Consider the four quantum numbers of an electron in an atom, n, l, ml, and ms. The energy of an electron in an isolated atom depends on:______a. l, ml, and msonly.b. n only.c. n and l only.d. n, l, and mlonly.e. all four quantum numbers.

Answer:

a)  the initial mass of O₂ is 373.92 gr

b) the mass leaked of O₂ is 104.26 gr

Explanation:

we can assume ideal gas behaviour of oxygen , then we can calculate the mass using the ideal gas equation

P*V = n*R*T ,

where P= absolute pressure , V= volume occupied by the gas , n = number of moles of gas , R= ideal gas constant = 8.314 J/mol K , T= absolute temperature

Initially

P = Pg + Pa ( 1 atm) = 3.20 *10⁵ Pa + 101325 Pa = 4.21*10⁵ Pa

where Pg= gauge pressure , Pa=atmospheric pressure

T = 39 °C= 312 K

V= 7.20 * 10⁻² m³

therefore

P*V = n*R*T → n = P*V/ (R*T)

replacing values

n = P*V/ (R*T) = 4.21*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*312 K) = 11.685 mol

since

m= n*M , where m= mass , n= number of moles , M= molecular weight of oxygen

then

m = n*M = 11.685 mol *32.0 g/mol = 373.92 gr of O₂

therefore the initial mass of O₂ is 373.92 gr

for the part B)

P₂= Pg₂ + Pa ( 1 atm) = 1.85*10⁵ Pa + 101325 Pa = 2.86*10⁵ Pa

T₂ = 20.9 °C= 293.9 K

V= 7.20 * 10⁻² m³

therefore

n₂ = P₂*V/ (R*T₂) = 2.86*10⁵ Pa*7.20 * 10⁻² m³/(8.314 J/mol K*293.9 K) = 8.427 mol

m₂ = n*M = 8.427 mol*32.0 g/mol = 269.66 gr of O₂

thus the mass leaked of oxygen is

m leaked = m - m₂ = 373.92 gr - 269.66 gr = 104.26 gr

Answer:

18.4 L

Explanation:

A chocolate sundae contains 500 Cal. Considering that 1 Cal = 1 kcal and 1 kcal = 4.184 kJ,  a chocolate sundae contains:

500 Cal × (4.184 kJ / 1 Cal) = 2.09 × 10³ kJ

Let's suppose we drink water at 10°C that increases its temperature to 37°C by absorbing this heat (Q). We can find the mass of water using the following expression.

Q = c × m × ΔT

where,

c: specific heat capacity of the water

m: mass

ΔT: change in the temperature

Q = c × m × ΔT

2.09 × 10³ kJ = (4.19 kJ/kg.°C) × m × (37°C - 10°C)

m = 18.4 kg

Considering the density of water to be of 1 kg/L, the volume of water is 18.4L.

Final answer:

To counteract the 500 kcal from a chocolate sundae, you would need to drink approximately 14.3 liters of ice-cold water. However, this is neither practical nor safe, and weight management should involve a balance of diet and exercise instead.

Explanation:

Drinking cold water does indeed require energy from the body to warm it up to body temperature, thus expending calories in the process.

To estimate the volume of cold water needed to negate the caloric intake of one chocolate sundae, we can use the given information that 250 mL of ice-cold water takes approximately 8,750 calories (or 8.75 kcal) to be heated up to body temperature.

If a chocolate sundae contains about 500 Cal (500 kcal), we would need to drink a significantly larger volume of cold water to burn those calories. Specifically, we can calculate this using a simple ratio:
500 kcal (chocolate sundae) / 8.75 kcal (250 mL water) = x / 250 mL

This gives us:

x = 250 mL × (500/8.75)

x = 14,285.71 mL

So, you would need to drink approximately 14.3 liters of ice-cold water to counteract the caloric intake from one chocolate sundae, which is not practical or safe.

