For a demonstration, a professor uses a razor blade to cut a thin slit in a piece of aluminum foil. When she shines a laser pointer (λ=680nm) through the slit onto a screen 5.5 m away, a diffraction pattern appears. The bright band in the center of the pattern is 7.8 cm wide. What is the width of the slit?

Answer:

Option E is correct.

The two objects rise to the same height.

Explanation:

Let the mass, velocity and height the bigger object will rise to be M, V and H respectively.

Let the mass, velocity and height the object will rise to be m, v and h respectively.

Note that V = v

Using the work energy theorem

The change in kinetic energy of a body between two points is equal to the work done on the body between those two points.

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(M)(V²) = (-MV²/2)

(The final kinetic energy = 0 J because the object comes to rest at the final point)

Work done on the bigger object = work done by all the forces acting on the body

But the only force acting on the body is the force of gravity (since the inclined plane is frictionless)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - MgH

(-MV²/2) = - MgH

H = (V²/2g)

For the small body,

Change in kinetic energy of the bigger object = (final kinetic energy) - (initial kinetic energy) = 0 - (1/2)(m)(v²) = (-mv²/2)

Workdone on the body = work done by the force of gravity in moving the body up a height of H = - mgh

(-mv²/2) = - mgh

h = (v²/2g)

Since V = v as given in the question (Both bodies have the same speeds)

H = h = (V²/2g) = (v²/2g)

Hope this Helps!!!

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]

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:

(a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

Explanation:

Suppose, A steel ball of mass 0.500 kg is fastened to a cord that is 50.0 cm long and fixed at the far end. The ball is then released when the cord is horizontal.

Given that,

Mass of steel block = 2.30 kg

Mass of ball = 0.500 kg

Length of cord = 50.0 cm

We need to calculate the initial speed of the ball

Using conservation of energy

[tex]\dfrac{1}{2}mv^2=mgl[/tex]

[tex]v=\sqrt{2gl}[/tex]

Put the value into the formula

[tex]u=\sqrt{2\times9.8\times50.0\times10^{-2}}[/tex]

[tex]u=3.13\ m/s[/tex]

The initial speed of the ball [tex]u_{1}=3.13\ m/s[/tex]

The initial speed of the block [tex]u_{2}=0[/tex]

(a). We need to calculate the speed of the ball after collision

Using formula of collision

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

Put the value into the formula

[tex]v_{1}=(\dfrac{0.5-2.30}{0.5+2.30})\times3.13[/tex]

[tex]v_{1}=-2.01\ m/s[/tex]

Negative sign shows the opposite direction of initial direction.

(b). We need to calculate the speed of the block after collision

Using formula of collision

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

Put the value into the formula

[tex]v_{2}=(\dfrac{2\times0.5}{0.5+2.30})\times3.13+0[/tex]

[tex]v_{2}=1.11\ m/s[/tex]

Hence, (a). The speed of the ball after collision is 2.01 m/s.

(b). The speed of the block after collision 1.11 m/s.

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:

Since at 20V most of the photons released are red then when the voltage keeps increasing the hotter the filament will be, therefore the color of light will be bright red.

Explanation:

The higher the energy the more the electrons in the molecules of the object will be excited, and when they de-excite to their ground states they release energy in the form of infrared light. The increase in voltage and higher temperatures make the object release brighter color and sometimes at the highest temperatures +1400 degrees Celsius, the color glows hot white.

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

To solve this problem we will apply the concepts related to the electric field in a ring. This concept is already standardized in the following mathematical expression, which relates the coulomb constant, the distance to the axis, the distance of the two points. Mathematically it is described as,

[tex]E = \frac{k_exq}{(x^2+r^2)^{(3/2)}} \hat{i}[/tex]

Here,

[tex]k_e[/tex] = Coulomb constant

q = Charge

x = Distance to the axis

r = Distance between the charges

Our values are given as,

[tex]r = 10.0cm = 10*10^{-2} m[/tex]

[tex]q = 75\mu C = 75*10^{-6} C[/tex]

[tex]x_a = 1.00cm[/tex]

[tex]x_b = 5.00cm[/tex]

[tex]x_c = 30cm[/tex]

[tex]x_d = 100cm[/tex]

So applying this to our 4 distances, we have

PART A)

[tex]E_a = \frac{(9*10^9)(1.00*10^{-2})(75*10^{-6})}{((1.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_a = 6.64*10^6N/C \hat{i}[/tex]

PART B)

