After fixing a flat tire on a bicycle you give the wheel a spin. If its initial angular speed was 6.36 rad/s and it rotated 14.7 revolutions before coming to rest, what was its average angular acceleration (assuming that the angular acceleration is constant)

Answer:

If the radius of the disk is much greater than the point where the electric field is calculated, then the field of the disk matches the field of the infinite sheet.

Explanation:

First, we have to calculate the electric field of the disk.

We should choose an infinitesimal area, 'da', on the disk and calculate the E-field of this small portion, 'dE'. Then we will integrate dE over the entire disk using cylindrical coordinates.

According to the cylindrical coordinates: da = rdrdθ

The small portion is chosen at a distance r from the axis. Let's find the dE at a point on the axis and a distance z from the center of the disk.

[tex]dE = \frac{1}{4\pi\epsilon_0}\frac{dQ}{r^2 + z^2}[/tex]

Here dQ can be found by the following relation: The charge density of the disk is equal to the total charge divided by the total area of the disk. The small portion of the disk will have the same charge density, therefore:

[tex]\frac{Q}{\pi R^2} = \frac{dQ}{da}\\dQ = \frac{Qda}{\pi R^2}[/tex]

Furthermore, we need to separate the vertical and horizontal components of dE, because it is a vector and cannot be integrated without separating the components. By symmetry, the horizontal components of dE will cancel out each other, leaving only the vertical components in the z-direction.

[tex]dE_z = dE\sin(\alpha) = dE \frac{z}{\sqrt{z^2+r^2}}\\dE_z = \frac{1}{4\pi\epsilon_0}\frac{Qzda}{\pi R^2(z^2+r^2)^{3/2}} = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\frac{rdrd\theta}{(z^2+r^2)^{3/2}}[/tex]

We have to use a double integral over the radius and the angle to find the total electric field due to a uniformly charged disk:

[tex]E = \int \int dE = \frac{1}{4\pi\epsilon_0}\frac{Qz}{\pi R^2}\int\limits^{2\pi}_0 {\int\limits^R_0 {\frac{1}{(z^2+r^2)^{3/2}}} \, rdr} \, d\theta\\E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[1 - \frac{1}{\sqrt{(R^2/z^2) + 1}}][/tex]

If the radius of the disk is much greater than the point z, R >> z, than the term in the denominator becomes very large, and the fraction becomes zero. In that case electric field becomes

[tex]E = \frac{1}{2\epsilon_0}\frac{Q}{\pi R^2}[/tex]

This is equal to the electric field of an infinite sheet.

As a result, the condition for the field of a disk to be equal to that of a infinite sheet is R >> z.

Answer:

Please see below as the answer is self - explanatory

Explanation:

a) 10³ = kilo (kilogram = 10³ grams, kilometer= 10³ meters)

Symbol : k.

b) 10⁻² = centi (it is the 100th part of a unit, like centimeter, or centigram) Symbol:  c.

c) .1 = deci (it is the tenth part (deci comes from the word that means "ten"in latin) of a unit: decimeter, decigram)

Symbol: d.

d) 10⁻3 = mili (it is a 1000th part of a unit: milimeter, miligram), the name comes from the word used in latin to mean "one thousand".

Symbol: m

e) 1,000,000 = 10⁶ = mega (megawatt)

Symbol: M

f= 0.000001 = 10⁻6 = micro (micrometer, microsecond),

Symbol: μ.

Answer:

wave speed= constant

frequency = increase

wavelength = decrease

Explanation:

Solution:

- The three basic parameters of a wave are speed, frequency and wavelength. These three parameters are related to each other by an expression:

                                             v = f * λ

Where,

- v is the speed of the wave in m/s.

- f frequency of the wave in Hz.

- λ wavelength of the wave in m

- We are asked how would each of these parameter change if we move the hand up and down faster. The hand moves from a crest to trough faster than before and back again. We can see that the time between a cycle has decreased; hence, frequency f increases. Consequently, we can see that wave speed v remains constant - the medium of transfer of wave energy - remains same. Then from our relation above if we hold speed constant and increase f then the wavelength λ would have to decrease.

The question deals with the calculation of the magnitude and direction of an electric field necessary to keep a charged object motionless. The two forces acting on the object, namely the gravitational force and the electric force, cancel out making it motionless. The electric field direction is upward as it must counteract the gravitational pull.

In this question, we're examining an object that stays motionless in a uniform electric field. This can be resolved using the equilibrium of forces acting on the object. Given that the object stays motionless, the gravitational force and the electric force on the object should balance each other.

The gravitational force (Fg) experienced by the object is the object mass (m) times the acceleration due to gravity (g), which equals 3.82g * 9.81 m/s². The electric force (Fe) is equal to the charge (q) times the electric field (E), which equals -16.5µC * E.