Answer:

2.1 m/s

Explanation:

Momentum is conserved, so:

m₁ u₁ + m₂ u₂ = (m₁ + m₂) v

(9.1 kg) (6.6 m/s) = (9.1 kg + 19.3 kg) v

v = 2.1 m/s

Answer:

0.003034 s

1.035 m

4.5 m

Explanation:

[tex]f[/tex] = frequency of the tone = 329.6 Hz

[tex]T[/tex] = Time period of the sound wave

we know that, Time period and frequency are related as

[tex]T =\frac{1}{f}\\T =\frac{1}{329.6}\\T = 0.003034 s[/tex]

[tex]v[/tex] = speed of the sound in the air = 341 ms⁻¹

wavelength of the sound is given as

[tex]\lambda =\frac{v}{f} \\\lambda =\frac{341}{329.6}\\\lambda = 1.035 m[/tex]

[tex]v[/tex] = speed of the sound in the water = 1480 ms⁻¹

wavelength of the sound in water is given as

[tex]\lambda =\frac{v}{f} \\\lambda =\frac{1480}{329.6}\\\lambda = 4.5 m[/tex]

Answer:

Explanation:

Given

Displacement is [tex]\frac{1}{3}[/tex] of Amplitude

i.e. [tex]x=\frac{A}{3}[/tex] , where A is maximum amplitude

Potential Energy is given by

[tex]U=\frac{1}{2}kx^2[/tex]

[tex]U=\frac{1}{2}k(\frac{A}{3})^2[/tex]

[tex]U=\frac{1}{18}kA^2[/tex]

Total Energy of SHM is given by

[tex]T.E.=\frac{1}{2}kA^2[/tex]

Total Energy=kinetic Energy+Potential Energy

[tex]K.E.=\frac{1}{2}kA^2 -\frac{1}{18}kA^2[/tex]

[tex]K.E.=\frac{8}{18}kA^2[/tex]

Potential Energy is [tex]\frac{1}{8}[/tex] th of Total Energy

Kinetic Energy is [tex]\frac{8}{9}[/tex] of Total Energy

(c)Kinetic Energy is [tex]0.5\times \frac{1}{2}kA^2[/tex]

[tex]P.E.=\frac{1}{4}kA^2[/tex]

[tex]\frac{1}{2}kx^2=\frac{1}{4}kA^2[/tex]

[tex]x=\frac{A}{\sqrt{2}}[/tex]                  

Answer:

Explanation:

Let the amplitude is A.

Displacement, x = one third of the amplitude = A/3

The total energy of the body executing Simple Harmonic Motion is given by

[tex]T=\frac{1}{2}KA^{2}[/tex]

(a) Kinetic energy of the particle executing SHM is given by

[tex]K=\frac{1}{2}K\left (A^{2} -x^{2} \right )[/tex]

[tex]K=\frac{1}{2}K\left (A^{2} -\frac{A}{9}^{2} \right )[/tex]

[tex]K=\frac{1}{2}\times \frac{8A^{2}}{9}[/tex]

So, the ratio of kinetic energy to the total energy is given by

K / T = 8 / 9

(b) Potential energy of a particle executing SHM is given by

[tex]U=\frac{1}{2}Kx^{2}[/tex]

[tex]T=\frac{1}{2}\times \frac{A^{2}}{9}[/tex]

So, the ratio of potential energy to the total energy is given by

U / T = 1 / 9

(c) Let at a displacement y the kinetic energy is equal to the potential energy

[tex]\frac{1}{2}\times K\times \left ( A^{2}-y^{2} \right )=\frac{1}{2}\times K\times y^{2}[/tex]

[tex]y=\frac{A}{\sqrt{2}}[/tex]

Answer:

Explanation:

C is center of mass of the bar ie middle point. Spring is attached with one end ie A of the bar . When this end is displaced by distance x ( small) a restoring force kx is produced which creates a torque