[tex]E_b = \frac{(9*10^9)(5.00*10^{-2})(75*10^{-6})}{((5.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_b = 24.1*10^6N/C \hat{i}[/tex]

PART C)

[tex]E_c = \frac{(9*10^9)(30.00*10^{-2})(75*10^{-6})}{((30.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_c = 6.39*10^6N/C \hat{i}[/tex]

PART D)

[tex]E_d = \frac{(9*10^9)(100.00*10^{-2})(75*10^{-6})}{((100.00*10^{-2})^2+(10*10^{-2})^2)^{(3/2)}} \hat{i}[/tex]

[tex]E_d = 0.664*10^6N/C \hat{i}[/tex]

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: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:

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:

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:

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:

So, Magnitude of the field (B) = 2.09 x 10^(-2) T.

It's in the negative direction since acceleration is in positive direction.

Explanation:

First of all, since acceleration is in positive x- direction, the magnetic field must be in negative y- direction.

We know that The magnitude of the Lorentz force F is; F = qvB sinθ

So, B = F/(qvsinθ)

F = ma.

Speed(v) = 1.00 x10^(7) m/s

acceleration (a) = 2.00 x10^(13) m/s^(2)

Mass of proton = 1.673 × 10^(-27) kilograms

q(elementary charge of proton) = 1.602×10^(−19)

Since right hand thumb rule, θ= 90°

So;B = [1.673 × 10^(-27) x 2.00 x10^(13)] / [ {1.602×10^(−19)} x {1.00 x10^(7)} x sin 90]

So,B = 2.09 x 10^(-2) T.

It's in the negative direction since acceleration is in positive direction.

The magnitude of the magnetic field that the proton experiences is 2.08 x 10^-4 Tesla. The direction of the magnetic field, determined by the right-hand rule, is in the negative y-direction.

The force experienced by a charged particle moving within a magnetic field, like the proton in your question, can be calculated using the Lorentz force law: F = qvBsinθ, where F is the force, q is the charge of the particle, v is its velocity, B is the magnetic field, and θ is the angle between the velocity and the magnetic field. Because the proton is moving perpendicular to the field, θ = 90°, and sinθ = 1. Therefore, F = qvB. To find the magnetic field, B, we can rearrange this formula to give B = F/qv.

Now, we know from Newton's second law that F = ma, so we can replace F in our formula with the proton's mass (m, which is 1.67 x 10^-27 kg for a proton) times the given acceleration (a), yielding B = ma/qv.

Substitute the given values for acceleration, proton's charge, and velocity into our equation, we get: B = (1.67 x 10^-27 kg)(2 x 10^13 m/s^2) / (1.60 x 10^-19 C)(1 x 10^7 m/s) = 2.08 x 10^-4 Tesla (T).

The direction of the magnetic field is given by the right-hand rule as discussed in Essential Knowledge 2.D.1. If you arrange your right hand such that your thumb points in the direction of the proton's velocity (positive z-direction) and your fingers curl in the direction of the force (positive x-direction), your palm points in the direction of the magnetic field, which would be in the negative y-direction.

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Answer:

22.5 m/s

Explanation:

Applying Newton's third law of motion

Momentum of the cannon and cart = momentum of the cannonball

MV = mv..................... Equation 1

Where M = mass of the cannon and the cart, V = Recoil velocity of the cannon and the cart, m = mass of the cannonball, v = velocity of the cannonball

make v the subject of the equation

v = MV/m............. Equation 2

Given: M = 150 kg, V = 1.5 m/s, m = 10 kg

Substitute into equation 2

v = 150(1.5)/10

v = 22.5 m/s

Hence the cannonball leave the cannon with a velocity of 22.5 m/s

Answer:

v2 = 22.5 m/s

Explanation:

Momentum is how hard to stop or turn a moving object . Generally, momentum measures mass in motion. Momentum is a vector quantity. Mathematically,

p = mass ×  velocity

The total momentum of an isolated system of  bodies remains constant.

mometum before = 0

mass of the canon (m1) = 150 kg

mass of the ball (m2) = 10 kg

velocity of the ball (v2) = ?

velocity of the cannon(v1) = 1.5 m/s

momentum after = momentum before

m2v2 + m1v1 = 0

10v2 = 150 × 1.5

10v2 = 225

divide both sides by 10

v2 = 225/10

v2 = 22.5 m/s

Answer:

[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/tex]

Explanation:

The initial momentum of the racquet ball is:

[tex]||\vec p || = (0.238\,kg)\cdot (12.4\,\frac{m}{s} )[/tex]

[tex]||\vec p || = 2.951\,\frac{kg\cdot m}{s}[/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:

Explanation:

note:

solution is attached due to error in mathematical equation. please find the attachment

Answer:

(a) 1.2 J

(b) 5 m/s

(c)  6.3 J

Explanation:

(a) Kinetic energy at A

Ek = 1/2mv².................. Equation 1

Where Ek = Kinetic energy at A, m = mass of the particle, v = velocity at A.