To find the electric field E, we equate these forces - this gives us

E = Fg / |q|,

where |q| means the absolute value of the charge. The direction of the electric field is taken as the direction of the force that a positive test charge would experience.

Thus, the electric field direction is upwards since the force needed to balance gravity must act against it.

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Answer:

 Q = 47.06 degrees

Explanation:

Given:

- The transmitted intensity I = 0.464 I_o

- Incident Intensity I = I_o

Find:

What angle should the principle axis make with respect to the incident polarization

Solution:

- The relation of transmitted Intensity I to to the incident intensity I_o on a plane paper with its principle axis is given by:

                                     I = I_o * cos^2 (Q)

- Where Q is the angle between the Incident polarized Light and its angle with the principle axis. Hence, Using the relation given above:

                                     Q = cos ^-1 (sqrt (I / I_o))

- Plug the values in:

                                     Q = cos^-1 ( sqrt (0.464))

                                     Q = cos^-1 (0.6811754546)

                                     Q = 47.06 degrees

Answer:

The same

Explanation:

Charges of the same sign repel, while those of different sign attract. So, the magnitude of both electrostatic forces is the same but in the opposite direction. On the other hand, when the force on the charge is exerted by an electric field: If the charge is positive, it experiences a force in the direction of the field; If the load is negative, it experiences a force in the opposite direction to the field. Therefore, the magnitude of both forces is the same but in the opposite direction.

Final answer:

The magnitude of the force is the same on both a negative and positive charge due to Coulomb's Law, but the forces act in opposite directions with attractions between opposite charges and repulsions between like charges.

Explanation:

The magnitude of the force experienced by a negative charge is the same as that experienced by a positive charge when they are acting upon each other. This is because the electric force between two charged particles is dictated by Coulomb's Law, which states that the magnitude of the force between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This law can be summarized by the equation:

F = k * |q₁ * q₂| / r²

where F is the magnitude of the force, k is Coulomb's constant, q₁ and q₂ are the amounts of the charges and r is the distance between the charges. Importantly, the law indicates only that the magnitudes of the forces are equal; however, the directions will be opposite due to the nature of attraction and repulsion between the charges. Thus, negative and positive charges attract each other, while like charges repel each other.

The force on the negatively charged object, using the formula F = qE, where q is the charge and E is the electric field, will be equal in magnitude but opposite in direction to the force on the positively charged object assuming the charges have the same magnitude. For example, if the electric field is directed eastward, a negative charge will experience a force to the west, while a positive charge will experience a force to the east.

Because of their different masses, a proton and an electron will experience different accelerations due to their different inertia, even though the forces acting on them are of the same magnitude.

The electrostatic force between the two charges is [tex]3.4\cdot 10^{-8}N[/tex]

Explanation:

The electrostatic force between two charges is given by Coulomb's law:

[tex]F=k\frac{q_1 q_2}{r^2}[/tex]

where:

[tex]k=8.99\cdot 10^9 Nm^{-2}C^{-2}[/tex] is the Coulomb's constant

[tex]q_1, q_2[/tex] are the two charges

r is the separation between the two charges

In this problem, we have the following data:

[tex]q_1 = q_2 = 1.00e[/tex] is the magnitude of the two charges, where

[tex]e=1.6\cdot 10^{-19}C[/tex] is the fundamental charge

[tex]r=8.19\cdot 10^{-11}m[/tex] is the separation between the two charges

Substutiting into the equation, we find the force:

[tex]F=(8.99\cdot 10^9)\frac{(1.00\cdot 1.6\cdot 10^{-19})^2}{(8.19\cdot 10^{-11})^2}=3.4\cdot 10^{-8}N[/tex]

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The charge on the dust particle is [tex]-2.07 x 10^{-14} C.[/tex]

The dust particle will obtain a charge due to the electric field produced by the charged carpet. Be that as it may, calculating its correct charge requires a few presumptions and steps:

1. Charge density:

To begin with, we ought to calculate the charge density [tex]\sigma[/tex] of the carpet:

[tex]\sigma[/tex] = total Charge / Range = -10 μC / (2.0 m x 4.0 m) = -2.5  μC/m²

2. Electric Field:

The charge thickness creates an electric field (E) over the carpet. Ready to utilize the equation:

E = [tex]\sigma[/tex] / ϵ0

here,  ϵ0 is the permittivity of free space which is equal to [tex]8.85 x 10^{-12[/tex]F/m

Electric field, E = (-2.5 μC/m² / [tex]8.85 x 10^{-12} F/m[/tex]) = ([tex]-2.83 x 10^{5} N/C[/tex])

3. dust particle Charge:

The dust particle will involve an electrostatic force due to the electric field. Since the molecule is suspended, the net force on it must be zero. This implies the electrostatic force must balance the gravitational force acting on the molecule.

