Octane (C8H18) is burned with dry air. The volumetric analysis of the products on a dry basis is as below. (Fig. 15–12) Determine (a) the air–fuel ratio, (b) the percentage of theoretical air used, and (c) the amount of H2O that condenses as the products are cooled to 25°C at 100 kPa

Answer:

a)radius of the arc of the teardrop at the top is 10.58m

b)T = mg

c) the centipetal acceleration at the top is 14.5m/s

Explanation:

Part A

Given that roller coaster is of tear drop shape

So the speed at the top is given as

v = 14.4 m/s

acceleration at the top is given as

a = 2 g

[tex]a = 2(9.8) = 19.6 m/s^2[/tex]

now we know the formula of centripetal acceleration as

[tex]a = \frac{v^2}{R}\\[/tex]

[tex]19.6 = \frac{14.4^2}{R}\\R = 10.58 m[/tex]

Part B

now

the total mass of the car and the ride is M

Let the force exerted by the track be n

By Newton law

[tex]n +Mg =\frac{Mv^2}{r} \\\\n=\frac{Mv^2}{r} -Mg\\\\=M(\frac{v^2}{r}-g )\\\\=M(2g-g)\\\\T=Mg[/tex]

Part C

If the radius of the loop is 21.4 m

speed is given by same v = 14.4 m/s

now the acceleration is given as

[tex]a = \frac{v^2}{R}[/tex]

[tex]a = \frac{14.4^2}{21.4} \\\\= 9.69 m/s^2[/tex]

Now for normal force at the top is given by force equation

[tex]F_n + mg = ma\\F_n = m(a-g)[/tex]

The force exerted by the rail is less than zero because acceleration is less than 9.69m/s²

So the normal force would have to point away from the centre, For safe ride this normal force must be positive i.e [tex]a \prec g[/tex]

[tex]\frac{v^2}{r} \prec \sqrt{g} \\\\v = \sqrt{rg} \\\\v = \sqrt{21.4 \times 9.8} \\\\v = 14.5m/s[/tex]

Answer:

a) If the currents are in the same direction, the magnetic field is zero at x = 0.298 m = 29.8 cm

That is, in between the wires, 29.8 cm from the 73.0 A wire and 10.2 cm from the 25.0 A wire.

b) If the currents are in opposite directions, the magnetic field is zero at x = 0.608 m = 60.8 cm

That is, along the positive x-axis, 60.8 cm from the 73.0 A wire and 20.8 cm from the 25.0 A wire.

Explanation:

The origin is at the 73.0 A wire and the 25.0 A wire is at x = 0.40 m

The magnetic field in a current carrying wire at a distance r from the wire is given by

B = (μ₀I/2πr)

μ₀ = magnetic constant = (4π × 10⁻⁷) H/m

a) If the currents are in the same direction, at what positions is the magnetic field equal to 0.

According to laws describing the direction.of magnetic fields, this position will be at some point between the two wires.

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire points into the plane of the book, moving in the negative x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (0.4 - x)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(0.4-x)]

(73/x) = [25/(0.4-x)]

73(0.4-x) = 25x

29.2 - 73x = 25x

73x + 25x = 29.2

98x = 29.2

x = (29.2/98) = 0.298 m

b) If the currents are in the opposite directions, at what positions is the magnetic field equal to 0?

According to laws describing the direction.of magnetic fields, this position will be at some point beyond the second wire (since we're initially concerned about the positive x-direction).

The magnetic field due to the 73.0 A wire points out of the book, at points along the positive x-axis while the magnetic field due to the 25.0 A wire (whose direction is now in the opposite direction to the current in the first wire) is also along the positive x-direction.

Hence,

For the 73.0 A wire, I₁ = 73.0 A, r₁ = x

For the 25.0 A wire, I₂ = 25.0 A, r₂ = (x - 0.4)

B = B₁ - B₂ = 0

(μ₀/2π) [(I₁/r₁) - (I₂/r₂)] = 0

(I₁/r₁) = (I₂/r₂)

(I₁/x) = [I₂/(x-0.4)]

(73/x) = [25/(x-0.4)]

73(x-0.4) = 25x

73x - 29.2 = 25x

73x - 25x = 29.2

48x = 29.2

x = (29.2/48) = 0.608 m

Hope this Helps!!!

Final answer:

The net magnetic field can be found zero at certain points between or outside two long parallel wires carrying currents, either in the same or opposite directions, by equating the magnetic fields produced by each and solving for the distance.

