The deflection plates in an oscilloscope are 10 cm by 2 cm with a gap distance of 1 mm. A 100 volt potential difference is suddenly applied to the initially uncharged plates through a 975 ohm resistor in series with the deflection plates. How long does it take for the potential difference between the deflection plates to reach 75 volts

Answer:

[tex]a=-0.33\ m/s^2[/tex]

Explanation:

Accelerated Motion

The acceleration of a moving body is defined as the relation of change of speed (or velocity in vector form) with the time taken. The formula is given by

[tex]\displaystyle a=\frac{\Delta v}{t}[/tex]

Or, equivalently

[tex]\displaystyle a=\frac{v_f-v_o}{t}[/tex]

Where vf and vo are the final and initial speeds respectively. The problem gives us these values: v0 = 3 m/s, vf = 1 m/s, t = 3 seconds. Computing a

[tex]\displaystyle a=\frac{1-3}{3}=-0.33\ m/s^2[/tex]

The negative sing of a indicates there is deceleration or decreasing speed

Final answer:

The skateboard's deceleration as it coasts uphill, changing speed from 3 m/s to 1 m/s in 3 seconds, is calculated as -0.67 m/s². This indicates a decrease in speed and is consistent with deceleration.

Explanation:

If a skateboard coasts uphill and experiences a change in speed from 3 m/s to 1 m/s in 3 seconds, to find the deceleration, we use the formula for acceleration, which is the change in velocity divided by the time taken for the change. Deceleration is simply acceleration in the opposite direction to the motion (negative acceleration).

The change in velocity ({\Delta v}) is 1 m/s - 3 m/s, which equals -2 m/s (the negative sign indicates a decrease in speed). The time ({t}) is 3 seconds. Thus, the deceleration is  {\Delta v / t} which is (-2 m/s) / (3 s) = -0.67 m/s². This negative sign signifies that it is indeed a deceleration.

Answer:

Explanation:

W (Ce) = 2.1 eV

W (Al) = 4.1 eV

W (Be) = 5 eV

W (K) = 2.3 eV

W (Pt) = 6.4 eV

W (Mg) = 3.7 eV

Part A:

Work function is directly proportional to the cut off frequency.

let f denotes the cut off frequency.

So, f (Be) > f (Be) > f (Al) > f (Mg) > f (K) > f (Ce)

Part B:

Maximum wavelength for the emission is inversely proportional to the cut off frequency

So, λ (Ce) > λ (K) > λ (Mg) > λ (Al) > λ (Be) > λ (Pt)

Part C:

E = 3.10 eV

Let K is the maximum kinetic energy

K = E - W

K (Ce) = 3.1 - 2.1 = 1 eV

K (Al) 3.1 - 4.1 = not possible

K (Be) = 3.1 - 5 = not possible

K (K)  = 3.1 - 2.3 = 0.8 eV

K (Pt) = .1 - 6.4 = not possible

K (Mg) = 3.1 - 3.7 = not possible

So, K (Ce) > K (K) > K (Al) = K (Be) = K (Pt) = K (Mg)

Metals can be ranked based on their work functions, which determine cutoff frequency, the maximum wavelength of light needed to free electrons, and maximum kinetic energy of emitted electrons under specific light. Lower work functions result in higher cutoff frequencies and longer required wavelengths, and more energetic emitted electrons when under 400 nm light.

The ranking of the metals, in terms of cutoff frequency, maximum wavelength of light, and maximum kinetic energy, is based on the work function of each metal, which is the minimum energy required to free an electron from the metal's surface. The cutoff frequency is inversely related to the work function, so the metal with the lowest work function (cesium, 2.1 eV) will have the highest cutoff frequency, and the metal with the highest work function (platinum, 6.4 eV) will have the lowest cutoff frequency.

Conversely, the maximum wavelength of light needed to free electrons is directly related to the work function, so cesium will require the longest wavelength and platinum will require the shortest wavelength. When illuminated with 400 nm (3.10 eV) light, the kinetic energy of the emitted electrons will be the difference between the energy of the light and the work function, so cesium (with the smallest work function) will have the most energetic electrons, and metals with a work function larger than 3.10 eV (like aluminum, beryllium, and platinum) will not emit electrons at all.

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The minimum width b that can be used for the beam is 4.32 in

To determine the minimum width (b) of the solid rectangular wood beam, we employ principles of structural engineering.