Suspicions:

The dust particle could be a circle with uniform charge dissemination.

Discussing resistance is unimportant.

Calculations:

Tidy molecule mass (m): 6.0 μg = [tex]6.0 x 10^{-9} kg[/tex]

Gravitational force (Fg): Fg = m * g (where g is increasing speed due to gravity,= 9.81 m/s²)

Electrostatic force (Fe): Fe = q * E (where q is the charge of the dust particle)

Likening the powers:

Fg = Fe

m * g = q * E

Tackling for q:

q = Fg / E = (m * g) / E = [tex](6.0 x 10^{-9} kg[/tex] * 9.81 m/s²) / ([tex]-2.83 x 10^{5} N/C[/tex]) = [tex]-2.07 x 10^{-14}[/tex] C

Subsequently, the charge on the dust particle is[tex]-2.07 x 10^{-14}[/tex] C.

Answer:

Work Done by the earth's gravitational force on the satellite as it travels from apogee to perigee is

W = F*D*Cos90° = 0

Explanation:

Although there is a change in the kinetic energy of the satellite at the apogee and perigee, the work done by the earth's gravitational force on the satellite is Zero.

W = F.D, F is the gravitational force, D is the displacement. Both F and D are vectors and perpendicular to each other. That is, the angle between F and D is 90°.

Answer:

0.08

Explanation:

The alpha particle suffers a head-on collision with the gold nucleus, so it retraces it path after the collision.

Let us take the masses of the particles in atomic mass units.

The initial momentum and kinetic energy of the gold nucleus is 0(since it is stationary). So, applying conservation of momentum and energy, we get the following two equations:

[tex]m_{1}u_{1}=m_{1}v_{1}+m_{2}v_{2}[/tex]      ..........(1)

[tex]\frac{1}{2}m_{1}u_{2}^{2}=\frac{1}{2} m_{1} v_{1}^{2} +\frac{1}{2} m_{2} v_{2}^{2}[/tex]       ..........(2)

where,

[tex]m_{1}[/tex] = mass of the alpha particle = 4 units

[tex]m_{2}[/tex] = mass of the gold nucleus = 197 units

[tex]u_{1}[/tex] = initial velocity of the alpha particle

[tex]v_{1}[/tex] = final velocity of the alpha particle

[tex]v_{2}[/tex] = final velocity of the gold nucleus

Now, we shall substitute the value of [tex]v_{2}[/tex] from equation (1) in equation (2). After some simplifications, we get,

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

Dividing both sides by [tex]u_1^2[/tex] and substituting [tex]x=\frac{v_1}{u_1}[/tex] and [tex]k=\frac{m_1}{m_2}[/tex] , we get,

[tex]1=x^2+k(1+x^2-2x)\\[/tex]

or, [tex]x^2(k+1)-2kx+(k-1)=0[/tex]

Here, [tex]k=\frac{m_1}{m_2}=\frac{4}{197}=0.02[/tex]

Therefore, [tex]x=\frac{2(0.02)\pm\sqrt{(2\times0.02)^2-(4\times1.02\times-0.98)} }{2\times1.02}[/tex]

or, [tex]x = 1, -0.96[/tex]

Our required solution is -0.96 because the final velocity([tex]v_1[/tex]) of the alpha particle will be a little less the initial velocity([tex]u_1[/tex]). The negative sign comes as the alpha particle reverses it's direction after colliding with the gold nucleus.

Fractional change in kinetic energy is given by,

[tex]\delta E=\frac{\frac{1}{2} m_1u_1^2-\frac{1}{2}m_1v_1^2 }{\frac{1}{2}m_1u_1^2 }=1-x^2=0.078\approx0.08[/tex]

The alpha particle can lose a significant amount of its kinetic energy in a head-on elastic collision with a gold nucleus due to the gold nucleus's much larger mass. The original kinetic energy of the alpha particle is converted to potential energy before being transferred mostly to the gold nucleus. Specific loss would depend upon the alpha particle's original kinetic energy.

The question pertains to the concept of elastic collisions, specifically between an alpha particle and a gold nucleus. In an elastic collision, both momentum and kinetic energy are conserved. However, while total energy is conserved, individual kinetic energies of colliding particles may change. Since the gold nucleus, which was initially at rest, is significantly more massive (197 units) than the alpha particle (4.0 units), the alpha particle can lose a significant amount of its kinetic energy in a head-on collision.

To calculate the fractional loss of kinetic energy for the alpha particle in this instance, we would use the principle of conservation of kinetic energy and momentum. The kinetic energy of an alpha particle before the collision is transformed into both kinetic and potential energy during the collision as it approaches the gold nucleus until its original energy is converted to potential energy.