Torque created = kx . l /2  , which creates angular acceleration α

moment of inertia I = ml²/12

torque = I. x α

kx . L /2 = I. x α

α = L kx / 2I

a = linear acceleration

= α x L/2

=  L² kx / 4I

= L²x 12 kx / 4 mL²

a  = 3kx/m

This shows that motion is SHM as acceleration is proportional to displacement x .

angular frequency ω² = 3k / m

frequency f = 1/2π √ 3k/m

= (1/6.28) x √ 3x 500/8

= 2.18 Hz

Final answer:

The frequency of small oscillations is approximately 3.98 Hz, and the smallest value of the spring constant for which oscillations will occur is 0 N/m.

Explanation:

(a)  To determine the frequency of small oscillations, we can use the formula: f = (1/2π) × (√(k/m)), where f is the frequency, k is the spring constant, and m is the mass of the bar. Plugging in the values, we get:
f = (1/2π) × (√(500/8)) ≈ 3.98 Hz

(b)  To find the smallest value of the spring constant k for which oscillations will occur, we need to consider critical damping. Critical damping occurs when the damping force is equal to the force exerted by the spring. The formula for critical damping is c = 2√(mk), where c is the damping coefficient, m is the mass, and k is the spring constant. Since we know that the damping coefficient is 0, we can solve for k to get:
k = (c^2)/(4m). Substituting in the values, we get:
k = (0^2)/(4 x 8) = 0 N/m

Answer:

[tex]x'=134.999983\ m[/tex]

[tex]y'=y=30\ m[/tex]

[tex]z'=z=55\ m[/tex]

Explanation:

Given:

Now from the equation of length contraction:

[tex]x'=x\times \sqrt{1-\frac{v_x^2}{c^2} }[/tex]

[tex]x'=135\times \sqrt{1-\frac{(1.5\times 10^{5})^2}{(3\times 10^8)^2} }[/tex]

[tex]x'=134.999983\ m[/tex]

Rest other values will remain unaffected since they are along the axis of motion. So,

[tex]y'=y=30\ m[/tex]

[tex]z'=z=55\ m[/tex]

Answer:

122.22 W/m²

Explanation:

Let the intensity of unpolarized light is Io.

from first polariser

I' = Io/2

From second polariser

I'' = I' Cos²30 = 3 Io/8

From third polariser

I''' = I'' Cos²30 = 9Io/32

According to the question

9Io/32 = 275

Io = 977.78 watt/m²

Now, from first polariser

I' = Io/2 = 977.78 / 2 = 488.89 W/m²

I'' = 488.89 x cos²60 = 122.22 W/m²

thus, the intensity of light is 122.22 W/m².

Answer:

Explanation:

Given

Intensity of light emerging out is [tex]I=275 W/m^2[/tex]

Polarizer axis are inclined at [tex]30^{\circ}] , [tex]60^{\circ}[/tex] , [tex]90^{\circ}[/tex]

If [tex]I_0[/tex] is the Intensity of Incoming light then

[tex]275=\frac{I_0}{2}\times \cos ^2{30}\times \cos^2 {30}[/tex]

as they are inclined to [tex]30^{\circ}[/tex]to each other

[tex]I_0=\frac{275}{9}\times 32[/tex]

[tex]I_0=977.77 W/m^2[/tex]

If middle Filter is removed then

[tex]I=0.5\cdot I_0\cos ^2{60}[/tex]

[tex]I=0.5\cdot 977.77\cdot \frac{1}{4}[/tex]

[tex]I=122.22 W/m^2[/tex]                                    

