A truck runs into a pile of sand, moving 0.80 m as it slows to a stop. The magnitude of the work that the sand does on the truck is 5.5×105J. Part A Determine the magnitude of the average force that the sand exerts on the truck

Answer:

Because it's used to find definite solutions to More complex quantum mechanical system. Harmonic oscillator helps in determining motion of small mass if strings.

Explanation:

The particle box is very precise when using it to calculate or find definite solutions in complex quantum mechanics in which particles might be trapped in a regions of electric potentials. Since we're dealing with electric potentials here, the harmonic oscillator is needed because it has to do squarely wit potential energy which is given as

V(x) = 0.5kx²

Where k is force constant and x is distance.

Final answer:

The particle in a box and the harmonic oscillator models are useful for understanding quantum mechanical systems. They can be used to represent various chemically significant systems, including blackbody radiators, atomic and molecular spectra, and optoelectronic devices.

Explanation:

The particle in a box and the harmonic oscillator are useful models for quantum mechanical systems because they provide insights into the behavior of particles in confined spaces and under the influence of potential energy. The particle in a box model describes a particle free to move in a small space surrounded by impenetrable barriers, and can be used to represent systems such as blackbody radiators, atomic and molecular spectra, and optoelectronic devices. The harmonic oscillator model, on the other hand, represents systems with periodic motion, such as molecular vibrations and wave packets in quantum optics.

Answer:

465.6 N/m

Explanation:

We are given that

F=12 N

[tex]y(t)=4.50cos\left \{(19.5s^{-1}t-\frac{\pi}{8}\right \}[/tex]

We have to find the spring constant of the spring.

[tex]F=mg[/tex]

Where [tex]g=9.8 m/s^2[/tex]

Using the formula

[tex]12=m\times 9.8[/tex]

[tex]m=\frac{12}{9.8}[/tex]kg

Compare the given equation with

[tex]y(t)=Acos(\omega t-\phi)[/tex]

We get [tex]\omega=19.5[/tex]

[tex]k=m\omega^2[/tex]

Using the formula

Spring constant,[tex]k=\frac{12}{9.8}\times (19.5)^2=465.6 N/m[/tex]

The spring constant (k) for the object executing simple harmonic motion is approximately 463.29 N/m, calculated using the provided angular frequency and the mass of the object.

To find the spring constant (k) of the spring for an object executing simple harmonic motion (SHM), we use the equation of motion provided:

  y(t) = 4.50 cm cos[(19.5 s-1)t - π/8]

From the equation of motion, we can see that the angular frequency (ω) is 19.5 s-1. In SHM, the angular frequency ω is related to the spring constant k and the mass m by the following equation:

  ω = sqrt(k/m)

To calculate k, we rearrange the equation to solve for k:

  k = mω2

First, we find the mass (m) by using the weight of the object (12.0 N):

  m = weight / g = 12.0 N / 9.81 m/s2 ≈ 1.22 kg

Then we calculate the spring constant (k):

  k = (1.22 kg)(19.5 s-1)2 ≈ 463.29 N/m

Therefore, the spring constant of the spring is approximately 463.29 N/m.

Answer:

(a) E = 1.25×10⁶ V/m

(b) v = 83.8×10⁶ m/s

Explanation:

Please see attachment below.

qV = electric potential energy of the electron. This is the energy converted into the kinetic energy of the motion of the electron through the plates. Kinetic energy is 1/2×mv². Where m = mass of electron = 9.11×10-³¹ kg.

q = 1.6×10-¹⁹ C.

So qV = 1/2mv² so rearranging and substituting the given quantities give the speed of the electrons.

Final answer:

The electric field strength between the plates is 1,250,000 N/C. Calculating the exit speed of an electron involves using the charge of the electron and the potential difference, but the accurate computation requires relativity due to the high velocity.

Explanation:

The student has asked two questions regarding the operation of an electron gun in an old TV picture tube: calculating the electric field strength between the plates and determining the exit speed of an electron.

Electric Field Strength Between the Plates

The electric field strength (E) between two parallel plates can be calculated using the formula E = V/d, where V is the potential difference between the plates, and d is the distance between them. With V = 20,000 volts (20 kV) and d = 0.016 meters (1.6 cm), E = 20,000 / 0.016 = 1,250,000 N/C.

Exit Speed of an Electron

To find the exit speed of an electron, we use the formula KE = eV, where KE is the kinetic energy of the electron, e is the charge of the electron (1.6 x 10⁻¹⁹ C), and V is the potential difference. The exit speed can then be derived from kinetic energy using the formula KE = 1/2 mv² leading to a theoretical calculation. However, the exit speed is significantly high, and thus, actually calculating it accurately involves considerations of relativity.