Upon collision, a good proportion of this energy is transferred to the gold atom, given its much larger mass. However, necessary calculations would require specific knowledge of the kinetic energy of the alpha particle before the collision, which may vary depending upon the specific nuclear decay process involved.

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Final answer:

To raise the boiling point of 2.85 kg of water by 2°C, one needs to add approximately 320.41 grams of sodium chloride (NaCl), calculated based on the boiling point elevation formula and considering NaCl's dissociation into ions.

Explanation:

To calculate how much NaCl is needed to increase the boiling point of 2.85 kg of water by 2°C, we use the boiling point elevation formula: ΔT = i*Kb*m, where ΔT is the change in boiling point, i is the van 't Hoff factor (which is 2 for NaCl because it dissociates into Na+ and Cl- ions), Kb is the ebullioscopic constant of water (0.52 °C kg/mole), and m is the molality of the solution. First, we solve for m knowing that we want to increase the boiling point by 2°C. With Kb = 0.52 °C kg/mole and i=2, we have 2°C = (2)*(0.52 °C kg/mol)*m. From here, m = 1.923 mol/kg.

Next, to find the mass of NaCl needed, we convert molality to moles of solute needed using the mass of the solvent (water) in kg, then multiply by the molar mass of NaCl. Since molality = moles of solute / kg of solvent, moles of NaCl = molality * kg of solvent = 1.923 mol/kg * 2.85 kg = 5.48 moles. The mass of NaCl required = moles * molar mass = 5.48 mol * 58.44 g/mol = 320.41 g.

Therefore, to increase the temperature of the boiling water by 2°C, we need to add approximately 320.41 grams of NaCl.

Answer:

a) [tex]t=5\ s[/tex]

b) [tex]s=61.5\ m[/tex]

Explanation:

Given:

acceleration of the car, [tex]a=-4.92\ m.s^{-1}[/tex]

initial velocity of the car, [tex]u=24.6\ m.s^{-1}[/tex]

final velocity of the car, [tex]v=0\ m.s^{-1}[/tex]

a)

Using eq. of motion:

[tex]v=u+a.t[/tex]

[tex]0=24.6-4.92\times t[/tex]

[tex]t=5\ s[/tex]

b)

Distance travelled before stopping:

[tex]s=u.t+\frac{1}{2} a.t^2[/tex]

[tex]s=24.6\times 5-0.5\times 4.92\times 5^2[/tex]

[tex]s=61.5\ m[/tex]

c)

The car takes deceleration in 5 seconds to stop and travels a distance of 61.5 meters.

a) To find the time it takes for the car to stop, we can use the equation v = u + at, where v is the final velocity, u is the initial velocity, a is the deceleration, and t is the time elapsed. Rearranging the equation to solve for t, we have t = (v - u) / a. Substituting the given values, we get t = (0 - 24.6) / -4.92 = 5 seconds.

b) To find the distance traveled during this time, we can use the equation s = ut + (1/2)at^2, where s is the distance, u is the initial velocity, a is the deceleration, and t is the time elapsed. Substituting the given values, we have s = 24.6(5) + (1/2)(-4.92)(5)^2 = 61.5 meters.

c) The graph of x vs t would be a straight line with a negative slope, representing the car's distance decreasing over time. The graph of v vs t would also be a straight line with a negative slope, representing the car's velocity decreasing over time.

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Answer:

λ = 397 nm

Explanation:

given,

Rydberg wavelength equation for Balmer series

[tex]\dfrac{1}{\lambda}=R(\dfrac{1}{n_f^2}-\dfrac{1}{n_i^2})[/tex]

R is the Rydberg constant, R = 1.097 x 10⁷ m⁻¹

n_i = initial energy level  

n_f = final energy level

where as for Balmer series n_f = 2

            n_i = 7

[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]

[tex]\dfrac{1}{\lambda}=(1.097\times 10^7)(\dfrac{1}{2^2}-\dfrac{1}{7^2})[/tex]

[tex]\dfrac{1}{\lambda}=2.5186\times 10^6[/tex]

[tex]\lambda = 3.97\times 10^{-7}[/tex]

Hence, the wavelength is equal to  λ = 397 nm

Answer: The continuation and the last part of the question is  (3) Determine the expression for E0 by substitution of r0 into the above equation for EN. What is the equation that represents the expression for E0?

Explanation:

The detailed steps and appropriate derivation and by differentiation is shown in the attachment.

The knowledge of differential calculus is applied.

Answer:

hₘₒₒₙ = 6.05 h

Rₘₒₒₙ = 6.05 R

Explanation:

Let θ be the angle of jump.

Let h and R be maximum height and horizontal range attained on earth respectively.