Answer:

average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal

density = 1.66 ×[tex]10^{-12}[/tex] kg/m³

Explanation:

given data

temperature of sun = 6000 K

concentration of atoms = [tex]10^{15}[/tex]  particles/m³

to find out

average pressure and density of its atmosphere

solution

first we get here average pressure of  atmosphere that is express as

average pressure = [tex]\frac{nRT}{V}[/tex]   .............1

put here value we get

average pressure = [tex]\frac{10^15*8.314*6000}{6*10^{23}}[/tex]

average pressure = 8.314 ×[tex]10^{-5}[/tex] pascal

and

density of atmosphere will be

density =  [tex]\frac{nM}{V}[/tex]   .............2

density = [tex]\frac{10^{15}*1.66*10^{-27}}{1}[/tex]

density = 1.66 ×[tex]10^{-12}[/tex] kg/m³

Answer:Time constant gets doubled

Explanation:

Given

L-R circuit is given and suppose R and L is the resistance and inductance of the circuit then current is given by

[tex]i=i_0\left [ 1-e^{-\frac{t}{\tau }}\right ][/tex]

where [tex]i_0[/tex] is maximum current

i=current at any time

[tex]\tau =\frac{L}{R}=time\ constant[/tex]

[tex]\tau '=\frac{2L}{R}=2\tau [/tex]

thus if inductance is doubled then time constant also gets doubled or twice to its original value.                                      

A. [tex]\( \iint_{S_2} F \cdot dS = 128 \)[/tex]

B. [tex]\( \iint_{S_2} F \cdot dS = 128 \)[/tex]

(A) If the magnitude of (F) is inversely proportional to the square of the distance from the origin:

We know that the flux integral over [tex]\( S_1 \)[/tex] is [tex]\( \iint_{S_1} F \cdot dS = 8 \)[/tex].

Since ( F ) is a radial force field, (F) and (dS) are parallel, and the dot product simplifies to [tex]\( |F| \cdot |dS| \)[/tex].

The surface area of a sphere is [tex]\( 4\pi r^2 \)[/tex], so the magnitude of [tex]\( dS \)[/tex] on [tex]\( S_1 \)[/tex] is [tex]\( |dS| = 16\pi \)[/tex].

Therefore, [tex]\( |F| \cdot |dS| = 8 \implies |F| \cdot 16\pi = 8 \)[/tex], which gives [tex]\( |F| = \frac{1}{2\pi} \)[/tex].

For [tex]\( S_2 \)[/tex], with radius 16, [tex]\( |dS| = 256\pi \)[/tex].

So, [tex]\( \iint_{S_2} F \cdot dS = |F| \cdot 256\pi = \frac{1}{2\pi} \cdot 256\pi = 128 \)[/tex].

(B) If the magnitude of (F) is inversely proportional to the cube of the distance from the origin:

For [tex]\( S_1 \)[/tex], [tex]\( |dS| = 16\pi \)[/tex] as before. So, [tex]\( |F| \cdot |dS| = 8 \implies |F| = \frac{1}{2} \)[/tex].

For [tex]\( S_2 \)[/tex], [tex]\( |dS| = 256\pi \)[/tex] as before.

So, [tex]\( \iint_{S_2} F \cdot dS = |F| \cdot 256\pi = \frac{1}{2} \cdot 256\pi = 128 \)[/tex].

To determine the convection heat transfer coefficient of air at the upper surface, we need to know the surface area of the plate which is not provided in the given question.

The convection heat transfer coefficient of air at the upper surface can be determined using Newton's law of cooling. According to Newton's law of cooling, the rate of heat transfer through convection is directly proportional to the temperature difference between the solid surface and the surrounding fluid and the surface area of the solid.

Therefore, the heat transfer rate can be calculated using the formula:

Q = h * A * (Ts - T∞)

Where:

In this case, we are given the following values:

To find the surface area, we need to know the dimensions of the plate. Once we have the surface area, we can solve for the convection heat transfer coefficient using the given formula. However, the surface area is not provided in the question, so we cannot determine the convection heat transfer coefficient without that information.