Answer:

(a) [tex]F_g=1.62*10^{-48}N[/tex]

(b) [tex]F_e=3.68*10^{-9}N[/tex]

Explanation:

(a) We use Newton's law of universal gravitation, in order to calculate the gravitational force between electron and proton:

[tex]F_g=-G\frac{m_1m_2}{r^2}[/tex]

Where G is the Cavendish gravitational constant, [tex]m_1[/tex] and [tex]m_2[/tex] are the masses of the electron and the proton respectively and r is the distance between them:

[tex]F_g=-6.67*10^{-11}\frac{N\cdot m^2}{kg^2}\frac{(9.11*10^{-31}kg)(1.67*10^{-27}kg)}{(2.5*10^{-10}m)^2}\\F_g=-1.62*10^{-48}N[/tex]

The minus sing indicates that the force is repulsive. Thus, its magnitude is:

[tex]F_g=1.62*10^{-48}N[/tex]

(b) We use Coulomb's law, in order to calculate the electric force between electron and proton, here k is the Coulomb constant and e is the elementary charge:

[tex]F_e=-k\frac{e^2}{r^2}\\F_e=-8.99*10^{9}\frac{N\cdot m^2}{C^2}\frac{(1.6*10^{-19}C)^2}{(2.5*10^{-10}m)^2}\\F_e=-3.68*10^{-9}N[/tex]

Its magnitude is:

[tex]F_e=3.68*10^{-9}N[/tex]

Answer:

The time of flight of the ball is 1.06 seconds.

Explanation:

Given [tex]\Delta x=7\ m[/tex]

[tex]\theta=45 \°[/tex]

Also, [tex]\Delta y=(3.5-2)=1.5\ m[/tex]

[tex]a_x=0\ and\ a_y=-9.81\ m/s^2[/tex]

Let us say the velocity in the x-direction is [tex]v_x[/tex] and in the y-direction is [tex]v_y[/tex]. And acceleration in the x-direction is [tex]a_x[/tex] and in the y-direction is [tex]a_y[/tex].

Also, [tex]\Delta x\ and\ \Delta y[/tex] is distance covered in x and y direction respectively. And [tex]t[/tex] is the time taken by the ball to hit the backboard.

We can write [tex]v_x=v_0cos(45)\ and\ v_y=v_0sin(45)[/tex]. Where [tex]v_0[/tex] is velocity of ball.

Now,

[tex]\Delta x=v_x\times t+\frac{1}{2}\times a_x\times t^2\\ \Delta x=v_x\times t+\frac{1}{2}\times 0\times t^2\\\Delta x=v_xt[/tex]

[tex]\Delta x=v_0cos(45)\times t\\7=v_0cos(45)\times t\\\\t=\frac{7}{v_0cos(45)}[/tex]

Also,

[tex]\Delta y=v_y\times t+\frac{1}{2}\times a_y\times t^2\\ 1.5=v_0sin(45)\times \frac{7}{v_0cos(45)}+\frac{1}{2}\times (-9.81)\times(\frac{7}{v_0cos(45)} )^2\\\\1.5=7-\frac{481}{(v_0)^2}\\ \\\frac{481}{(v_0)^2}=5.5\\\\(v_0)^2=\frac{481}{5.5}\\ \\(v_0)^2=87.45\\\\v_0=\sqrt{87.45}=9.35\ m/s[/tex].

Plugging this value in

[tex]t=\frac{7}{v_0cos(45)}\\ \\t=\frac{7}{9.35\times 0.707}\\ \\t=\frac{7}{6.611}[/tex]

[tex]t=1.06\ seconds[/tex]

So, the time of flight of the ball is 1.06 seconds.

The electric potential at 5 meters on the x-axis due to a uniformly distributed charge of 8.473 nC along the x-axis from -1m to 1m is approximately 51.84 volts.

To determine the electric potential at a point on the x-axis created by a uniformly distributed charge, we can use the formula for electric potential due to continuous charge distribution. For a continuous charge distribution along the x-axis, the electric potential (V) at a point (P) is given by the integral of [tex]\(k \cdot \frac{dq}{r}\), where \(k\)[/tex] is Coulomb's constant, (dq) is the small charge element, and \(r) is the distance between the charge element and the point (P).

In this scenario, we integrate the expression across the entire distribution of charge from -1m to 1m along the x-axis to find the total potential at point 5m on the x-axis. Since the charge is uniformly distributed, we can represent (dq) as [tex]\(\lambda dx\),[/tex]where [tex]\(\lambda\)[/tex] is the charge per unit length and \(dx\) is the small element of length along the axis.

By integrating this expression and plugging in the given values into the formula, we obtain the result of approximately 51.84 volts for the electric potential at the point 5m away from the charge distribution on the x-axis.

This value signifies the work done per unit charge to bring a positive test charge from infinity to the specified point, considering the influence of the continuous charge distribution along the x-axis.

"".