Explanation:

The problem involves finding the locations where the net magnetic field is zero due to currents carried by two long parallel wires. The Biot-Savart Law or Ampère's Law can be used for such problems; however, in high school problems, we typically use the formula for the magnetic field due to a long straight wire, B = (μ0 * I) / (2π * r), where B is the magnetic field, μ0 is the permeability of free space, I is the current, and r is the distance from the wire.

For wires with currents in the same direction, the magnetic fields due to each wire will point in opposite directions in the region between the wires and in the same direction outside the wires. To find locations where the net magnetic field is zero, we can equalize the magnetic fields coming from each wire and solve for the distance r from one of the wires where the fields cancel out.

For wires with currents in opposite directions, the magnetic fields due to the wires will be in the same direction in the region between the wires and in opposite directions outside of the wires. Again, we can find locations of zero net magnetic field by equalizing the magnitudes of the magnetic fields and solving for r.

Answer:

The tennis ball has more kinetic energy

Explanation:

Recall the formula for kinetic energy: [tex]K=\frac{1}{2} \,m\,v^2[/tex] , so we can estimate it for each case and compare the results:

For the baseball:

[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.14\,(41)^2\,\,J\\K=117.67\,\,J[/tex]

For the tennis ball:

[tex]K=\frac{1}{2} \,m\,v^2\\K=\frac{1}{2} \,0.058\,(67)^2\,\,J\\K=130.181\,\,J[/tex]

Therefore, the tennis ball has more kinetic energy

To solve this problem we will apply the concepts related to the double slit experiment. Here we test a relationship between the sine of the deviation angle and the distance between slit versus wavelength and the bright fringe order. Mathematically it can be described as,

[tex]dsin\theta = m\lambda[/tex]

Here,

d = Distance between slits

m = Any integer which represent the order number or the number of repetition of the spectrum

[tex]\lambda[/tex] = Wavelength

[tex]\theta[/tex] = Angular deviation

Replacing with our values we have,

[tex](6.93*10^{-6}) sin\theta = (3)(491*10^{-9})[/tex]

[tex]\theta = sin^{-1} (\frac{(3)(491*10^{-9}}{6.93*10^{-6}) })[/tex]

Part A)

[tex]\theta = 0.2141rad[/tex]

PART B)

[tex]\theta = 0.2141rad(\frac{360\°}{2\pi rad})[/tex]

[tex]\theta = 12.27\°[/tex]

Answer

Given,

Revolution of wheel = 23 rev.

                                  = 23 x 2π = 46 π

Time = 3 s

final angular velocity = 16 rad/s

angular acceleration of wheel = ?

Now, Calculating the initial angular speed of the wheel

Angular displacement = [tex]\dfrac{1}{2}[/tex](initial velocity + final velocity) x time.

[tex]46 π = \dfrac{\omega_o+16}{2}\times 3[/tex]

[tex]\omega_0 = 80.29\ rad/s[/tex]

now, angular acceleration

[tex]\alpha = \dfrac{\omega-\omega_0}{t}[/tex]

[tex]\alpha = \dfrac{16-80.29}{3}[/tex]

[tex]\alpha = -21.43\ rad/s^2[/tex]

Hence, the angular acceleration of wheel is negative means wheel is decelerating.

The angular acceleration of the wheel is negative (-).

The angular acceleration would be the temporal ratio during which the angular speed changes and therefore is commonly denoted by alpha (α) as well as written throughout radians/sec.

According to the question,

Revolutions, 23 rev or,

                      23 × 2π = 46π

Time, 3 seconds

Final angular velocity, 16 rad/s

We know the formula,

→ Angular displacement = [tex]\frac{1}{2}[/tex] (Initial velocity + Final velocity) × Time

By substituting the values,

                                    46 = [tex]\frac{\omega_o +16}{2}[/tex] × 3

                                     [tex]\omega_o[/tex] = 80.29 rad/s

hence,

The angular acceleration will be:

→ α = [tex]\frac{\omega - \omega_o}{T}[/tex]

      = [tex]\frac{16-80.29}{3}[/tex]

      = -21.43 rad/s²

Thus the above response is correct.

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Answer:

Current

Explanation:

A wire loop with a current flowing through it is also within a uniform magnetic field. While the loop may have a net force equal to zero, in most cases, it will have an induced current on it.