The maximum bending moment (M) for a simply supported beam with a uniformly distributed load (\(w_0\)) occurs at the center and is given by [tex]\(M = \frac{w_0 L^2}{8}\).[/tex]

The section modulus (S) for a rectangular cross-section is [tex]\(S = \frac{b \times h^2}{6}\).[/tex]

The bending stress (\(σ_b\)) is given by[tex]\(σ_b = \frac{M}{S}\)[/tex].

Setting \(σ_b\) equal to the allowable bending stress [tex](\(1.95 \, \text{ksi}\))[/tex]and using the specified aspect ratio (h/b = 1.75), we can solve for b.

This ensures that the wood beam meets structural safety criteria.

The calculation yields a minimum width b of 4.32 in, ensuring that the beam is structurally sound, withstanding the specified uniformly distributed load and adhering to the aspect ratio constraint in compliance with the allowable bending stress.

Answer:

The angle through which each bicycle wheel rotates is 35.03 rad.

Explanation:

Given;

the radius of the circular path, R = 9.30 m

the angular displacement of the bicycle, θ = 1.130 rad

the radius of the bicycle wheel, r = 0.3

S = θR

where;

s is the distance of the circular path

S = 1.13 x 9.3 = 10.509 m

The angle (in radians) through which each bicycle wheel of radius 0.300 m rotates is given as;

θr = 10.509 m

θ = 10.509 / 0.3

θ = 35.03 rad.

Therefore, the angle through which each bicycle wheel of radius 0.300 m rotates is 35.03 rad.

The magnitude of the net force on the driver during a turn on a banked curve, when no friction is required, is equal to the driver's mass multiplied by gravity and the factor x, which represents the perceived increase in weight. Therefore, the magnitude of the net force is m × g × x.

If a sports car moves around a banked curve at a constant speed such that no friction is needed, this means that the net force is providing the necessary centripetal force for the turn. According to the problem, the driver feels as though she weighs x times her actual weight. This perception of increased weight is due to the normal force provided by the banked road, which has both vertical and horizontal components.

Using Newton's second law, the net force on the driver can be expressed as net force = mass × acceleration. In this case, the centripetal acceleration is due to the net horizontal force, which is the horizontal component of the normal force.

The vertical component of the normal force is balancing the driver's actual weight, and the horizontal component of the normal force equals the centripetal force necessary for circular motion. Therefore, the normal force experienced by the driver (which is responsible for the feeling of increased weight) is FN = m × g × x. Since this is the net force and it is providing the centripetal force, the magnitude of the net force on the driver is also FN = m × g × x.

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Answer:

Explanation:....

The range of the projectile is 4675 m. The maximum height reached by the projectile is 2124 m. The speed of the projectile at impact is 234 m/s.

(a) Range is defined as the maximum horizontal distance covered by a projectile during its motion. It is given by the formula:

[tex]R = \frac{u^2 sin2\theta}{g}[/tex],

where u = initial speed of the projectile = 230 m/s

θ = angle of projection = 60°

g = acceleration due to gravity = 9.8 m/s²

Using these in formulas, we get:

[tex]R = \frac{(230 \hspace{0.8mm} m/s)^2 \hspace{0.8mm} sin2(60)}{9.8 \hspace{0.8mm} m/s^2}[/tex]

or, R = 4674.77 m ≈ 4675 m

(b) The maximum height reached by the projectile is given by the formula:

[tex]H = \frac{u^2 \hspace{0.8mm} sin^2\theta}{2g}[/tex]

using the numeric values, we get:

[tex]H = \frac{(230 \hspace{0.8mm}m/s)^2 sin^2(60)}{2 \times (9.8 \hspace{0.8mm} m/s^2)}[/tex]

or, H = 2024.23 m

However, since the projectile was fired 100 m above the ground, hence, the maximum height would be:

h = 100 m + H = 100 m + 2024.23 m = 2124.23 m ≈ 2124 m

(c) The speed of the projectile comprises of two parts, a horizontal velocity and a vertical velocity.