Answer:

0.82 mm

Explanation:

The formula for calculation an [tex]n^{th}[/tex] bright fringe from the central maxima is given as:

[tex]y_n=\frac{n \lambda D}{d}[/tex]

so for the distance of the second-order fringe when wavelength [tex]\lambda_1[/tex] = 745-nm can be calculated as:

[tex]y_2 = \frac{n \lambda_1 D}{d}[/tex]

where;

n = 2

[tex]\lambda_1[/tex] = 745-nm

D = 1.0 m

d = 0.54 mm

substituting the parameters in the above equation; we have:

[tex]y_2 = \frac{2(745nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]

[tex]y_2[/tex] = 0.00276 m

[tex]y_2[/tex] = 2.76 × 10 ⁻³ m

The distance of the second order fringe when the wavelength [tex]\lambda_2[/tex] = 660-nm is as follows:

[tex]y^'}_2 = \frac{2(660nm*\frac{10^{-9m}}{1.0nm}(1.0m) }{0.54 (\frac{10^{-3m}} {1.0mm})}[/tex]

[tex]y^'}_2[/tex] = 1.94 × 10 ⁻³ m

So, the distance apart the two fringe can now be calculated as:

[tex]\delta y = y_2-y^{'}_2[/tex]

[tex]\delta y[/tex] = 2.76 × 10 ⁻³ m - 1.94 × 10 ⁻³ m

[tex]\delta y[/tex] = 10 ⁻³ (2.76 - 1.94)

[tex]\delta y[/tex] = 10 ⁻³ (0.82)

[tex]\delta y[/tex] = 0.82 × 10 ⁻³ m

[tex]\delta y[/tex] =  0.82 × 10 ⁻³ m [tex](\frac{1.0mm}{10^{-3}m} )[/tex]

[tex]\delta y[/tex] = 0.82 mm

Thus, the distance apart the second-order fringes for these two wavelengths = 0.82 mm

Answer:

Minimum capacitance = 200 μF

Explanation:

From image B attached, we can calculate the current flowing through the capacitors.

Thus;

Since V=IR; I = V/R = 5/500 = 0.01 A

Maximum charge in voltage is from 5V to 4.9V. Thus, each capacitor will have 2.5V. Hence, change in voltage(Δv) for each capacitor will be ; Δv = 0.05 V

So minimum capacitance will be determined from;

i(t) = C(dv/dt)

So, C = i(t)(Δt/Δv) = 0.01[0.001/0.05]

C = 0.01 x 0.0002 = 200 x 10^(-6) F = 200 μF

Final answer:

To find the minimum capacitance for the cardiac pacemaker circuit to maintain the pulse amplitude between 4.9V and 5.0V over 1ms, use the discharging formula for a capacitor and solve for the capacitance with the given resistance and time.

Explanation:

The problem involves designing a cardiac pacemaker circuit where two capacitors charged to 2.5V in parallel are switched to series to deliver a pulse of 5V across a 500-ohm resistor, which models the heart. The question asks for the minimum value of the capacitance needed to ensure the voltage does not drop below 4.9V over the 1ms duration of the pulse.

To solve for the capacitance (C), we use the formula for the discharging of a capacitor through a resistor (Q = Q0e-(t/RC)), where Q is the charge at time t, Q0 is the initial charge, R is the resistance, and C is the capacitance. The minimum voltage (V min) we want after 1ms is 4.9V, which will be across two capacitors in series, effectively having a 5.0V drop across two capacitances when fully charged. The effective capacitance when two identical capacitors are in series is half that of one capacitor, so we need to calculate for C based on an effective voltage drop from 5.0V to 4.9V across a single capacitor, equivalent to the heart resistance of 500 ohms, over 1ms.

We rearrange the discharging formula to solve for C:

V = V 0 e -(t/RC)

Now we solve for C:

C = -t / (R*ln(V/V 0 )

Plugging in the values:

C = -0.001s / (500Ω*ln(4.9V/5.0V))

This calculation will give us the minimum capacitance value to ensure the pulse amplitude stays between 4.9V and 5.0V during the 1ms pulse duration.

Answer:a

Explanation:

Given

Potential difference [tex]V=110\ V[/tex]

Power of bulb A [tex]P_A=60\ W[/tex]

Power of bulb B [tex]P_B=100\ W[/tex]

If voltage is same for both the bulbs then Power is given by

[tex]P=\dfrac{V^2}{R}[/tex]

[tex]P_A=\dfrac{(110)^2}{R_A}[/tex]

[tex]60=\dfrac{110^2}{R_A}[/tex]

[tex]R_A=201.66\ \Omega[/tex]

similarly

[tex]R_B=\dfrac{110^2}{100}[/tex]

[tex]R_B=121\ \Omega[/tex]

[tex]R_A>R_B[/tex]

so current in bulb A is smaller than B

Thus the 60 W bulb has a greater resistance.

Thus option (a) is correct

The correct statement is that a) the 60 W bulb has a greater resistance and smaller current than the 100 W bulb, since both bulbs are operated at the same voltage but the 60 W bulb has lower power.

To determine the correct statement about the resistance and current for two light bulbs operated at a potential difference of 110 V, we need to consider the power ratings of the bulbs and know that power (P) is the product of current (I) and voltage (V), P = IV. Furthermore, by using Ohm's law (V = IR), we can infer that Power can also be expressed as P = I2R or P = V²/R, which relates power to resistance (R).