Let hₘₒₒₙ and Rₘₒₒₙ be the maximum height and horizontal range on the moon respectively

The range for a projectile is given as

R = v₀(x)T = v₀ cos(θ) T

T = (2v₀ sinθ)/g

Range, R = (v₀ cos θ)(2v₀ sinθ)/g = v₀²(2sinθcosθ)/g = v₀² (sin2θ)/g

The maximum range occurs at θ = 45°

Maximum range R = v₀²/g = v₀²/9.8 = 0.102v₀²

On the moon, g = 1.62 m/s²

Maximum range, Rₘₒₒₙ = v₀²/gₘₒₒₙ = v₀²/1.62 = 0.617v₀²

Rₘₒₒₙ = 6.05 R

Maximum Height of a projectile is given as = (v₀² Sin²θ)/2g

θ = 45°; sin 45° = (√2)/2; sin²45° = 2/4 = 1/2

h = v₀²(1/2)/2g = v₀²/4g

On earth, g = 9.8 m/s²

h = v₀²/(4×9.8) = v₀²/39.2 = 0.0255v₀²

On the moon, gₘₒₒₙ = 1.62 m/s²

hₘₒₒₙ = v₀²/(4×1.62) = v₀²/6.48 = 0.154v₀²

hₘₒₒₙ = 6.05 h

Answer:

B. +5.75 m/s

Explanation:

When there are two bodies, a and b, whose velocities measured by a third observer (in this case, the ground) are [tex]V_a[/tex] and [tex]V_b[/tex] respectively, the relative velocity of B with respect to A is given by:

[tex]V_{ba}=V_b-V_a[/tex]

Thus, the velocity of the girl relative to the lawnmower is:

[tex]V_{ba}=6.5\frac{m}{s}-0.75\frac{m}{s}\\V_{ba}=5.75\frac{m}{s}[/tex]

I took the test and got B) +5.75 m/s correct

Answer:

6.8 m/s2

Explanation:

Let g = 9.8 m/s2. The total weight of both the rope and the mouse-robot is

W = Mg + mg = 1*9.8 + 2*9.8 = 29.4 N

For the rope to fails, the robot must act a force on the rope with an additional magnitude of 43 - 29.4 = 13.6 N. This force is generated by the robot itself when it's pulling itself up at an acceleration of

a = F/m = 13.6 / 2 = 6.8 m/s2

So the minimum magnitude of the acceleration would be 6.8 m/s2 for the rope to fail

The robot must have a minimum acceleration of 16.6 m/s^2 for the rope to fail.

To determine the minimum magnitude of acceleration the robot should have for the rope to fail, we need to consider the forces acting on the rope. The weight of the robot is 2.0 kg multiplied by the acceleration, which we need to find. The tension in the rope is equal to the weight of the rope plus the weight of the robot. Since the rope breaks if the tension exceeds 43 N, we can set up the equation:

Tension = Weight of rope + Weight of robot
43 N = (1.0 kg)(9.8 m/s^2) + (2.0 kg)(acceleration)

Solving for the acceleration, we get:

acceleration = (43 N - 9.8 N) / 2.0 kg = 16.6 m/s^2

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Answer:

s_y = 9.82 m

Explanation:

Given:

- Initial velocity v_i = 43 m/s

- Angle with the horizontal Q = 25 degree

- Initial distance s_o = 1 m

- The distance of the center field fence x_f = 122 m

Find:

- How high above the ground is the ball when it reaches the center field fence

Solution:

- The time taken for the ball to reach the fence t_f:

                             s_x = S(0) + v_x,o*t

                             122 = 0 + (43*cos(25))*t

                              t = 122 / (43*cos(25)) = 3.1305 s

- Compute the height of the ball when it reaches the fence:

                              s_y = S(0) + v_y,o*t + 0.5*g*t^2

                              s_y = 1 + 43*sin(25)*3.1305 - 0.5*(9.81)*(3.1305)^2

                             s_y = 9.82 m

Answer and Explanation

• Heterotopic protons are those that when substituted by the same substituent, are structurally different. They are not similar, diastereotopic or enantiotopic.

• Diastreotopic protons refers to two protons in a molecule which, if replaced by the same substituent, would generate compounds that are diastereomers. Diastereotopic groups are often, but not always, identical groups attached to the same atom in a molecule containing at least one chiral center.

For example, the two hydrogen atoms of the C3 carbon in (S)-2-bromobutane are diastereotopic (shown in the attached image). Replacement of one hydrogen atom with a bromine atom will produce (2S,3R)-2,3-dibromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the diastereomer (2S,3S)-2,3-dibromobutane.

• Homotopic protons in a compound are equivalent protons. Two protons A and B are homotopic if the molecule remains the same (including stereochemically) when the protons are interchanged with some other atom (substituent) while the remaining parts of the molecule stay fixed. Homotopic atoms are always identical, in any environment.

For example, ethane, the two H atoms on C1 and C2 carbons on the same side (as shown in the attached image) are homotopic as they exhibit the phenomenon described above.