Answer:

F'=(8/3)F

Explanation:

to find the change in the force you take into account that the electric force is given by:

[tex]F=k\frac{q_1q_2}{(18.0u)^2}=k\frac{q_1q_2}{324.0}u^2[/tex]

However, if q1'=1/3*q, q2'=2*q2 and the distance is halved, that is 18/=9.o unit:

[tex]F'=k\frac{q_1'q_2'}{(9u)^2}=k\frac{(1/3)q_1(2)q_2}{81.0u^2}=\frac{2}{3}k\frac{q_1q_2}{81.0u^2}[/tex]

if you multiply this result by 4 and divide by 4 you get:

[tex]F'=\frac{8}{3}k\frac{q_1q_2}{324.0u^2}=\frac{8}{3}F[/tex]

hence, the new force is 8/3 of the previous force F.

Note: The values of the current and the radii are not given, substitute whatever the value of the current and radius are to the given solution to obtain the magnitude.

Answer:

magnitude of the B-field at the smaller coil due to the larger coil is [tex]B = \frac{7200\mu_{0}I }{2a}[/tex]

Explanation:

N₁ = 7200 turns

N₂ = 920 turns

a = 75.50 cm = 0.755 m

b = 1.00 cm = 0.01 m

From the given data, b<<a, and it is at the center of the larger coil,

so we are safe to assume that the magnetic field at the smaller coil is constant.

The formula for magnetic field due to circular loop at the center of a coil is given by:

[tex]B = \frac{N\mu_{0}I }{2R}[/tex]

Therefore, magnetic field in the smaller coil  due to the larger coil of radius a will be given by:

[tex]B = \frac{7200\mu_{0}I }{2a}[/tex]

Where [tex]\mu_{0} = 4\pi * 10^{-7} Wb/A-m[/tex]

I = current in the larger coil

The magnitude of the B-field at the smaller coil due to the larger coil will be "4π × 10⁻⁷ Wb/A-m".

According to the question,

Number of turns, N₁ = 7200 turns

                             N₂ = 920 turns

Radius, a = 75.50 cm or,

                = 0.755 m

             b = 1.00 cm or,

                = 0.01 m

We know the formula,

Magnetic field, B = [tex]\frac{N \mu_0 I}{2R}[/tex]

here, μ₀ = 4π × 10⁻⁷ Wb/A-m

By substituting the values,

                             = [tex]\frac{7200 \mu_0 I}{2a}[/tex]

Thus the above answer is correct.

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Answer:

[tex]18.8\times 10^{-3} T[/tex]

Explanation:

We are given that

Number of turns,N=500

Radius,r=0.04 m

Length of solenoid,L=40 cm=[tex]\frac{40}{100}=0.4 m[/tex]

1 m=100 cm

Current,I=12 A

We have to find the magnitude of magnetic field near the center of the solenoid.

Number of turns per unit length,n=[tex]\frac{500}{0.4}=1250[/tex]

Magnetic field near the center of the solenoid,B=[tex]\mu_0 nI[/tex]

Where [tex]\mu_=0=4\pi\times 10^{-7}Tm/A[/tex]

[tex]B=4\pi\times 10^{-7}\times 1250\times 12=18.8\times 10^{-3} T[/tex]

[tex]B=18.8\times 10^{-3} T[/tex]

Answer:

Chloroplasts convert light energy into sugars that can be used by the cells. This conversion creates 'food' for the plant.

Explanation:

Answer:

The energy of the photon is  [tex]x = 2.86 eV[/tex]

Explanation:

From the question we are told that  

      The first orbit is [tex]n_1 = 5[/tex]

       The second orbit is [tex]n_2 = 2[/tex]

According to  Bohr model

     The energy of difference of the electron as it moves from on orbital to another is mathematically represented as

              [tex]\Delta E = k [\frac{1}{n^2 _1} + \frac{1}{n^2 _2} ][/tex]

  Where k is a constant which has a value of [tex]k = -2.179 *10^{-18} J[/tex]

       So

              [tex]\Delta E = - 2.179 * 10^{-18} [\frac{1}{5^2 _1} + \frac{1}{2^2 _2} ][/tex]

                     [tex]= 4.576 *10^{-19}J[/tex]

Now we are told from the question that

         [tex]1 eV = 1.602 * 10^{-19} J[/tex]

so      x eV  =  [tex]= 4.576 *10^{-19}J[/tex]

  Therefore

                [tex]x = \frac{4.576*10^{-19}}{1.602 *10^{-19}}[/tex]

                   [tex]x = 2.86 eV[/tex]

The energy of the photon produced by an electron in a Hydrogen atom moving from the 5th orbit to the 2nd orbit is 2.85 electron Volts as calculated via the Rydberg formula.

The energy of the photon produced by the transition of an electron in a hydrogen atom from the 5th orbit to the 2nd orbit can be calculated using Rydberg's formula.