Both bulbs operate at the same voltage (110 V), which means their power differences are due to differences in current and resistance.

For bulb A (60 W): P = V²/R, so R = V²/P = (110 V)2/60 W.

For bulb B (100 W): R = V²/P = (110 V)²/100 W.

Comparing these two, bulb A has higher resistance because it has lower power for the same voltage. Also, since P = IV, and bulb A has a lower power at the same voltage, it must also have a smaller current. Therefore, the correct answer is: a. The 60 W bulb has a greater resistance and smaller current than the 100 W bulb.

Answer:

[tex]\frac{d_f}{d} =0.9999983[/tex]

Explanation:

Given:

Stress induced due to the applied force:

[tex]\sigma=\frac{F}{a}[/tex]

[tex]\sigma=\frac{7000}{0.01}[/tex]

[tex]\sigma=700000\ Pa=0.7\ MPa[/tex]

Now the longitudinal strain:

[tex]\epsilon=\frac{\sigma}{E}[/tex]

[tex]\epsilon=\frac{0.7}{128\times 10^3}[/tex]

[tex]\epsilon=5.468\times 10^{-6}[/tex]

Now from the relation of Poisson's ratio:

[tex]\mu=\frac{\nu}{\epsilon}[/tex]

where:

[tex]\nu=[/tex] lateral strain

[tex]0.3=\frac{\nu}{5.468\times 10^{-6}}[/tex]

[tex]\nu=1.6406\times 10^{-6}[/tex] ..................(1)

Now we find the diameter of the wire:

[tex]a=\pi.\frac{d^2}{4}[/tex]

[tex]0.01=\pi\times \frac{d^2}{4}[/tex]

[tex]\frac{0.04}{\pi} =d^2[/tex]

[tex]d=0.1128\ m=112.8\ mm[/tex]

When the tensile load is applied its diameter decreases:

The lateral strain is also given as,

[tex]\nu=\frac{\Delta d}{d}[/tex]

[tex]1.6406\times 10^{-6}=\frac{\Delta d}{112.8}[/tex]

[tex]\Delta d=0.000185\ mm[/tex]

Now the final diameter will be:

[tex]d_f=d-\Delta d[/tex]

[tex]d_f=112.8-0.000185[/tex]

[tex]d_f=112.799815\ mm[/tex]

Now the ratio:

[tex]\frac{d_f}{d} =\frac{112.799815}{112.8}[/tex]

[tex]\frac{d_f}{d} =0.9999983[/tex]

Answer:

123.30 m

Explanation:

Given

Speed, u = 22 m/s

acceleration, a = 1.40 m/s²

time, t = 7.30 s

From equation of motion,

                       v = u + at

where,

v is the final velocity

u is the initial velocity

a is the acceleration

t is time  

                       V = at + U

using equation  v - u = at to get line equation for the graph of the motion of the train on the incline plane

                       [tex]V_{x} = mt + V_{o}[/tex]      where m is the slope

Comparing equation (1) and (2)

[tex]V = V_{x}[/tex]

a = m    

[tex]U = V_{o}[/tex]

Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.

          a = -  1.40 m/s²

Sunstituting a = -  1.40 m/s² and  u = 22 m/s

                        [tex]V_{x} = -1.40t + 22[/tex]

                            [tex]V_{x} = -1.40(7.30) + 22[/tex]

                             [tex]V_{x} = -10.22 + 22[/tex]

                             [tex]V_{x} = 11. 78 m/s[/tex]

The speed of the train at 7.30 s is 11.78 m/s.

The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.

           Area of triangle +  Area of rectangle

          [tex][\frac{1}{2} * (22 - 11.78) * (7.30)] + [(11.78 - 0) * (7.30)][/tex]

                           = 37.303 + 85.994

                           = 123. 297 m

                           ≈ 123. 30 m

Answer:

123.297 m

Explanation:

A train, traveling at a constant speed of 22.0 m/s,

v = 22.0 m/s

the train slows down with a constant acceleration of magnitude 1.40 m/s².

[tex]a_s[/tex] = -1.4 m/s²

How far has the train traveled up the incline after 7.30 s

t =7.30 s

We can calculate the distance traveled up the incline after 7.30 s by using the formula:

[tex]x_f =x_i+v_xt+\frac{1}{2}a_st^2[/tex]

where;

[tex]x_f[/tex] = the distance traveled up

[tex]x_i[/tex] = 0

[tex]v_x[/tex] = speed of the train

[tex]a_s[/tex] = deceleration

t = time

Substituting our data; we have:

[tex]x_f = 0+(22m/s)(7.30s)+\frac{1}{2}(-1.4m/s^2)(7.30s)[/tex]

[tex]x_f =16.06 -37.303[/tex]

[tex]x_f[/tex] = 123.297 m

Answer:

95.5ms-2

Explanation:

First we obtain the time taken for the motion from the equation of motion. Secondly we obtain the horizontal velocity of the stone. This can now be used to calculate the centripetal acceleration. Note that the length of the chord is the radius of the circle around which the stone moves.