• Enantiotopic protons are two protons in a molecule which, if one or the other were replaced (by the same substituent), would generate a chiral compound. The two possible compounds resulting from that replacement would be enantiomers.

For example, in the attached image to this answer, the two hydrogen atoms attached to the second carbon in butane are enantiotopic. Replacement of one hydrogen atom with a bromine atom will produce (R)-2-bromobutane. Replacement of the other hydrogen atom with a bromine atom will produce the enantiomer (S)-2-bromobutane.

Hope this helps!!!

Final answer:

The charge enclosed by the cubical box with a total electric flux of 1840 N·m2/C is calculated using Gauss's law and is found to be 16.29 nC.

Explanation:

The question deals with the concept of electric flux and its relation to the enclosed charge using Gauss's law, which is a fundamental principle in electromagnetism. According to Gauss's law, the total electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space (ε0).

The formula for Gauss's law in integral form is Φ = Q / ε0, where Φ is the electric flux, Q is the charge enclosed, and ε0 is the electric constant (approximately 8.854 x 10-12 C2/N·m2). Given that the total electric flux from a cubical box is 1840 N·m2/C, and using the value of ε0, the enclosed charge (Q) can be calculated.

To find the charge, we rearrange the equation as Q = Φ·ε0 and substitute the given values to get Q = 1840 N·m2/C × 8.854 x 10-12 C2/N·m2, resulting in Q = 1.629 x 10-8 C or 16.29 nC (nanocoulombs).

The charge enclosed by the cubical box,is approximately 1.63 × 10⁻⁸ C .

To determine the charge enclosed by a cubical box, we can use Gauss's Law. According to Gauss's Law, the electric flux (Φ) through a closed surface is given by:

Φ = Q / ε₀

where:

Rearranging this formula to solve for Q, we get:

Q = Φ × ε₀

Substitute the given values:

Q = 1840 N·m²/C × 8.854 × 10⁻¹² C²/N·m²

Q ≈ 1.63 × 10⁻⁸ C

Therefore, the charge enclosed by the cubical box is approximately 1.63 × 10⁻⁸ C.

Answer:

0.072 seconds

Explanation:

t = Time taken

u = Initial velocity = 100 km/h

v = Final velocity

s = Displacement = 1 m

a = Acceleration

[tex]v^2-u^2=2as\\\Rightarrow a=\dfrac{v^2-u^2}{2s}\\\Rightarrow a=\dfrac{0^2-(\dfrac{100}{3.6})^2}{2\times 1}\\\Rightarrow a=-385.802469136\ m/s^2[/tex]

[tex]v=u+at\\\Rightarrow t=\dfrac{v-u}{a}\\\Rightarrow t=\dfrac{0-\dfrac{100}{3.6}}{-385.802469136}\\\Rightarrow t=0.072\ s[/tex]

The time taken is 0.072 seconds

Answer:

C.

Measure the wear on his treads before and after riding a certain number of laps.

Answer:

Measure the wear on his treads before and after riding a certain number of laps.

Explanation:

By riding in a circular motion the inside of the tire will be in contact with the road more than the outside of the tire. Thus, to see if the constant circular motion had any effect on his tires David should measure the tread depth on both the inside and the outside of the tires before the experiment and measure the inside and the outside of the tires (at the same location on the tires) after the experiment. Then he can compare the tread loss on the inside of the tire to the tread loss on the outside of the tire.

Answer:

x = 2 x 10¹⁶ Km

Explanation:

distance between two star,d = 10⁹ Km

separation between them, θ = 10⁻⁵ radians

focal length of the eyepiece = 1.5 cm = 0.015 m

focal length of the objective = 3 m

observer distance from star, x = ?

we know,

[tex]tan \theta = \dfrac{d}{x}[/tex]

for small angle

[tex]\theta = \dfrac{d}{x}[/tex].......(1)

angular magnification of telescope

[tex]M = \dfrac{f_{objective}}{f_{eyepiece}}=\dfrac{3}{0.015} = 200[/tex]

Angular magnification of the telescope is also calculated by

[tex]M = \dfrac{observed\ angle}{original\ angle}[/tex]

[tex]M = \dfrac{\theta_0}{\theta}[/tex]

now,

[tex]\dfrac{\theta_0}{\theta}=200[/tex]

[tex]\dfrac{\theta_0}{200}=\theta[/tex]

from equation (1)

[tex]\dfrac{\theta_0}{200}=\dfrac{d}{x}[/tex]

[tex]x=\dfrac{200d}{\theta_0}[/tex]

[tex]x=\dfrac{200\times 10^9}{10^{-5}}[/tex]

x = 2 x 10¹⁶ Km

Distance between the observer and the star is x = 2 x 10¹⁶ Km

Answer:

d=894 m

Explanation:

Given that

initial velocity ,u= 35 m/s

Acceleration ,a= 38 m/s²

time ,t= 6 s

Given that at t= 0 s ,x= 0 m

We know that

[tex]d=ut+\dfrac{1}{2}at^2 [/tex]

d=Displacement

Now by putting the values

[tex]d=35\times 6+\dfrac{1}{2}\times 38\times 6^2 [/tex]

d=894 m

Therefore the particle position after 6 sec will be 894 m.