Rydberg's formula for energy is given as E = 13.6 * (1/n1^2 - 1/n2^2), where n1 and n2 are the initial and final energy levels respectively, and E is the energy difference in eV (electron volts). Here n1 = 2 and n2 = 5.

So, substituting these values in the formula E = 13.6 * (1/2^2 - 1/5^2) = -13.6 * (-0.21) = 2.85 eV. Notice that the energy is negative which signifies a transition down to a lower energy level (which is exothermic), however, we are interested in the magnitude of the energy which is 2.85 eV.

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Answer:

v = 1176.23 m/s

y = 741192.997 m = 741.19 km

Explanation:

Given

M₀ = 9 Kg  (Initial mass)

me = 0.225 Kg/s   (Rate of fuel consumption)

ve = 1980 m/s    (Exhaust velocity relative to rocket, leaving at atmospheric pressure)

v = ? if t = 20 s

y = ?

We use the equation

v = ∫((ve*me)/(M₀ - me*t)) dt - ∫g dt     where t ∈ (0, t)

⇒   v = - ve*Ln ((M₀ - me*t)/M₀) - g*t

then we have

v = - 1980 m/s*Ln ((9 Kg - 0.225 Kg/s*20 s)/(9 Kg)) - (9.81 m/s²)(20 s)

v = 1176.23 m/s

then we apply the formula

y = ∫v dt = ∫(- ve*Ln ((M₀ - me*t)/M₀) - g*t) dt

⇒   y = - ve* ∫ Ln ((M₀ - me*t)/M₀) dt - g*∫t dt

⇒   y = - ve*(Ln((M₀ - me*t)/M₀)*t + (M₀/me)*(M₀  - me*t - M₀*Ln(M₀ - me*t))) - (g*t²/2)

For t = 20 s   we have

y = Ln((9 Kg - 0.225 Kg/s*20 s)/9 Kg)*(20 s) + (9 Kg/0.225 Kg/s)*(9 Kg  - 0.225 Kg/s*20 s - 9 Kg*Ln(9 Kg - 0.225 Kg/s*20 s)) - (9.81 m/s²*(20 s)²/2)

⇒   y = 741192.997 m = 741.19 km

The graphs are shown in the pics.

Following are the solution to the given points:

Given:

[tex]\to v_o =1980\ \frac{m}{s}\\\\[/tex]

[tex]\to a=g= -9.8 \frac{m}{s} \ \text{(Diection downwords )}[/tex]

Solution:

Using formula:

[tex]\to V= V_o +at\\\\[/tex]

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

For point a:

[tex]\to V= V_o +at\\\\[/tex]

        [tex]=1980+ (-9.8) (20) \\\\[/tex]

        [tex]= 1980-196 \\\\ =1784\ \frac{m}{s}\\\\[/tex]

The velocity at the conclusion of a [tex]20[/tex] second is [tex]\bold{1784\ \frac{m}{s}}\\\\[/tex].

For point b:

Using formula:

[tex]\to v^2-v^2_0= 2ad \\\\[/tex]

[tex]\to d=\frac{v^2_0 -V^2}{2a}\\\\[/tex]

       [tex]= \frac{(1980)^2 - (1784)^2}{2\times 9.8} \\\\ =\frac{3920400-3182656}{19.6}\\\\= \frac{737744}{17.6}\\\\ =37640\ m \\\\ = 37.64\ km\\\\[/tex]

In [tex]20[/tex] seconds, the total distance traveled is [tex]401737 \ \ m (or \ 401.737\ \ km)[/tex].  

For point c:

Please find the attached file.

Answer:

Color

Explanation:

The most obvious property of a mineral, its color, and is unfortunately also the least diagnostic.

Answer:

color

Explanation:

Answer: (a) Smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].

(b) Smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].

Explanation:

(a) When both the spheres have same charge then force is repulsive in nature as like charges tend to repel each other.

Therefore, total charge on the two non-conducting spheres will be calculated as follows.

        [tex]Q_{1} + Q_{2} = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]

                      = [tex]9 \times 10^{-5} C[/tex]

Therefore, force between the two spheres will be calculated as follows.

        F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]

       12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]

       [tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]

or,     [tex]Q_{1}(9 \times 10^{-5} - Q_{1}) = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]9 \times 10^{-5}Q_{1} - Q^{2}_{1} = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} + 0.104 \times 10^{-9} = 0[/tex]

        [tex]Q_{1} = 11.7 \times 10^{-5} C, 2.7 \times 10^{-5} C[/tex]

This means that smaller charge is [tex]2.7 \times 10^{-5} C[/tex] and larger charge is [tex]11.7 \times 10^{-5} C[/tex].