All details are contained in the detailed step-by-step solution attached to this answer.

Answer:

102.52 W

Explanation:

Power: This ca be defined as the rate at which energy is used up.

The S.I unit of power is Watt(W).

From the question,

P = E/t.............. Equation 1

Where P = average power of the snake head, E = energy of the snake, t = time.

But,

E = 1/2mv²............ Equation 2

Where m = mass of the head, v = velocity of the snake head.

Substitute equation 2 into equation 1

P = 1/2mv²/t ............ Equation 3

Given: m = 170 g = 0.17 kg, v = 28 m/s, t = 0.65 s.

Substitute into equation 3

P = 1/2(0.17)(28²)/0.65

P = 102.52 W.

Hence the average power of the snake = 102.52 W

Answer:

(a). The final velocity of the first glider is −2.15 m/s.

(b). The final velocity of the second glider is -1.67 m/s.

Explanation:

Given that,

Mass of glider = 0.140 kg

Speed = 0.900 m/s

Mass of another glider = 0.297 kg

Speed =2 .25 m/s

Suppose, Find the magnitude of the final velocity of the first glider.

Find the magnitude of the final velocity of the second glider.

(a). We need to calculate the final velocity of the first glider

Using formula of collision

[tex]v_{1}=\dfrac{u_{1}(m_{1}-m_{2})+2m_{2}u_{2}}{m_{1}+m_{2}}[/tex]

Put the value into the formula

[tex]v_{1}=\dfrac{0.900(0.140+0.297)+2\times(-0.297)\times2.25}{0.140+0.297}[/tex]

[tex]v_{1}=-2.15\ m/s[/tex]

(b). We need to calculate the final velocity of the second glider

Using formula of collision

[tex]v_{2}=\dfrac{u_{2}(m_{2}-m_{1})+2m_{1}u_{1}}{m_{1}+m_{2}}[/tex]

Put the value into the formula

[tex]v_{1}=\dfrac{-2.25(0.297+0.140)+2\times(0.140)\times0.900}{0.140+0.297}[/tex]

[tex]v_{1}=-1.67\ m/s[/tex]

Hence,  (a). The final velocity of the first glider is −2.15 m/s.

(b). The final velocity of the second glider is -1.67 m/s.

Answer with Explanation:

We are given that

Mass of glider=m=0.14 kg

Initial speed of first glider, u=0.9 m/s

m'=0.297 kg

Initial speed of second glider, u'=-2.25m/s

a.We have to find the magnitude of final velocity of 0.140 kg glider.

The collision is elastic then the final velocity of 0.140 kg glider

[tex]v=\frac{(m-m')u+2m'u'}{m+m'}[/tex]

Substitute the values

[tex]v=\frac{(0.140-0.297)\times 0.9+2\times 0.297\times (-2.25)}{0.140+0.297}=-3.38m/s[/tex]

Magnitude of final velocity of 0.140 kg glider=3.38 m/s

b.[tex]v'=\frac{u'(m'-m)+2mu}{m+m'}[/tex]

[tex]v'=\frac{-2.25(0.297-0.140)+2\times 0.140\times 0.9}{0.140+0.297}=-0.23 m/s[/tex]

Hence, the magnitude of final velocity of 0.297 kg glider=0.23 m/s

Answer:

The magnetic field will be [tex]\large{\dfrac{1.4 \times 10^{-4}}{d}} T[/tex], '2d' being the distance the wires.

Explanation:

From Biot-Savart's law, the magnetic field ([tex]\large{\overrightarrow{B}}[/tex]) at a distance '[tex]r[/tex]' due to a current carrying conductor carrying current '[tex]I[/tex]' is given by

[tex]\large{\overrightarrow{B} = \dfrac{\mu_{0}I}{4 \pi}} \int \dfrac{\overrightarrow{dl} \times \hat{r}}{r^{2}}}[/tex]

where '[tex]\overrightarrow{dl}[/tex]' is an elemental length along the direction of the current flow through the conductor.

Using this law, the magnetic field due to straight current carrying conductor having current '[tex]I[/tex]', at a distance '[tex]d[/tex]' is given by

[tex]\large{\overrightarrow{B}} = \dfrac{\mu_{0}I}{2 \pi d}[/tex]

According to the figure if '[tex]I_{t}[/tex]' be the current carried by the top wire, '[tex]I_{b}[/tex]' be the current carried by the bottom wire and '[tex]2d[/tex]' be the distance between them, then the direction of the magnetic field at 'P', which is midway between them, will be perpendicular towards the plane of the screen, shown by the [tex]\bigotimes[/tex] symbol and that due to the bottom wire at 'P' will be perpendicular away from the plane of the screen, shown by [tex]\bigodot[/tex] symbol.