Answer:

Explanation:

a)

[tex]l[/tex] = length of each blade = 3 m

[tex]m[/tex] = mass of each blade = 120 kg

[tex]L[/tex] = length of the rod

[tex]M[/tex] = mass of the rod

Length of the rod is given as

[tex]L = 2 l = 2 (3) = 6 m[/tex]

[tex]w_{0}[/tex] = Angular velocity at t = 0, = 0 rad/s

[tex]w_{30}[/tex] = Angular velocity at t = 30, = 1200 rpm = 125.66  rad/s

[tex]\Delta t[/tex] = time interval = 30 - 0 = 30 s

angular acceleration is given as

[tex]\alpha = \frac{w_{30} - w_{0}}{\Delta t} = \frac{125.66 - 0}{30} = 4.2 rad/s^{2}[/tex]

Angular velocity at t = 10 s is given as

[tex]w_{10} = w_{0} + \alpha t\\w_{10} = 0 + (4.2) (10)\\w_{10} = 42 rad/s[/tex]

Angular velocity at t = 20 s is given as

[tex]w_{20} = w_{0} + \alpha t\\w_{20} = 0 + (4.2) (20)\\w_{10} = 84 rad/s[/tex]

Mass of the rod is given as

[tex]M = 2m = 2(120) = 240 kg[/tex]

Moment of inertia of the propeller is given as

[tex]I = \frac{ML^{2} }{12} =  \frac{(240)(6)^{2} }{12} = 720 kgm^{2}[/tex]

Angular momentum at t = 10 s is given as

[tex]L_{10} = I w_{10} = (720) (42) = 30240 kgm^{2}/s[/tex]

Angular momentum at t = 20 s is given as

[tex]L_{20} = I w_{20} = (720) (84) = 60480 kgm^{2}/s[/tex]

b)

Torque on propeller is given as

[tex]\tau = I \alpha \\\tau = (720) (4.2)\\\tau = 3024 Nm[/tex]

Answer:

(a) 4800 kg m^2/s^2 , 9600 kg m^2/s^2

(b) 480 Nm

Explanation:

length of each blade = 3 m

total length, L = 2 x 3 = 6 m

mass of each blade = 120 kg

total mass, m = 2 x 120 = 240 kg

initial angular velocity, ωo = 0 rad/ s

final angular velocity, ω = 1200 rpm = 1200 / 60 = 20 rps = 125.6 rad/s

t = 30 s

Let the angular acceleration is α.

α = (ω - ωo)/t = 120 / 30 = 4 rad/s^2

(a) moment of inertia. I = mL^2 / 12

I = 240 x 6 / 12 = 120 kg m^2

Let ω' be the angular velocity at the end of 10 s

use first equation of motion

ω' = 0 + αt

ω' = 4 x 10 = 40 rad/s

Angular momentum at t = 10 s

L = I x ω = 120 x 40 = 4800 kg m^2/s^2

Let ω'' be the angular velocity at the end of 20 s

use first equation of motion

ω'' = 0 + αt

ω'' = 4 x 20 = 80 rad/s

Angular momentum at t = 20 s

L = I x ω = 120 x 80 = 9600 kg m^2/s^2

(b) Torque, τ = I x α

τ = 120 x 4 = 480 Nm

To find the maximum values that four bricks of length L can be on the verge of falling while stacked on top of each other in equilibrium, we need to consider the moments of each brick and the conditions for equilibrium. The moments of each brick can be calculated by multiplying the mass of the brick by the distance from the axis of rotation, which is half the length of the brick. By setting the sum of the moments equal to zero, we can find the maximum values for the mass of each brick in terms of L.

When four identical and uniform bricks are stacked on top of each other such that part of each brick extends beyond the one beneath, the stack is in equilibrium and on the verge of falling. To find the maximum values in terms of L, we need to consider the moment of each brick and the conditions for equilibrium. The moment is the product of the mass of the brick and the distance from the axis of rotation, which is half the length of the brick.