The horizontal velocity is given as [tex]v_x[/tex]. It is constant throught the motion and its magnitude can be determined by the formula:

[tex]v_x = v cos\theta = 230 \hspace{0.8 mm} m/s \times cos(60)[/tex]

or, [tex]v_x = 115 \hspace{0.8mm} m/s[/tex]

The vertical velocity is given as [tex]v_y[/tex]. It is not constant throught the motion. At the highest point, the magnitude of vertical velocity is zero. Using the kinetic equation,

[tex]v_y^2[/tex] = u² + 2gs, where s = h = 2124 m, we get:

[tex]v_y^2[/tex] = 0 + (2 × 9.8 m/s² × 2124 m)

or, [tex]v_y^2[/tex] = 41630.4 m²/s²

or, [tex]v_y[/tex] = 204.03 m/s ≈ 204 m/s

Now, the magnitude of speed will be equal to the resultant of both the horzontal and vertical velocities.

[tex]v = \sqrt{v_x^2 + v_y^2}[/tex]

or, [tex]v = \sqrt{204^2 + 115^2} \hspace{0.8 mm} m/s[/tex]

or, v = 234.21 m/s ≈ 234 m/s

Answer:

Therefore,

Current required is , I

[tex]I = 0.0289\ Ampere[/tex]

Explanation:

Given:

Turns = N = 1140

length of solenoid = l = 0.415 m

Magnetic Field,

[tex]B = 1.00\times 10^{-4}\ T[/tex]

To Find:

Current , I = ?

Solution:

If N is the number of turns in the length, the total current through the rectangle is NI. Therefore, Ampere’s law applied to this path gives

[tex]\int {B} \, ds= Bl=\mu_{0}NI[/tex]

Where,  

B = Strength of magnetic field

l = Length of solenoid

N = Number of turns

I = Current

[tex]\mu_{0}=Permeability\ in\ free\ space=4\pi\times 10^{-7}\ Tm/A[/tex]

Therefore,

[tex]I =\dfrac{Bl}{\mu_{0}N}[/tex]

Substituting the values we get

[tex]I =\dfrac{1.00\times 10^{-4}\times 0.415}{4\times 3.14\times 10^{-7}\times 1140}=0.0289\ Ampere[/tex]

Therefore,

Current required is , I

[tex]I = 0.0289\ Ampere[/tex]

Answer:

2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.

Explanation:

Microwaves are a form of electromagnetic radiation. Most people are familiar with this type of waves because they are used in microwave ovens. When compared to visible light, microwaves have lower frequency, same speed and longer wavelength than visible light. The prefix "micro" is used to indicate that microwaves are smaller (shorter wavelengths) than radio waves.

"The correct option is 2. Microwaves have lower frequency, same speed, and longer wavelength than visible light.

To understand why this option is correct, let's consider the relationship between frequency, wavelength, and speed for electromagnetic waves, which is given by the equation:

[tex]\[ c = f \times \lambda \][/tex]

 1. Speed of light (in a vacuum) is constant for all electromagnetic waves, including microwaves and visible light. Therefore, the speed of microwaves and visible light is the same.

2. Microwaves have a lower frequency than visible light. The frequency of microwaves typically ranges from 300 MHz to 300 GHz, while the frequency of visible light ranges from approximately 430 THz to 750 THz. Since microwaves have a lower frequency, they also have a longer wavelength according to the equation [tex]\( c = f \times \lambda \)[/tex].

3. Since the speed of all electromagnetic waves is the same in a vacuum, and microwaves have a lower frequency, they must have a longer wavelength to maintain the constant speed. This is consistent with the relationship , where a lower frequency  requires a longer wavelength  to keep the speed constant.

In summary, microwaves have a lower frequency and longer wavelength than visible light, but they travel at the same speed. This makes option 2 the correct choice."

Answer:

W = N!/(n0! * n1!)

Explanation:

Let n0 = number of particles in the lowest energy state

n1 = number of particles in the excited energy state.

Using this, we can say that N = n0 + n1

From this we can then express the weight, W of the close system by finding the factorials of each particles

W = N!/(n0! * n1!)

Hence, the weight W is expressed as W = N!/(n0! * n1!)

Answer:

a. 55.6v

b. [tex]1.61*10^{-7}J[/tex]

Explanation:

Data given

charge 1=+4.10nC

charge 2 =-2.40nC

distance, r= 55cm =0.55m

a. the electric potential at the mid point is the sum of the potential due to individual charge.