Given [tex]\large{I_{t} = 19.5 A}[/tex] and [tex]\large{I_{B} = 12.5 A}[/tex]

Therefore, the magnetic field ([tex]\large{B_{t}}[/tex]) at 'P' due to the top wire

[tex]B_{t} = \dfrac{\mu_{0}I_{t}}{2 \pi d}[/tex]

and the magnetic field ([tex]\large{B_{b}}[/tex]) at 'P' due to the bottom wire

[tex]B_{b} = \dfrac{\mu_{0}I_{b}}{2 \pi d}[/tex]

Therefore taking the value of [tex]\mu_{0} = 4\pi \times 10^{-7}[/tex] the net magnetic field ([tex]\large{B_{M}}[/tex]) at the midway between the wires will be

[tex]\large{B_{M} = \dfrac{4 \pi \times 10^{-7}}{2 \pi d} (I_{t} - I_{b}) = \dfrac{2 \times 10^{-7}}{d} = \dfrac{41.4 \times 10 ^{-4}}{d}} T[/tex]

Answer:

4.14 cm.

Explanation:

Given,

For Coil 1

radius of coil, r₁ = 5.6 cm

Magnetic field, B₁ = 0.24 T

For Coil 2

radius of coil, r₂ = ?

Magnetic field, B₂ = 0.44 T

Using formula of maximum torque

[tex]\tau_{max}= NIAB[/tex]

Since both the coil experience same maximum torques

now,

[tex] NIA_1B_1 = NIA_2B_2[/tex]

[tex]A_1B_1 = A_2B_2[/tex]

[tex] r_1^2 B_1 = r_2^2 B_2[/tex]

[tex] 5.6^2\times 0.24= r_2^2\times 0.44[/tex]

[tex]r_2 = 4.14\ cm[/tex]

Radius of the coil 2 is equal to 4.14 cm.

Answer:

[tex]9.56\cdot 10^{-7} C[/tex]

Explanation:

A parallel-plate capacitors consist of two parallel plates charged with opposite charge.

Since the distance between the plates (1 cm) is very small compared to the side of the plates (19 cm), we can consider these two plates as two infinite sheets of charge.

The electric field between two infinite sheets with opposite charge is:

[tex]E=\frac{\sigma}{\epsilon_0}[/tex]

where

[tex]\sigma=\frac{Q}{A}[/tex] is the surface charge density, where

Q is the charge on the plate

A is the area of the plate

[tex]\epsilon_0 = 8.85\cdot 10^{-12}F/m[/tex] is the vacuum permittivity

In this problem:

- The side of one plate is

L = 19 cm = 0.19 m

So the area is

[tex]A=L^2=(0.19)^2=0.036m^2[/tex]

Here we want to find the maximum charge that can be stored on the plates such that the value of the electric field does not overcome:

[tex]E=3\cdot 10^6 N/C[/tex]

Substituting this value into the previous formula and re-arranging it for Q, we find the charge:

[tex]E=\frac{Q}{A\epsilon_0}\\Q=EA\epsilon_0 = (3\cdot 10^6)(0.036)(8.85\cdot 10^{-12})=9.56\cdot 10^{-7} C[/tex]

Answer:

Explanation:

Moment of inertia of propeller

= (1/12) m l²

=( 1 / 12) x 102 x 2.18²

= 40.4 kgm²

torque = 1950 N-m

angular acceleration

=  torque / moment of inertia

= 1950 / 40.4

= 48.26 rad / s²

Answer:

a = -1.51 m/s^2

Explanation:

Given:

Vi= 55 m/s

Vf= 0 m/s

S = 1km =1000 m

a = ? m/s^2

Sol:

3rd equation of motion

2as = vf^2 - vi^2

a = (55)^2 / 2x1000

a = -1.51 m/s^2

-'ve sign shows the deceleration.

-1.5m/s²

Consider one of the equations of motion as follows;

v² = u² + 2as     -------------------------(i)

Where;

v = final velocity of moving body (airplane in this case)

u = initial velocity of the body

a = acceleration of motion of the body

s = distance covered by the body

From the question;

v = 0             [since the airplane comes to a stop]

u = 55m/s

s = 1.00km = 1000m

Substitute these values into equation (i) as follows;

0² = 55² + 2a(1000)

0 = 3025 + 2000a

Collect like terms;

2000a = -3025

Solve for a;

a = [tex]\frac{-3025}{2000}[/tex]

a = -1.5m/s²

Therefore the acceleration that allows the plane to come to a stop in 1.00km is -1.5m/s².

The negative sign shows that the plane is actually decelerating.

Answer:

37.42 m/s

Explanation:

We know that apparent frequency, [tex]\bar f[/tex] is given by

[tex]\bar f=f\frac {V}{V-V_s}[/tex] where f is the given frequency in this case 392, V is the speed of sound in air which is given as 343 and [tex]V_s[/tex] is the speed of car which is unknown, \bar f is given as 440 Hz

[tex]440=392\times \frac {343}{343-V_s}\\343-V_s=392\times \frac {343}{440}=305.5818182\\V_s=343-305.5818182=37.41818182\approx 37.42 m/s[/tex]

Explanation:

The misunderstanding here is that the thing that turns on the light bulb is not the same electrons near the light switch. So, the electrons near the switch is not moving all the way across the circuit instantly. The electrons are distributed across the wire. When the light switch is turned on, the circuit is connected and there is a potential difference between the bulb and the source. This potential difference creates an electric field, and free electrons move under the influence of this electric field according to Coulomb's Law. When they start to move the electrons closest to the bulb causes the bulb to glow.