Let's assume the bricks have a length L and that they are stacked vertically with their centers of mass aligned. The distance from the axis of rotation to the center of mass is L/2 for each brick. The moment of the top brick is then (mass of the brick) * (L/2). The moment of the second brick is (mass of the brick) * (3L/2), considering the length of the first brick as the distance from the axis of rotation. Similarly, the moment of the third brick is (mass of the brick) * (5L/2), and the moment of the fourth brick is (mass of the brick) * (7L/2).

For the stack to be in equilibrium, the sum of the moments must be zero. Therefore, we have the equation:

(mass of the first brick) * (L/2) + (mass of the second brick) * (3L/2) + (mass of the third brick) * (5L/2) + (mass of the fourth brick) * (7L/2)

By simplifying this equation and substituting the mass of each brick with a constant value (let's say M), we can find the maximum values for the mass of each brick such that the stack is in equilibrium on the verge of falling, given a certain length L.

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Answer

The answer and procedures of the exercise are attached in the following archives.

Explanation  

You will find the procedures, formulas or necessary explanations in the archive attached below. If you have any question ask and I will aclare your doubts kindly.  

The probability of finding a particle in its first excited state within the center 20% of an infinite square well of length L is approximately 11.8%. This involves integrating the square of the wavefunction over the specific interval. The key steps are defining the potential energy and solving the Schrödinger equation.

To find the probability of locating a particle in its first excited state within the center 20% of a one-dimensional box of length L, we follow these steps:

Define the Potential Energy, V: Inside the box (0 ≤ x ≤ L), V(x) = 0. Outside the box, V(x) = ∞.

Solve the Schrödinger Equation: The normalized wavefunction for the first excited state (n=2) is ψ2(x) = √(2/L) * sin(2πx/L).

The center 20% of the box is the interval from 0.4L to 0.6L. We calculate the probability of finding the particle in this region by integrating the square of the wavefunction:

Prob(center20%) = ∫0.4L0.6L |ψ2(x)|² dx = ∫0.4L0.6L ">2/L * sin²(2πx/L) dx.

Using integration techniques, the result is:

Prob(center20%) = 2 * [0.1 - (sin(0.4π))/(π)]. This computes to approximately 0.118 or 11.8%.

Final answer:

The intensity of the laser beam can be calculated from the given power and beam area, which is needed to assess safety for retinal exposure. The max electric and magnetic fields depend on the intensity and constants from electromagnetic wave equations. The average energy density is the ratio of beam intensity to the speed of light.

Explanation:

Calculating Intensity and Assessing Laser Safety:

(a) To find the intensity of the laser beam, we use the formula: I = P / A, where P is the power and A is the area of the beam. The power is given as 1.5 mW, which is 1.5 x 10-3 W. The area, given the diameter of the beam is 0.64 mm, can be calculated using the area of a circle, πr2 (where r is the radius in meters). So, the intensity is:

I = (1.5 x 10-3 W) / (π x (0.64 x 10-3 m / 2)2) = Intensity Value in W/m2.

(b) To determine if the laser is safe for the retina, we compare its intensity to the damage threshold of the retina, which is 100 W/m2. If the calculated intensity is less than this value, then the laser is considered safe for direct viewing.

(c) The maximum values of the electric (E) and magnetic (B) fields can be found using the following relationships, which are derived from the electromagnetic wave equations: E = cB, where c is the speed of light, and B = √(2I/ε0c), where ε0 is the vacuum permittivity.

(d) The average energy density of the beam can be found using: u = E / c, where u is the energy density and E is the intensity of the beam.