The electric potential is expressed as  

[tex]V=\frac{kq}{r}\\[/tex]

since we are interested in the electric potential at the mid point, we have

[tex]V=\frac{kq_1}{r/2}-\frac{kq_2}{r/2}\\ V=\frac{2k}{r}(q_1-q_2)\\ V=\frac{2*9*10^9}{0.55}(4.1-2.4)*10^{-9}\\ V=55.6v[/tex]

Hence the electric potential at the mid-point is 55.6v

b. to calculate the potential energy, we use the formula below

[tex]U=\frac{kq_1q_2}{r} \\U=\frac{9*10^9 *4.10*10^{-9}*2.4*10^{-9}}{0.55}\\ U=1.61*10^{-7}J[/tex]  

Explanation:

The given data is as follows.

       Speed of the bullet (v) = 500 m/s

     Distance during one RC time constant (d) = 1 mm = [tex]1 \times 10^{-3} m[/tex]

      Capacitance (C) = 600 [tex]\mu F[/tex] = [tex]600 \times 10^{-6} F[/tex]

Hence, formula for speed of the bullet is as follows.

                v = [tex]\frac{d}{t}[/tex]

or,           t = [tex]\frac{d}{v}[/tex]

Time constant for RC circuit is as follows.

          t = RC

        R = [tex]\frac{t}{C}[/tex]

            = [tex]\frac{d}{vC}[/tex]

            = [tex]\frac{1 \times 10^{-3}}{500 \times 600 \times 10^{-6}}[/tex]

            = [tex]3.33 \times 10^{-3} ohm[/tex]

Thus, we can conclude that resistance in the flash tube is [tex]3.33 \times 10^{-3} ohm[/tex].

Answer:

The magnetic flux links to its turns = [tex]7.6 \times10^{-3}[/tex] Wb.

Explanation:

Given :

Radius of circular coil = [tex]18 \times 10^{-2}[/tex] m

Number of turns = 25

Magnetic field = [tex]3 \times10^{-3}[/tex] T

Magnetic flux (Φ) is a measure of the magnetic field lines passes through a given area. The unit of magnetic flux is weber (Wb).

We know that,

⇒    Φ = [tex]BA[/tex]

Where [tex]B =[/tex] ext. magnetic field, [tex]A =[/tex] area of loop or coil.

But here given in question, we have turns of wire so our above eq. modified as follows.

⇒   Φ = [tex]NBA[/tex]

Where [tex]N =[/tex] no. of turns.

∴    Φ = [tex]25 \times 3 \times 10^{-3} \pi (18 \times10^{-2} )^{2}[/tex]

     Φ = [tex]7.6 \times 10^{-3} Wb[/tex]

Thus, the magnetic flux links to its turns = [tex]7.6 \times 10^{-3} Wb[/tex]

Answer:

b. Both stars will have the same shift.

Explanation:

It's a very simple problem to solve. Star 1 is approaching toward Earth with a speed v, so let's assume that the change in Doppler Shift is +F and Star 2 is moving away so the change in Doppler shift is -F. But it's time to notice the speed of both stars and that is same but only directions are different. speed is the main factor here. The magnitude of both shifts is F as we can see and + and - are showing there direction of motion. So, because of same amount of speed, both stars will have same shift magnitude. (Just the directions are different)

Answer:

Option (b)

Explanation:

As the actor swings on a steel cable, the cable stretches due to the weight of the actor. So, the stress produced by the weight of the actor is longitudinal stress and the strain is also longitudinal in nature.

The modulus of elasticity associated to the normal stress and the longitudinal strain is young's modulus.

Answer: 1.176×10^-3 s

Explanation: The time constant formulae for an RC circuit is given below as

t =RC

Where t = time constant , R = magnitude of resistance = 21 ohms , C = capacitance of capacitor = 56 uf = 56×10^-6 F

t = 56×10^-6 × 21

t = 1176×10^-6

t = 1.176×10^-3 s

Given Information:

Internal resistance of battery = 0.460

Resistance = 21.0 Ω

Capacitance = 56.0 µF

Required Information:

time constant = τ = ?

Answer:

τ = 0.0012

Explanation:

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

τ = Req*C

Where Req is the equivalent resistance and C is the capacitance.

Req = R + r

Where R is the resistance of the resistor and r is the internal resistance of the battery.

Req = 21.0 + 0.460 = 21.460 Ω

τ = Req*C

τ = 21.46*56x10⁻⁶

τ = 0.0012

A capacitor approximately charges to 63% in one τ and about 99% in 5τ

Answer:

Length of the sides of the square loop is given by

s = √[(τ)/(NIB sin θ)]

Explanation:

The torque, τ, experienced by a square loop of area, A, with N number of turns around the loop and current of I flowing in the wire, with a magnetic field presence, B, and the plane of the loop tilted at angle θ to the x-axis, is given by

τ = (N)(I)(A)(B) sin θ

If everything else is given, the length of a side of the square loop, s, can be obtained from its Area, A.