So, the important factor here is not the drift velocity of the electrons but the number of electrons and the strength of the electric field.

Answer:

[tex] v = \frac{2 V}{3}= 0.667 v[/tex]

Since we have identical diodes we can use the equation:

[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]

And replacing we have:[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]

Since we know that 1 mA is drawn away from the output then the real value for I would be

[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]

And for this case the value for [tex] v_D[/tex] would be:

[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]

And the output votage on this case would be:

[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]

And the net change in the output voltage would be:

[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]

Explanation:

For this case we have the figure attached illustrating the problem

We know that the equation for the current in a diode id given by:

[tex] I_D = I_s [e^{\frac{V_D}{V_T}} -1] \approx I_S e^{\frac{V_D}{V_T}}[/tex]

For this case the voltage across the 3 diode in series needs to be 2 V, and we can find the voltage on each diode [tex] v_1 + v_2 + v_3= 2[/tex] and each voltage is the same v for each diode, so then:

[tex] v = \frac{2 V}{3}= 0.667 v[/tex]

Since we have identical diodes we can use the equation:

[tex] I_D =I= I_S e^{\frac{V_D}{V_T}}[/tex]

And replacing we have:

[tex]I = 10^{-14} A e^{\frac{0.667}{0.025}}= 3.86x10^{-3} A = 3.86 mA[/tex]

Since we know that 1 mA is drawn away from the output then the real value for I would be

[tex] I_D = I = 3.86 mA -1 mA= 2.86 mA[/tex]

And for this case the value for [tex] v_D[/tex] would be:

[tex] V_D = V_T ln (\frac{I_D}{I_T})= 0.025 ln (\frac{0.0029}{10^{-14}})= 0.660 V[/tex]

And the output votage on this case would be:

[tex] V = 3 V_D = 3 *0.660 V = 1.98 V[/tex]

And the net change in the output voltage would be:

[tex] \Delta V = |2 v-1.98 V| = |0.02 V |= 20 m V[/tex]

Answer:

0.0187 eV

Explanation:

Given that:

diameter of the hydrogen gas (λ) = 1.48 ×10⁻¹⁰ m'

mass of the hydrogen gas = 3.35 ×10⁻²⁷ kg

We need to determine the momentum first before calculating the kinetic energy.

So momentum of the hydrogen gas molecule is written as;

[tex]p =\frac{h}{ \lambda}[/tex]

[tex]p = \frac{96.626*10^{-34}J.s}{1.48*10^{-10}m}[/tex]

[tex]p=4.477*10^{-24} kg.m/s[/tex]

NOW, the kinetic energy of the hydrogen gas molecule is calculated as follows by using the formula:

[tex]k_o = \frac{p^2}{2m}[/tex]

[tex]k_o =\frac{(4.477*10^{-24}kg.m/s)^2}{2(3.35*10^-{27})kg}[/tex]

[tex]k_o=2.9916*10^{-21}J(\frac{1eV}{1.6*10^{-19}J})[/tex]

[tex]K_o=0.0187eV[/tex]

Answer:

x= 0.077 m from the wire carrying 5.0 A current.

Explanation:

       [tex]B =\frac{\mu_{0} * I}{2*\pi*d}[/tex]

       [tex]\frac{\mu_{0} * I_{1}}{2*\pi*x} - \frac{\mu_{0} * I_{2} }{2*\pi*(d-x)} = 0[/tex]

        [tex]\frac{I_{1} }{x} = \frac{I_{2} }{(d-x)} \\ \frac{5.0A}{x(m)} = \frac{8.0A}{(0.2m-x(m))}[/tex]

        ⇒ [tex]x =\frac{1m}{13} = 0.077 m = 7.7 cm[/tex]

Answer:

(B) is a volume in space.

Explanation:

A system is the space chosen form the universe for the study and analysis.

There are 3 kinds of systems :

1) Open system:

This system can interact to surroundings in terms of both mass and energy transfer through its boundaries.

2) Closed system:

This type of system can only interact to the surrounding in terms of energy transfer through its boundary.

3) Isolated system:

This type of system cannot interact with the surrounding by any means and its energy remains constant with time under the observation.

It has insulated boundary.

A system in physics often represents a fixed set of physical interactions, whereas a control volume is a designated region in space chosen to study the balance of mass, energy, and momentum. Option (A) best defines these concepts.

In physics, a system typically refers to a set of physical interactions that are being studied. This could include anything from a single particle to a complex machine. A control volume, on the other hand, is a specific volume in space. The essence of its definition is that it is a region in space chosen for study of the balance of mass, energy, and momentum. The boundaries of the control volume can be real or imagined. The control volume can also move and deform with time.