Given: m = 0.6 kg, v = 2.0 m/s

Substitute into equation 1

Ek = 1/2(0.6)(2²)

Ek = 0.6(2)

Ek = 1.2 J.

(b)

Speed at point B

Ek' = 1/2mv'²............... Equation 2

Make v' the subject of the equation,

v' = √(2Ek'/m).................. Equation 3

Where, Ek' = kinetic energy at B, v' = velocity at B.

Given: Ek' = 7.5 J, m = 0.6 kg,

Substitute into equation 3

v' = √[(2×7.5)/0.6]

v' =√(15/0.6)

v' = √25

v' = 5 m/s.

(c)

Wt =  Δ kinetic energy from A to B

Where Wt = total work done as the particle moves from A to B.

Wt = 1/2m(v'²-v²)

Wt = 1/2(0.6)(5²-2²)

Wt = 0.3(25-4)

Wt = 0.3(21)

Wt = 6.3 J

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]

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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Answer:

A plane of infinite length and width.

Explanation:

For the case of the sphere, we should consider spherical coordinates.

You can move around the sphere without changing your distance from the center of the sphere, however you can alter your azimuthal angle and polar angle, since they are symmetric in spherical coordinate system.

r --> Cannot change

Φ --> Free to change

θ --> Free to change

For the case of the infinite cylinder, we should consider cylindrical coordinates.

You can change your height and angular coordinate, but you cannot change your distance from the axis of the cylinder.

r --> Cannot change

θ --> Free to change

z --> Free to change

For the case of the infinite plane, we should consider cartesian coordinates.

Since the length and width of the plane is infinite, we cannot recognize whether we are getting closer or further away.

x --> Free to move

y --> Free to move

z --> Free to move

Therefore, in the case of infinite plane you have the most freedom to move about without your view of the object changing.

Final answer:

An infinite plane allows the most freedom of movement without changing the view, due to its infinite symmetry. In contrast, a sphere and a cylinder have more limited symmetrical properties, resulting in a restricted freedom of movement to keep the view unchanged.

Explanation:

When comparing the freedom to move above a sphere, an infinitely long cylinder, and an infinite plane, the plane provides the most freedom without changing your view of the object. On a sphere, any move results in a different orientation or distance from the surface. With the cylinder, movement along the axis does not change the view, but any other direction does. However, the infinite plane allows movement in any direction on a two-dimensional plane without changing the viewed geometry or orientation.

From a mathematical perspective, this question revolves around symmetry and geometric invariance under transformation. The infinite plane exhibits infinite symmetry; therefore, no matter how you move parallel to its surface, the view remains unchanged. Conversely, a sphere has rotational symmetry around its center, and a cylinder along its axis, limiting the directions in which one can move without altering the perceived distance or orientation.

Answer:

a) 2 x10^7 eV

b) 2 x10^4 keV

c) 20 MeV

d) 0.02 Gev

e) 3.2 x 10^-12J

Explanation:

The potential difference = 20 x 10^6 V

The charge on the proton = 1.6 x10^-19

The work done to move the proton will be basically the proton will acquire if it accelerates.

Kinetic energy gained = ΔVq = 20 x10^6 x 1.6 x 10^-19

                                                 =3.2 x 10^-12J or 2 x10^7 eV

2 x10^7 eV = 2 x10^4 keV = 20 MeV = 0.02 Gev

Explanation:

Below is an attachment containing the solution.

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.

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:

The stream function for this potential flow is -4696.8

Explanation:

Given that,

Circulation = 8000 m²/s

Radius = 40 m

Pressure = 2 kPa

Suppose the determine the stream function for this potential flow

We need to calculate the stream function

Using formula of stream function

[tex]\Psi=-(\dfrac{\Gamma}{2\pi})ln(r)[/tex]

Where, Γ = circulation

r = radius

Put the value into the formula

[tex]\Psi=-\dfrac{8000}{2\pi}\times ln(40)[/tex]

[tex]\Psi=-4696.8[/tex]

Hence, The stream function for this potential flow is -4696.8

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:

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))

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:

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