(b)  When force is attractive in nature then it means both the charges are of opposite sign.

Hence, total charge on the non-conducting sphere is as follows.

      [tex]Q_{1} + (-Q_{2}) = 90 \mu \times \frac{10^{-6}C}{1 \muC}[/tex]

      [tex]Q_{1} - Q_{2} = 9 \times 10^{-5} C[/tex]

Now, force between the two spheres is calculated as follows.

    F = [tex]k\frac{Q_{1}Q_{2}}{r^{2}}[/tex]

    12 N = [tex]\frac{(9 \times 10^{9} Nm^{2}/C^{2})Q_{1}Q_{2}}{(0.28 m^{2})}[/tex]

   [tex]Q_{1}Q_{2} = 0.104 \times 10^{-9} C^{2}[/tex]

   [tex]Q_{1}(Q_{1} - 9 \times 10^{-5}) = 0.104 \times 10^{-9} C^{2}[/tex]

    [tex]Q^{2}_{1} - 9 \times 10^{-5}Q_{1} = 0.104 \times 10^{-9} C^{2}[/tex]

        [tex]Q_{1} = -11.4 \times 10^{-5}, 9.1 \times 10^{-5}[/tex]

Hence, smaller charge is [tex]-11.4 \times 10^{-5}[/tex] and larger charge is [tex]9.1 \times 10^{-5}[/tex].

(a). The charges on one sphere is [tex]0.89\mu C[/tex]  and [tex]1.18 \mu C[/tex] on other sphere.

(b). If the force were attractive then charge on both sphere will be in opposite sign.

The electrostatic force is given as,

                 [tex]F=k\frac{Q_{1}Q_{2}}{r^{2} }[/tex]

Where ,

Given that, [tex]F=12N,r=28cm=0.28m,Q_{1}+Q_{2}=90[/tex]

substitute values in above relation.

             [tex]12=\frac{9*10^{9} *Q_{1}(90-Q_{1})}{(0.28)^{2} } \\\\Q_{1}^{2}-(90*10^{-6})Q_{1} +1.05*10^{-10}=0\\ \\Q_{1}=8.88*10^{-5}=0.89 \mu C\\ \\Q_{2}=1.18*10^{-6} =1.18\mu C[/tex]

If the force were attractive then charging on both sphere will be in opposite sign.

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Answer:

See explaination and attachment please.

Explanation:

Potential energy is defined as the energy possessed by a body by virtue of its position relative to others, stresses within itself, electric charge, and other factors.

Kinetic energy on the other hand is defined as the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity.

These understand will foster our knowledge to the best way of solving the question.

Please kindly check attachment for a step by step approach on the graph.

Final answer:

In a bungee jumping scenario, the stretching of the cord can be expressed in terms of length and height. Gravitational potential energy dominates early in the fall, while elastic potential energy is more significant later. Graphs can visually represent how potential energy changes with height during the jump.

Explanation:

a) Stretching of the cord: The stretching ?L of the cord can be expressed as ?L = d - z. The total potential energy U can be represented as U = UG + US, where UG is the gravitational potential energy and US is the elastic potential energy.

b) Largest potential energy: For large z (early in the fall), UG is largest. For small z (late in the fall), US is largest.

c) Graphs: Sketch a graph of UG(z) on the left plot, US(z) on the center plot, and U(z) = UG(z) + US(z) on the rightmost plot.

Answer:

What is the average velocity of the car over the first 0.25m?  

⇒ 0.11 m/s

What is the average velocity of the car over the second 0.25m?  

⇒ 0.28 m/s

Explanation:

Just did this problem! :)

Answer:

⇒ 2.23

⇒ 0.90

⇒ 0.28

Explanation:

Just did the problem (⌐■_■)

Answer:

Explanation:

a )  In this case A rests on B and both rest on horizontal surface

Both moves together .

Total reaction due to weight of both of them

R = 1.2 + 3.6 = 4.8 N

friction force by horizontal surface = μ R , μ is coefficient of friction .

= .3 x 4.8 = 1.44 N

Force equal to frictional force will be required to put both of them in uniform motion .

b ) If A is held stationary , friction force will arise at both , the upper and lower surface of B .

At upper surface friction force =  μ x weight of A

= .3 x 1.2 = .36 N

At lower surface friction force = μ x weight of A +B

= .3 x 4.8

= 1.44

Total frictional force on B

= 1.8 N  N .