To answer the question, option (A) would be most accurate. A system is a fixed amount of matter, meaning it represents a particular set of physical elements. A control volume is a region in space which can encompass these elements.

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Answer:

a. 2000W

b.0.2W

c. Bulb 1

Explanation:

Data given:

bulb 1 rating =20w,12v

bulb 2 rating =20w,120v.

we calculate the resistance for each bulb

[tex]bulb 1: R=\frac{V^2}{P}\\ bulb 1: R=\frac{12^2}{20}\\ R=7.2 ohms[/tex]

for bulb 2

[tex]bulb 2: R=\frac{V^2}{P}\\ bulb 2: R=\frac{120^2}{20}\\ R=720 ohms[/tex]

when miss connection occur we need to calculate the dissipated power from wash bulb

a. for bulb 1 with R=7.2ohms and V=120v,

the power is calculated as

[tex]P=\frac{v^2}{R}\\ P=\frac{120^2}{7.2}\\ P=2000W[/tex]

b. for bulb 2 with R=720 ohms and V=12v,

the power is calculated as

[tex]P=\frac{v^2}{R}\\ P=\frac{12^2}{720}\\ P=0.2W\\[/tex]

c. Bulb 1 is more likely to burn out because it is operating above the rated power.

Answer:

[tex]v=7.62\ m.s^{-1}[/tex]

Explanation:

Given:

where:

[tex]x=[/tex] distance in the +ve x-direction

We know:

[tex]F=m.a\\\Rightarrow a=\frac{F}{m}[/tex]

Now force change in force on the body:

[tex]F(x)=18-0.53(x'-x)[/tex]

[tex]F=18-0.53\times (14-0)[/tex]

[tex]F=10.58\ N[/tex]

Now the acceleration due to the force:

[tex]a=\frac{F}{m}[/tex]

[tex]a=\frac{10.58}{5.1}[/tex]

[tex]a=2.0745\ m.s^{-2}[/tex]

Now using equation of motion:

[tex]v^2=u^2+2a.x'[/tex]

[tex]v^2=0^2+2\times 2.0745\times 14[/tex]

[tex]v=7.62\ m.s^{-1}[/tex]

The density given implies that the total electric field will have a uniform magnitude.

Density simply means the degree of compactness that can be found in a substance.

It should be noted that for an infinite sheet of charge, the electric field will be perpendicular to the surface.

In this case, it doesn't depend on the distance of points and is uniform all through. Therefore, the total electric field will have a uniform magnitude.

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Answer:

Explanation:

dV = k (dq)/d

dV = k (sigma da)/d

dV = k (5 C r (r dr d(phi)))/d

d = sqrt(r^2 + x^2)

dV = 5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)

==> V = int{5 k C (r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi

==> V = 5 k C int{(r^2/sqrt(r^2+x^2)) dr d(phi)} ; from r=0 to r=R and phi=0 to phi=2pi

==> V = 5 k C (2 pi) int{(r^2/sqrt(r^2+x^2)) dr} ; from r=0 to r=R

==> V = 5 k C (2 pi) (1/2) (R sqrt(R^2+x^2) - x^2 ln(sqrt(r^2+x^2) + r) + x^2 Ln(x))

The change in potential energy as the object rises from 20m to 40m is 200 Joules. This result is the same regardless of the coordinate system used.

The change in potential energy, Δ P E, can be calculated using the formula Δ P E = P E f i n a l − P E i n i t i a l. Here, P E f i n a l is the final potential energy when the object is at 40m, and P E i n i t i a l is the initial potential energy when the object is at 20m. To get these values, we use the formula for gravitational potential energy: P E = m g h, where m is mass, g is the acceleration due to gravity, and h is the height above ground.

So, P E f i n a l = m g h = 1kg * 10 m/s² * 40m = 400 J (Joules), and P E i n i t i a l = m g h = 1kg * 10 m/s² * 20m = 200 J. Hence, Δ P E = P E f i n a l − P E i n i t i a l = 400 J - 200 J = 200 J. Regardless of their coordinate systems, both Liz and Connor should get the same Δ P E, assuming they've correctly identified the initial and final heights.

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The change in potential energy (ΔPE) as measured by both scientists would be 200J. This is calculated using the formula ΔPE = PEfinal - PEinitial, and substituting given values for mass, gravity, and height.

In this scenario, the change in potential energy (ΔPE) is determined by the difference in heights and the weight of the object. We calculate ΔPE with the formula ΔPE = PEfinal − PEinitial. Given that PE is calculated as m*g*h where m is mass, g is gravity (which we assume as 10m/s² here), and h is height, we can substitute these values in. Initially, the object is at 20m, so PEinitial is m*g*h = 1kg*10m/s²*20m = 200 J (joules). Finally, it rises to 40m, so PEfinal = 1kg*10m/s²*40m = 400 J. Hence, ΔPE is PEfinal−PEinitial = 400J – 200J = 200J. Therefore, the change in potential energy, as measured by both Liz and Connor, would be 200J regardless of their coordinate systems.

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