A = s²

τ = (N)(I)(A)(B) sin θ

A = (τ)/(NIB sin θ)

s² = (τ)/(NIB sin θ)

s = √[(τ)/(NIB sin θ)]

In this question, τ = 0.076 N.m, I = 1.70 A

But we still need the following to obtain a numerical value for the length of a side of the square loop.

N = number of turnsof wire around the loop

B = magnetic field strength

θ = angle to which the plane of the loop is tilted, measured with respect to the x-axis.

The question is incomplete! The complete question along with answer and explanation is provided below

Question:

A wire loop with 40 turns is formed into a square with sides of length s. The loop is in the presence of a 2.0 T uniform magnetic field B that points in the negative y direction. The plane of the loop is tilted off the x-axis by 15°. If 1.70 A of current flows through the loop and the loop experiences a torque of magnitude 0.0760 N.m, what are the lengths of the sides s of the square loop, in centimeters?

Given Information:

Number of turns = N = 40 turns

Torque = τ = 0.0760 N.m

Current = I = 1.70 A

Magnetic field = B =  2 T

θ = 15°

Required Information:

Length of the sides of square loop = s = ?

Answer:

s = 4.64 cm

Explanation:

τ = NIABsin(θ)

Where N is the number of turns, I is the current flowing through the square loop, A is the area of square loop, B is the magnetic field, and θ is the angle between square loop and magnetic field strength with respect to the x-axis.

Re-arranging the equation to find out A

A = τ/ NIBsin(θ)

A = 0.0760/40*1.70*2*sin(15)

A = 0.00215 m²

We know that area of a square is

A = s²

Taking square root on both sides yields

s = √A

s = √0.00215 = 0.0464 m

in centimeters

s = 4.64 cm

The fundamental frequency of the string will remain at 250 Hz, even when the string is doubled in length and the tension is quadrupled, due to the relationship between string length, tension, and frequency.

When a transverse standing wave on a string is vibrating at its fundamental frequency, the length of the string and the tension within it play crucial roles in determining that frequency. The fundamental frequency (f) is given by f = (1/2L) * sqrt(T/μ), where L is the length of the string, T is the tension, and μ is the linear mass density of the string.

In the original scenario, the string vibrates at a fundamental frequency of 250 Hz. If the same piece of string is made twice as long and the tension is increased fourfold, the new fundamental frequency F' can be found using the formula mentioned above.

Considering the string is now twice as long (L' = 2L), and the tension is four times greater (T' = 4T), we can substitute these into the original equation for fundamental frequency: f' = (1/2L') * sqrt(T'/μ). By inserting these new values, we can determine that the new fundamental frequency f' will be the same as the original frequency f, because the doubling of the string length and the quadrupling of the tension will have cancelling effects.

Final answer:

The fundamental frequency of a string that is twice as long and with four times the tension would be half the original frequency; therefore, the new fundamental frequency would be 125 Hz.

Explanation:

The student is asking about how the fundamental frequency of a string changes when its length is doubled and the tension is increased by four times. The fundamental frequency f of a string fixed at both ends can be described by the equation f = √(T/μ)/(2L), where T is the tension, μ (mu) is the mass per unit length, and L is the length of the string. When we double the length of the string, the new length is 2L. When we quadruple the tension, the new tension is 4T. Substituting these values into the equation, the new frequency f' becomes f' = √(4T/μ)/(2 × 2L) = √(T/μ)/2L = f/2.

Therefore, the new fundamental frequency of the longer string with four times the tension is half the original frequency, which is 125 Hz in this case.

Answer:

Explanation: see attachment below

Final answer:

The magnitude of force on the wire segment along the y-axis is 3.0N and directed along the negative x-axis, while there's no force on the segment along the x-axis.

Explanation:

To find the magnitude of the force on the wire, we can use the formula F = I * (L x B), where F is the magnetic force, I is the current, L is the length vector of the wire segment, and B is the magnetic field. The force will be perpendicular to both the current direction and the magnetic field direction, according to the right-hand rule.

