

Acceleration due to gravity = 9.8 [tex]m/s^2[/tex]

To determine the coefficient of static friction between the large block and smaller block:

A force of static friction can be defined as the frictional force that resists the relative motion of two (2) surfaces.

Hence, a force of static friction is a frictional force that keeps an object at rest or stationary rather than being in relative motion.

Mathematically, the force of static friction is given by the formula;

[tex]Fs = uFn[/tex]

Where;

For these block systems, the forces acting on them is given by:

[tex]uma - mg = 0\\\\uma = mg\\\\ua =g\\\\u=\frac{g}{a}[/tex]

Substituting the parameters into the formula, we have;

[tex]u=\frac{9.8}{13}[/tex]

u = 0.754

Read more: https://brainly.com/question/13754413

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:

[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.

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

Read more at; https://brainly.com/question/13799890

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