So 1.8 N force will be required to put B in uniform keeping A stationary.

To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest. In both scenarios, the friction force between block B and the surface is given by the equation F_friction = μ * N, where μ is the coefficient of friction and N is the normal force. The normal force is the sum of the weights of the blocks in scenario a) and the weight of block B in scenario b).

To find the magnitude of the horizontal force F necessary to drag block B to the left at constant speed, we need to consider two scenarios: a) when block A rests on block B and moves with it, and b) when block A is held at rest.

a) When block A rests on block B and moves with it, the friction force between the blocks and the surface is the force needed to overcome the friction. The friction force is given by the equation:
F_friction = μ * N

Here, μ is the coefficient of kinetic friction and N is the normal force between the blocks and the surface. In this case, the normal force N is the sum of the weights of both blocks, so N = m_A * g + m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_A * g + m_B * g)

b) When block A is held at rest, the friction force between block B and the surface is the force needed to overcome the friction. The friction force is the product of the coefficient of static friction and the normal force:
F_friction = μ * N

Here, μ is the coefficient of static friction and N is the normal force between block B and the surface. The normal force N is equal to the weight of block B, so N = m_B * g. Substituting this value into the equation, we get:
F_friction = μ * (m_B * g)

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Answer:

[tex]-16.6 rad/s^2[/tex]

Explanation:

The torque exerted on a rigid body is related to the angular acceleration by the equation

[tex]\tau = I \alpha[/tex] (1)

where

[tex]\tau[/tex] is the torque

I is the moment  of inertia of the body

[tex]\alpha[/tex] is the angular acceleration

Here we have a solid sphere: the moment of inertia of a sphere rotating about is centre is

[tex]I=\frac{2}{5}MR^2[/tex]

where

M = 240 g = 0.240 kg is the mass of the sphere

[tex]R=\frac{2.50}{2}=1.25 cm = 0.0125 m[/tex] is the radius of the sphere

Substituting,

[tex]I=\frac{2}{5}(0.240)(0.0125)^2=1.5\cdot 10^{-5} kg m^2[/tex]

The torque exerted on the sphere is

[tex]\tau = Fr[/tex]

where

F = -0.0200 N is the force of friction

r = 0.0125 m is the radius of the sphere

So

[tex]\tau=(-0.0200)(0.0125)=-2.5\cdot 10^{-4} Nm[/tex]

Substituting into (1), we find the angular acceleration:

[tex]\alpha = \frac{\tau}{I}=\frac{-2.5\cdot 10^{-4}}{1.5\cdot 10^{-5}}=-16.6 rad/s^2[/tex]

Answer:

See the explaination for the details.

Explanation:

Gauss Law states that the total electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity. The electric flux in an area is defined as the electric field multiplied by the area of the surface projected in a plane and perpendicular to the field.

According to the Gauss law, the total flux linked with a closed surface is 1/ε0 times the charge enclosed by the closed surface.

Please kindly check attachment for the step by step explaination of the answer.

In the building industry, linear measurements are expressed in units like miles, feet, and inches, or in the metric system, meters and millimeters. The choice of unit depends on the precision needed for the task, with smaller units used for finer details and larger ones for general dimensions.

Linear Measurements in the Buildings Industry

In the building industry, linear measurements are typically expressed in units that provide the most practical and precise information for the specific dimension being measured. Linear dimensions refer to measurements that can be expressed using linear units such as miles, feet, or inches; and in the metric system, meters or millimeters (mm). These measurements are essential for architects and builders to communicate the sizes of different components of a structure, like the length of a wall or the size of a window. Typically, for smaller measurements such as the dimensions of timber or the size of rooms, feet or meters are used, while millimeters might be preferred for finer details.

For example, when measuring a specific length like the width of a staircase, the builders would commonly use millimeters in the metric system to ensure precision. Similarly, to describe to a European the dimensions of "two-by-four" lumber used in the US, conversions from inches to centimeters and from feet to meters are required. The key is to use the most appropriate unit of measurement that ensures clarity and precision for constructing a building accurately.

When dealing with linear measurements, especially in scientific contexts or in detailed building plans, precise units like millimeters, centimeters, and meters are often used. The context will dictate whether a larger unit such as kilometers or smaller units like micrometers are more appropriate.