Final answer:

The position of the particle at t = 6 seconds, with an initial velocity of 35 m/s and a constant acceleration of 38 m/s², is 894 meters from the start.

Explanation:

The question asks us to calculate the position of a particle moving in a straight line at t = 6 seconds, given an initial velocity of 35 m/s and a constant acceleration of 38 m/s². To find the position, we can use the kinematic equation:

x = v0t + ½at²

where x is the position, v0 is the initial velocity, a is the acceleration, and t is the time. Plugging in our values we get:

x = (35 m/s)(6 s) + ½(38 m/s²)(6 s)²

x = 210 m + ½(38 m/s²)(36 s²)

x = 210 m + 684 m

x = 894 m

Therefore, the position of the particle at t = 6 s is 894 meters from the starting point.

Answer:

197.263157895 m/s

169.491525424 m/s

Explanation:

x Denotes position

t Denotes time

Average velocity is given by

[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-63}{4.75}\\\Rightarrow v_a=197.263157895\ m/s[/tex]

The average velocity is 197.263157895 m/s

[tex]v_a=\dfrac{x_2-x_1}{t_2}\\\Rightarrow v_a=\dfrac{1000-0}{5.9}\\\Rightarrow v_a=169.491525424\ m/s[/tex]

The average velocity is 169.491525424 m/s

The magnitude of the average velocity of the rocket during the 4.75-second part of the flight is 197.3 m/s, while for the first 5.90 seconds of the flight, the average velocity is 169.5 m/s.

To find the magnitude of the average velocity of the rocket, we use the formula average velocity = displacement / time.

(a) The displacement during the 4.75-second part of the flight is 1.00 km - 63 m = 937 m (we converted km to m to keep units consistent). Hence, the average velocity in this part of the flight is 937 m / 4.75 s = 197.3 m/s.

(b) For the first 5.90 seconds of flight, the displacement is 1.00 km = 1000 m (the height above the ground) while time is 5.90 s. Therefore, for this duration, the average velocity is 1000 m / 5.90 s = 169.5 m/s.

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Q: How much heat must be absorbed by 125 g of ethanol to change its temperature from 21.5 oC to 34.8 oC?  The specific heat of ethanol is 2.44 J/(gC).

Answer:

4056.5 J

Explanation:

The formula for the specific heat capacity of ethanol is given as

Q = cm(t₂-t₁)..................... Equation 1

Where q = quantity of heat, c = specific heat capacity of ethanol, m = mass of ethanol, t₁ = initial temperature of ethanol, t₂ = final temperature of ethanol.

Given: m = 125 g, t₁ = 25.5 °C, t₂ = 34.8 °C

Constant; c = 2.44 J/g.°C

Substitute into equation 1

Q = 125(2.44)(34.8-21.5)

Q = 125(2.44)(13.3)

Q = 4056.5 J.

Hence the amount of heat absorbed = 4056.5 J

Answer:

W = -1.97KJ, Q = 1.97KJ, Delta U = 0

Delta U = -1.03KJ, Q = -1.03KJ, Delta H = -1.72KJ

Explanation:

The deatiled step by step calculation using the ideal gas equation (Pv =nRT), The first law of thermodynamics ( dQ =dW + dU) as applied is as shown in the attached file.

In the first step, q = -157.29 R mol and w = -2.00 x 10^4 V Pa. In the second step, q = -141.45 R mol and w = 0. The total heat transfer (q_total) is -298.74 R mol and the total work done (w_total) is -2.00 x 10^4 V Pa.

The first step in the process is an isothermal expansion. In an isothermal process, the temperature remains constant, which means the change in internal energy (∆U) is zero. Since ∆U = q + w, this means that q = -w. We can calculate q using the equation q = nCv,m∆T, where n is the number of moles, Cv,m is the molar heat capacity at constant volume, and ∆T is the change in temperature. In this case, q = -w = nCv,m∆T = (1.65 mol)(3R/2)(-35.6 + 14.5) = -157.29 R mol.

The work done during an expansion or contraction process can be calculated using the equation w = -P∆V, where P is the external pressure and ∆V is the change in volume. In this case, the volume doubles, so ∆V = 2V, and the pressure is constant at 1.00 x 10^4 Pa. Therefore, w = -P∆V = -(1.00 x 10^4 Pa)(2V) = -2.00 x 10^4 V Pa.