Answer

Initial radius of the artery is (1.1 cm) / 2 = 0.55 cm

final radius of the artery is (0.90 cm) / 2 = 0.45 cm

initial velocity of the blood is 17 cm/s

Using  equation of continuity is

                                      A₁v₁=A₂v₂

                                 π r₁² x v₁ = π r₂² x v₂

                                  r₁² x v₁ = r₂² x v₂

                                  0.55² x 17 =0.45² x v₂

                                        v₂=25.39 cm/s

Bernoulli's equation is

[tex]P_1 - P_2 = \dfrac{1}{2}\rho (v_2^2-v_1^2)[/tex]

rho is the density of blood = 1060 kg/m^3

[tex]P_1 - P_2 = \dfrac{1}{2}\times 1060 \times (0.254^2-0.17^2)[/tex]

[tex]P_1 - P_2 =18.87\ Pa[/tex]

The student uses the continuity equation to correctly solve the first part, but for pressure drop, they need to apply Bernoulli's principle. Using the equation P1 + 1/2ρv1² = P2 + 1/2ρv2², with ρ being the blood's density and v1 and v2 the speeds, they can find the pressure drop. However, real-life complications due to blood viscosity and turbulence might affect the results.

The subject of this question is Physics (specifically, fluid dynamics). The concept being applied here is that of continuity and Bernoulli's principle. The continuity equation for fluid states that the mass flow rate must be constant throughout the length of the pipe. This translates to: Area 1 × Speed 1 = Area 2 × Speed 2. You correctly set up this equation to find Speed 2, which is the speed of the blood flow within the plaque region.

For the pressure drop, we need to use Bernoulli's principle, which describes that the sum of the pressure, kinetic energy per unit volume, and potential energy per unit volume is constant in a non-viscous, steady-flowing system. Applying Bernoulli's Equation: P1 + 1/2ρv1² = P2 + 1/2ρv2². Solving this equation gives us the pressure difference within the plaque region.

Keep in mind that you might have to use the change in blood speed, and also the density of blood within the equations to find the accurate pressure drop. Also, Bernoulli's principle applies to ideal situations and there could be changes in real-life situations due to the viscosity of blood and turbulence caused by plaque.

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Answer:

Impulse, |J| = 0.6716 kg-m/s

Force, F = 63.35 N

Explanation:

It is given that,

Mass of the baseball, m = 0.146 kg

Initial speed of the ball, u = 15.3 m/s

Final speed of the ball, v = 10.7 m/s

To find,

(a) The magnitude of this impulse.

(b) The magnitude of the average force of the glass on the ball.

Solution,

(a) Impulse of an object is equal to the change in its momentum. It is given by :

[tex]J=m(v-u)[/tex]

[tex]J=0.146\ kg(10.7-15.3)\ m/s[/tex]

J = -0.6716 kg-m/s

or

|J| = 0.6716 kg-m/s

(b) Another definition of impulse is given by the product of force and time of contact.

t = 0.0106 s

[tex]J=F\times \Delta t[/tex]

[tex]F=\dfrac{J}{\Delta t}[/tex]

[tex]F=\dfrac{0.6716\ kg-m/s}{0.0106\ s}[/tex]

F = 63.35 N

Hence, this is the required solution.

The impulse direction is opposite to the baseball's initial direction, and its magnitude is 0.6716 kg·m/s. The average force magnitude exerted by the window on the baseball is 63.36 N.

The direction of the impulse imparted by the window to the baseball is opposite to the baseball's initial direction of motion. This is because impulse is equal to the change in momentum, and since the window slows the ball down, it is applying a force in the opposite direction of the ball's initial velocity.

To calculate the magnitude of the impulse (I), use the formula

I = change in momentum = m(vf - vi)
where m is the mass of the baseball, vf is the final velocity, and vi is the initial velocity. The mass m = 0.146 kg, vi = 15.3 m/s, and vf = 10.7 m/s.

I = (0.146 kg)(10.7 m/s - 15.3 m/s)
I = (0.146 kg)(-4.6 m/s)
I = -0.6716 kg·m/s.

The negative sign indicates the impulse is in the opposite direction of the ball's initial motion, but since the question asks for the magnitude, we take the absolute value: 0.6716 kg·m/s.