Answer: The riders are subjected to 11.5 revolutions per minute

Explanation: Please see the attachments below

Answer:

Explanation:

find the solution below

Answer:

[tex]\theta = 0.195^0[/tex]

Explanation:

wavelength [tex]\lambda = 648 nm \ = 648*10^{-9}m[/tex]

d = 0.190 mm = 0.190 × 10⁻³ m

D = 1.91 m

By using the formula:

[tex]dsin \theta = n \lambda\\\\\theta = sin^{(-1)}(\frac{n \lambda}{d})\\\\\\\theta = sin^{(-1)}(\frac{1*648*10^{-9}}{0.190*10^{-3}})[/tex]

[tex]\theta = 0.195^0[/tex]

The first maximum will appear at an angle [tex]\theta = 0.195^0[/tex] from the beam axis

Complete Question:

In the same configuration of the previous problem 3, four long straight wires are perpendicular to the page, and their cross sections form a square of edge length a = 13.5 cm. Each wire carries 7.50 A, and the currents are out of the page in wires 1 and 4 and into the page in wires 2 and 3.

a) Draw a diagram in a (x,y) plane of the four wires with wire 4 perpendicular to the origin. Indicate the current's directions.

b) Draw a diagram of all magnetic fields produced at the position of wire 3 by the other three currents.

c) Draw a diagram of all magnetic forces produced at the position of wire 3 by the other three currents.

d) What are magnitude and direction of the net magnetic force per meter of wire length on wire 3?

Answer:

force, 1.318 ₓ 10⁻⁴

direction, 18.435°

Explanation:

The attached file gives a breakdown step by step solution to the questions

Answer:

The  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

Explanation:

Given;

velocity of wave on the string with lower tension, v₁ = 35.2 m/s

the fundamental frequency of the string, F₁ = 258 Hz

velocity of wave on the string with greater tension;

[tex]v_1 = \sqrt{\frac{T_1}{\mu }[/tex]

where;

v₁ is the velocity of wave on the string with lower tension

T₁ is tension on the string

μ is mass per unit length

[tex]v_1 = \sqrt{\frac{T_1}{\mu} } \\\\v_1^2 = \frac{T_1}{\mu} \\\\\mu = \frac{T_1}{v_1^2} \\\\ \frac{T_1}{v_1^2} = \frac{T_2}{v_2^2}\\\\v_2^2 = \frac{T_2v_1^2}{T_1}[/tex]

Where;

T₁ lower tension

T₂ greater tension

v₁ velocity of wave in string with lower tension

v₂ velocity of wave in string with greater tension

From the given question;

T₂ = 1.1 T₁

[tex]v_2^2 = \frac{T_2v_1^2}{T_1} \\\\v_2 = \sqrt{\frac{T_2v_1^2}{T_1}} \\\\v_2 = \sqrt{\frac{1.1T_1*(35.2)^2}{T_1}}\\\\v_2 = \sqrt{1.1(35.2)^2} = 36.92 \ m/s[/tex]

Fundamental frequency of wave on the string with greater tension;

[tex]f = \frac{v}{2l} \\\\2l = \frac{v}{f} \\\\thus, \frac{v_1}{f_1} =\frac{v_2}{f_2} \\\\f_2 = \frac{f_1v_2}{v_1} \\\\f_2 =\frac{258*36.92}{35.2} \\\\f_2 = 270.6 \ Hz[/tex]

Beat frequency = F₂ - F₁

                          = 270.6 - 258

                          = 12.6 Hz

Therefore, the  beat frequency when each string is vibrating at its fundamental frequency is 12.6 Hz

The fundamental frequency for the second string is found to be 272 Hz, resulting in a beat frequency of 14 Hz.

To find the beat frequency when two strings are vibrating at their fundamental frequencies, follow these steps:

Therefore, the beat frequency when both strings vibrate at their fundamental frequencies is 14 Hz.

Answer:

When the midpoint of the magnet is in the plane of the loop, the induced current in the loop viewed from above is essentially zero.

Explanation:

At some point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop. As the magnet descends and gets close to the entrance of the coil, some of the magnetic field from the magnet threads the top few imaginary disks of the coil. The time changing flux induces some EMF and subsequently some induced current.

As the magnet continues to fall, and enters the coil, more of its magnetic field is threading imaginary disks in the coil, so as it moves, the time rate of change of total flux increases, so the EMF goes up. Note that the field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.

As the bar moves through the plane of the coil, the North end first, flux is added by the motion of the magnet and flux is removed by the motion of South end.

At some point, the bar reaches the middle of the coil. At this point, the amount of flux added to the top half of the coil by a small motion of the magnet is equal to the amount of flux removed from the bottom half. Therefore, at this point the EMF is zero. And hence, there is no induced current observed in the wire at this point.

Hope this Helps!!!

When viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.