In the second step, the gas is cooled at constant volume, so no work is done (w = 0). The heat transfer (q) can be calculated using the same equation as before, q = nCv,m∆T. In this case, q = (1.65 mol)(3R/2)(-35.6 - 14.5) = -141.45 R mol.

Putting it all together, for the first step, q = -w = -157.29 R mol and for the second step, q = -141.45 R mol. The total heat transfer for the overall process is the sum of the heat transfers for each step, so q_total = q1 + q2 = (-157.29 R mol) + (-141.45 R mol) = -298.74 R mol. As for the total work done (w_total), it is the sum of the work done in the first step and the work done in the second step, so w_total = w1 + w2 = (-2.00 x 10^4 V Pa) + 0 = -2.00 x 10^4 V Pa.

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Answer:

C. Charge on the capacitor

Explanation:

Read further: Capacitors consist of two parallel conductive

plates (usually a metal) which are prevented

from touching each other (separated) by an

insulating material called the “dielectric”. When

a voltage is applied to these plates an

electrical current flows charging up one plate

with a positive charge with respect to the

supply voltage and the other plate with an

equal and opposite negative charge.

Then, a capacitor has the ability of being able

to store an electrical charge Q (units in

Coulombs ) of electrons. When a capacitor is

fully charged there is a potential difference,

p.d. between its plates, and the larger the area

of the plates and/or the smaller the distance

between them (known as separation) the

greater will be the charge that the capacitor

can hold and the greater will be its

Capacitance.

The capacitors ability to store this electrical

charge ( Q ) between its plates is proportional

to the applied voltage, V for a capacitor of

known capacitance in Farads.

The correct option is c which is charge on the capacitance. When a parallel-plate capacitor is disconnected from the battery, and the separation between its plates is doubled, the charge on the capacitor remains constant. Changes in plate separation affect capacitance and voltage, but not the existing charge on the plates.

The question asks which of the following remains constant when the battery is disconnected from a parallel-plate capacitor and the separation between the plates is doubled: voltage across the capacitor, capacitance of the capacitor, or the charge on the capacitor. The key to answering this question lies in understanding how capacitors work and the relationship between charge (Q), capacitance (C), and voltage (V).

Capacitance is given by C = εA/d, where A is the area of the plates, d is the separation between the plates, and ε is the permittivity of the material between the plates. When the battery is disconnected, the external voltage source is removed, but the charge on the plates does not have a path to dissipate. Therefore, the charge on the capacitor remains constant, even when the plate separation is changed. Doubling the separation would affect the capacitance and the voltage across the capacitor but not the charge.

Answer:

She must be launched with a speed of 74.2 m/s.

Explanation:

Hi there!

The equations of the horizontal component of the position vector and the vertical component of the velocity vector are the following:

x = v0 · t · cos θ

vy = v0 · sin θ + g · t

x = horizontal distance traveled at time t.

v0 = initial velocity.

t = time.

θ = launching angle.

vy = vertical component of the velocity vector at time t.

g = acceleration due to gravity (-9.8 m/s²).

To just cross the 520-m gap, the maximum height of the flight must be reached halfway of the gap at 260 m horizontally (see attached figure).

When she is at the maximum height, her vertical velocity is zero. So, when x = 260 m, vy = 0. Using both equations we can solve the system for v0:

x = v0 · t · cos θ

Solving for v0:

v0 = x/ (t · cos θ)

Replacing v0 in the second equation:

vy = v0 · sin θ + g · t

0 = x/(t·cos(56°)) · sin(56°) + g · t

0 = 260 m · tan (56°) / t - 9.8 m/s² · t

9.8 m/s² · t = 260 m · tan (56°) / t

t² = 260 m · tan (56°) / 9.8 m/s²

t = 6.27 s

Now, let's calculate v0:

v0 = x/ (t · cos θ)

v0 = 260 m / (6.27 s · cos(56°))

v0 = 74.2 m/s

She must be launched with a speed of 74.2 m/s.

Answer:

it must be launched at a speed of 74.2 m/s

Explanation:

I really hope this helps

Answer:

[tex]6\times 10^{10}\ Hz[/tex]

Explanation:

d = Slit gap = 5 mm

Slit distance = 16 cm

c = Speed of light = [tex]3\times 10^8\ m/s[/tex]

[tex]\lambda[/tex] = Wavelength

We have the relation

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

Here, [tex]\theta=90[/tex]

So

[tex]d=\lambda\\\Rightarrow \lambda=5\ mm[/tex]

Frequency is given by

[tex]f=\dfrac{c}{\lambda}\\\Rightarrow f=\dfrac{3\times 10^8}{5\times 10^{-3}}\\\Rightarrow f=6\times 10^{10}\ Hz[/tex]

The frequency is [tex]6\times 10^{10}\ Hz[/tex]