The wire is in two segments, one along the x-axis and one along the y-axis. First, we calculate the force on the section of the wire along the x-axis: Fx = I * Lx * B, which is 0 since the magnetic field is in the z-direction and Lx is in the x-direction, making the angle between them 90 degrees. So, there's no force for this segment.

Second, we calculate the force on the section of the wire along the y-axis: Fy = I * Ly * B. The force experienced by this segment can be calculated using the formula with the values provided (I=25A, Ly=3.0m, B=40mT). Therefore, Fy = 25A * 3.0m * 40mT = 3000mN or 3.0N, which will be directed along the negative x-axis, under the right-hand rule.

Answer:

MA = 178 Nm

MB = 72 Nm

MR = 249 Nm

For the moments MA the lever arm was taken as r1 + r2 which is the distance of the point of application of the force at point A to point D.

For the moment MB, the distance of the point of application of force is r1 - r2 which is the distance from the outer ring to the inner ring.

The couple moment is given by F × r1. Which is basically a sum of the moments of both forces applied on the wheel.

Explanation:

See the attachment for detail of the calculation.

The moment MA for the force at point A is 124.55 N*m, the moment MB for the force at point B is 53 N*m, and the resulting couple moment MR is 71.55 N*m.

To calculate the moments MA and MB about point D for the forces applied at points A and B, we can use the formula:

M = F * r

Where M is the moment, F is the force, and r is the radius. For point A, the radius is r1 = 0.470 m and the force is F = 265 N. So, MA = 265 N * 0.470 m = 124.55 N*m (counterclockwise).

Similarly, for point B, the radius is r2 = 0.200 m and the force is F = 265 N. So, MB = 265 N * 0.200 m = 53 N*m (clockwise).

To determine the resulting couple moment MR, we can subtract the clockwise moment (MB) from the counterclockwise moment (MA). MR = MA - MB = 124.55 N*m - 53 N*m = 71.55 N*m (counterclockwise).

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Answer:

Width of the slit will be equal to 1.47 mm

Explanation:

We have given wavelength of the light [tex]\lambda =550nm=550\times 10^{-9}m[/tex]

Distance D = 8 m

Distance between first minimum dark fringe and the central maximum is 2 mm

So [tex]x=3\times 10^{-3}m[/tex]

We have to find the width of the slit

For the first order wavelength is equal to [tex]\lambda =\frac{x}{D}\times a[/tex], here a width of slit

So [tex]a=\frac{\lambda D}{x}=\frac{550\times 10^{-9}\times 8}{3\times 10^{-3}}=1466.666\times 10^{-6}=1.47mm[/tex]

So width of the slit will be equal to 1.47 mm

Answer:

Explanation:

velocity of electron V = 4 x 10³ m/s

magnetic field B = 1.25 T .

magnetic force = 1.4 x 10⁻¹⁶ N.

If direction of velocity makes angle θ with magnetic field

magnetic force = magnetic field x charge on electron x velocity x sinθ

1.4 x 10⁻¹⁶ = 1.25 x 1.6 x 10⁻¹⁹ x 4 x 10³ x sinθ

1.4 x 10⁻¹⁶ = 8  x 10⁻¹⁶ x sinθ

sinθ  = 1.4 / 8

= .175

θ = 10 degree

or 180 - 10

= 170 degree

because

sinθ = sin (180 - θ)

Answer:

Check Explanation.

Explanation:

Momentum before collision = (2)(2) + (2)(0) = 4 kgm/s

a) Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Momentum after collision = (sum of the masses) × (common velocity) = (2+2) × (1) = 4 kgm/s

Which is equal to the momentum before collision, hence, momentum is conserved.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Momentum after collision = (2)(0.5) + (2)(1.5) = 1 + 3 = 4.0 kgm/s

This is equal to the momentum before collision too, hence, momentum is conserved.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Momentum after collision = (2)(0) + (2)(2) = 0 + 4 = 4.0 kgm/s

This is equal to the momentum before collision, hence, momentum is conserved.

b) Kinetic energy is normally conserved in a perfectly elastic collision, if the two bodies do not stick together after collision and kinetic energy isn't still conserved, then the collision is termed partially inelastic.

Kinetic energy before collision = (1/2)(2.00)(2.00²) + (1/2)(2)(0²) = 4.00 J.