The given problem is based on the concept of electric flux and induced emf. The electric flux is dependent on induced emf, which is produced to current. Since,  point far away from the plane of the loop of the wire, there's no flux and hence, no induced emf, and no current in the loop.

Note: - The field lines above and below the magnet's midpoint point in the same direction from the the North pole to the south.

Thus, we conclude that when viewed from the above, there is no induced current observed in the wire at this point due to zero EMF.

Learn more about the electromagnetic induction here:

https://brainly.com/question/12536260

Answer:

The total mechanical energy of the system would stay the same

Explanation:

The law of conservation energy states:

In a closed system, (a system that isolated from its surroundings) the total energy of the system is conserved.

For instance, the air resistance is negligible. So non conservative force is acting on the system and so the energy is conserved for the system.

The total initial energy must be equal to the total final energy , if the energy is conserved.

Therefore, when the pebbles was launched at 45 degrees to the horizontal the total mechanical energy stay the same.

Answer:

F'=1/9*F0

Explanation:

F0 is the gravitational force between the particles. When the distance is triplicated we have that

[tex]F'=G\frac{m_1m_2}{(3r)^{2}}[/tex]

where r is the distance before the particles are separated, m1 and m2 are the masses their masses and G is the Canvendish's constant.

By some algebra we have

[tex]F'=\frac{1}{9}G\frac{m_1m_2}{r^2}=\frac{1}{9}F_0[/tex]

hope this helps!!

Answer:

F₁ = [tex]\frac{1}{9}[/tex]F₀

Explanation:

Newton's law of universal gravitation states that the force of attraction or repulsion, F, between two particles of masses M₁ and M₂ is directly proportional to the product of these particles and inversely proportional to the square of the distance, r, between the two particles. i.e

F ∝ M₁M₂ / r²

F = GM₁M₂ / r²            --------------------(i)

Where;

G is the constant of proportionality.

From equation (i), since the force is inversely proportional to the square of the distance, holding other variables constant, the equation can be reduced to;

F = k / r²

This implies that;

Fr² = k          -------------------(ii)

Now, according to the question;

F = F₀

Substitute this into equation (ii) as follows;

F₀ r² = k       ----------------(iii)

Also, when the distance of separation, r, is trippled i.e r becomes 3r;

F = F₁

Substitute these values into equation (ii) as follows;

F₁(3r)² = k

9F₁r² = k                ---------------(iv)

Substitute the value of k in equation (iii) into equation (iv) as follows;

9F₁r² = F₀ r²             --------------(v)

Cancel r² on both sides of equation (v)

9F₁ = F₀

Now make F₁ subject of the formula

F₁ = [tex]\frac{1}{9}[/tex]F₀

Therefore, the new force F₁ = [tex]\frac{1}{9}[/tex]F₀

Answer:

Resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]

Explanation:

We have given a pupil of cat  width of the slit d = 0.540 mm = [tex]0.540\times 10^{-3}m[/tex]

Average wavelength of the light [tex]\lambda =471nm=471\times 10^{-9}m[/tex]

We have to find the angular resolution for horizontally separated mice  

Angular resolution is given by [tex]\Theta =\frac{\lambda }{d}=\frac{471\times 10^{-9}}{0.540\times 10^{-3}}=872.22\times 10^{-6}radian[/tex]

So resolution of the light will be equal to [tex]872.22\times 10^{-6}radian[/tex]

Answer:

The value of the constant force is [tex]\bf{296.88~N}[/tex].

Explanation:

Given:

Mass of the merry-go-round, [tex]m = 210~Kg[/tex]

Radius of the horizontal disk, [tex]r = 1.5~m[/tex]

Time required, [tex]t = 2.00~s[/tex]

Angular speed, [tex]\omega = 0.600~rev/s[/tex]

Torque on an object is given by

[tex]\tau = F.r = I.\alpha~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~(1)[/tex]

where [tex]I[/tex] is the moment of inertia of the object, [tex]\alpha[/tex] is the  angular acceleration and [tex]F[/tex] is the force on the disk.

The moment of inertia of the horizontal disk is given by

[tex]I = \dfrac{1}{2}mr^{2}[/tex]

and the angular acceleration is given by

[tex]\alpha = \dfrac{2\pi \omega}{t}[/tex]

Substituting all these values in equation (1), we have

[tex]F &=& \dfrac{I\alpha}{r}\\&=& \dfrac{\pi m r \omega}{t}\\&=& \dfarc{\pi(210~Kg)(1.5~m)(0.600~rev/s)}{2.00~s}\\&=& 296.88~N[/tex]