Scenario A

After collision, Mass A sticks to Mass B and they move off with a velocity of 1 m/s

Kinetic energy after collision = (1/2)(2+2)(1²) = 2.0 J

Kinetic energy lost = (kinetic energy before collision) - (kinetic energy after collision) = 4 - 2 = 2.00 J

Kinetic energy after collision isn't equal to kinetic energy before collision. This collision is evidently totally inelastic.

Scenario B

They bounce off of each other and move off in the same direction, mass A moves with a speed of 0.5 m/s and mass B moves with a speed of 1.5 m/s

Kinetic energy after collision = (1/2)(2)(0.5²) + (1/2)(2)(1.5²) = 0.25 + 3.75 = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

Scenario C

Mass A comes to rest after collision and mass B moves off with a speed of 2 m/s

Kinetic energy after collision = (1/2)(2)(0²) + (1/2)(2)(2²) = 4.0 J

Kinetic energy lost = 4 - 4 = 0 J

Kinetic energy after collision is equal to kinetic energy before collision. Hence, this collision is evidently elastic.

c) An impossible outcome of such a collision is that A stocks to B and they both move off together at 1.414 m/s.

In this scenario,

Kinetic energy after collision = (1/2)(2+2)(1.414²) = 4.0 J

This kinetic energy after collision is equal to the kinetic energy before collision and this satisfies the conservation of kinetic energy.

But the collision isn't possible because, the momentum after collision isn't equal to the momentum before collision.

Momentum after collision = (2+2)(1.414) = 5.656 kgm/s

which is not equal to the 4.0 kgm/s obtained before collision.

This is an impossible result because in all types of collision or explosion, the second law explains that first of all, the momentum is always conserved. And this evidently violates the rule. Hence, it is not possible.

Answer:

Explanation:

The system can be modeled as,

Using magnitude relationship for imbalance system.

Check attachment for solution

Final answer:

The maximum angle of inclination θ for which a block will remain stationary on an inclined plane, given a coefficient of static friction of 0.4, can be found using the arctangent function resulting in approximately 21.8 degrees.

Explanation:

To find the maximum angle of inclination θ at which a block will remain stationary on an inclined surface, we can relate the coefficient of static friction (0.4 in this case) to the angle using the tangent function. When static friction has reached its maximum value, the block is on the verge of sliding, and the maximum force of static friction is equal to the component of the block's weight parallel to the incline. This force can be described by the equation μsN = mg sin(θ), where μs is the static friction coefficient, N is the normal force, m is the mass of the block, g is the acceleration due to gravity, and θ is the angle of inclination. Since the normal force is equal to mg cos(θ), substituting this into the equation and solving for θ gives us θ = tan-1(μs). Plugging in the given coefficient of static friction, the maximum angle θ can be found.

Using the given coefficient of 0.4, we can calculate the angle: θ = tan-1(0.4). Therefore, the maximum angle θ, for the block to remain stationary, is approximately 21.8 degrees.

Final answer:

The maximum angle of inclination θ at which a block will remain stationary on an inclined plane with a coefficient of static friction of 0.4 is approximately 21.8 degrees.

Explanation:

The maximum angle of inclination θ at which a block will remain stationary on an inclined plane with a coefficient of static friction of 0.4 is determined by using the relationship between the angle of inclination and the coefficient of static friction. This relationship is given by the equation tan(θ) = μ, where μ is the coefficient of static friction. To find the angle where the block starts to slide, we take the inverse tangent (arctan or tan-1) of the coefficient of static friction. Therefore, the maximum angle θ is tan-1(0.4).

Calculating this, we get θ = tan-1(0.4) ≈ 21.8°. Hence, for angles of inclination less than or equal to 21.8 degrees, the block will not slide down the incline due to static friction.

Given Information:

frequency = 1240 Hz

width = a = 1.13 m

speed of sound = c = 344 m/s

Required Information:

angle = θ = ?

Answer:

θ = 14.18 rad

Explanation:

We can find out the angle relative to the centerline perpendicular to the doorway by using the following relation

sin(θ) = λ/a

Where λ is the wavelength of the sound wave and a is width

λ = c/f

Where c is the speed of the sound and f is the frequency

λ = 344/1240

λ = 0.277

sin(θ) = λ/a

θ =sin⁻¹(λ/a)

θ =sin⁻¹(0.277/1.13)

θ =sin⁻¹(0.277/1.13)

θ = 14.18 rad

Explanation:

Since, it is given that the magnet drops and falls lengthwise towards the canter of the ring. As a result, change in magnetic flux will occur which tends to induce an electric current in the ring.

Therefore, a magnetic field is also produced by the ring itself which will actually oppose or repel the magnet.  

Thus, we can conclude that the falling magnet be repelled by the ring due to the magnetic interaction of the magnet and the ring.

Given Information:

Velocity of water flow = 2 m/s

Temperature of water = 17° C

Heat dissipation = 2500 W

Area of copper plate = 0.04 m²

Required Information:

Temperature of copper plate  = ?

Answer:

[tex]T_{p} = 27[/tex]° [tex]C[/tex]

Explanation:

Each chip dissipates 25 W so 100 chips will dissipate 25*100 = 2500 W

Area of copper plate = 0.2*0.2 = 0.04 m²

According to the convection rate equation

[tex]T_{p}= T_{w} + \frac{q}{hA}[/tex]

Where Tp is the temperature of copper plate, Tw is the temperature of water, q is the the heat dissipation of chips, A is the area of copper plate and h is the convection coefficient

The convection coefficient is given by turbulent flow correlation

[tex]h = Nu_{L}(k/L) =0.037Re_{L}^{4/5}P_{r}^{1/3}(k/L)[/tex]

Where Nu is Nusselt number, Re is Reynolds number, Pr = 5.2 is Prandtl number and k = 0.620 W/m.K

[tex]Re_{L}= uL/v[/tex]

Where u = 2 m/s and L = 0.2 m and  v = 0.96x10⁻⁶m² /s

[tex]Re_{L}= 2*0.2/0.96x10^{-6}[/tex]

[tex]Re_{L}= 416666.66[/tex]

[tex]h = 0.037(416666.66)^{4/5}(5.2)^{1/3}(0.620/0.2)[/tex]

[tex]h = 6223.89[/tex] [tex]W/m^{2}K[/tex]

[tex]T_{p}= 17 + \frac{2500}{(6223.89)0.04}[/tex]

[tex]T_{p} = 27[/tex]° [tex]C[/tex]

Therefore, the temperature of the copper plate is 27° C

Answer:

the answer the correct one is B

Explanation:

The interference phenomenon is described by

           d sin θ = m λ

The spectrum is recorded on a screen, so we can use trigonometry

          tan θ = y / L

In this experiment the angles are very small, so

           tan θ = sin θ /cos θ = sin θ

  We replace

         d y / L = m La m

        y = m Lam L / d

When the slits approach the screen the value of L decreases,

Therefore the value of the separation between the slits must also decrease

When reviewing the answer the correct one is B

Answer:

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

Explanation:

The electric field due to electric potential at a distance Δs is given by

E=ΔV/Δs

We have to find the magnitude electric field in the membrane

Ecell= -ΔV/Δs

[tex]E_{cell}=-\frac{V_{in}-V_{out}}{s} \\E_{cell}=\frac{V_{out}-V_{in}}{s}\\E_{cell}=\frac{0.070V}{8*10^{-9}m } \\E_{cell}=8.8*10^{6}V/m[/tex]

The magnitude of the electric field inside the membrane is 8.8×10⁶V/m

The magnitude of the electric field inside the membrane, if it were empty, would be approximately [tex]\( 8.75 \times 10^6 \) V/m.[/tex]

To find the magnitude of the electric field inside the membrane, we can use the formula for electric field strength ( E ) due to a uniform field between two parallel plates, which is given by:

[tex]\[ E = \frac{V}{d} \][/tex]

where ( V ) is the potential difference across the plates (in volts) and ( d ) is the separation between the plates (in meters). In this case, the potential difference ( V ) is given as 70 mV, which we need to convert to volts:

[tex]\[ V = 70 \text{ mV} = 70 \times 10^{-3} \text{ V} \][/tex]

The thickness of the membrane ( d ) is given as 8 nm, which we need to convert to meters:

[tex]\[ d = 8 \text{ nm} = 8 \times 10^{-9} \text{ m} \][/tex]

Now we can plug these values into the formula for the electric field:

[tex]\[ E = \frac{70 \times 10^{-3} \text{ V}}{8 \times 10^{-9} \text{ m}} \][/tex]

[tex]\[ E = \frac{70}{8} \times 10^{(3 - (-9))} \text{ V/m} \][/tex]

[tex]\[ E = 8.75 \times 10^{6} \text{ V/m} \][/tex]