The magnitude of the average force (Favg) exerted on the ball can be found using the formula

Favg = I/t
where t is the contact time. For a contact time of 0.0106 s, we have:
Favg = 0.6716 kg·m/s / 0.0106 s
Favg = 63.36 N.

Answer:

[tex]t=2.044\ hr[/tex]

Explanation:

From the schematic we can visualize the situation and the position of the rays falling on the floor.

Considering the given data from the lowest edge of the mirror.

Now applying the trigonometric ratio to known sides in the first instance:

[tex]tan\ \theta_1=\frac{1.86}{3.8}[/tex]

[tex]\theta_1=26.08^{\circ}[/tex]

Applying the trigonometric ratio to known sides in the second instance:

[tex]tan\ \theta_2=\frac{1.86}{1.22}[/tex]

[tex]\theta_2=56.74^{\circ}[/tex]

Now by the law of reflection we know that the angle of incidence is equal to the angle of reflection. So the sun would have been at the same angle on the opposite side of the normal.

Hence the change in angle of the sun with respect to the mirror (also the earth)

[tex]\Delta \theta=\theta_2-\theta_1[/tex]

[tex]\Delta \theta=56.74-26.08[/tex]

[tex]\Delta \theta=30.66^{\circ}[/tex]

Now the time past for this change:

[tex]t=\frac{\Delta \theta}{\omega}[/tex]

[tex]t=\frac{30.66}{15}[/tex]

[tex]t=2.044\ hr[/tex]

Answer:

The moment (torque) is given by the following equation:

[tex]\vec{\tau} = \vec{r} \times \vec{F}\\\vec{r} \times \vec{F} = \left[\begin{array}{ccc}\^{i}&\^j&\^k\\r_x&r_y&r_z\\F_x&F_y&F_z\end{array}\right] = \left[\begin{array}{ccc}\^{i}&\^j&\3k\\0.23&0.04&0\\150&260&0\end{array}\right] = \^k((0.23*260) - (0.04*150)) = \^k (53.8~Nm)[/tex]

Explanation:

The cross-product between the distance and the force can be calculated using the method of determinant. Since the z-components are zero, it is easy to calculate.

To solve this problem it is necessary to apply the concepts related to the magnetic flow of a coil and take into account the angles for each case.

It is also necessary to delve into part C, the concept of electromotive force (emf) which is defined as the variation of the magnetic flux as a function of time.

By definition the magnetic flux is determined as:

[tex]\phi = NBA cos\theta[/tex]

Where

N = Number of loops [tex]\rightarrow[/tex] We will calculate the value for each of the spins

B = Magnetic Field

A = Cross-sectional Area

[tex]\theta =[/tex] Angle between the perpendicular cross-sectional area and the magnetic field.

PART A) The magnetic flux through the coil after it is rotated is as follows:

[tex]\phi_i = NBA cos\theta[/tex]

[tex]\phi_i = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(0)[/tex]

[tex]\phi_i = 7.2*10^{-8}T\cdot m^2[/tex]

PART B) For the second case the angle formed is perpendicular therefore:

[tex]\phi_f = NBA cos\theta[/tex]

[tex]\phi_f = (1turns)(6*10^{-5}T)(12*10^{-4}m^2)cos(90)[/tex]

[tex]\phi_f = 0[/tex]

PART C) The average induced emf of the coil is as follows:

[tex]\epsilon = - (\frac{\phi_f-\phi_i}{dt})[/tex]

[tex]\epsilon = -(\frac{0-7.2*10^{-8}}{0.04})[/tex]

[tex]\epsilon = 1.8*10^{-6}V[/tex]

Answer:

In this equation the scale used must be Kelvin

Explanation:

The absolute temperature is the value of the measured temperature relative to a scale starting at Absolute Zero (0 K or -273.15 °C). It is one of the main parameters used in thermodynamics and statistical mechanics. In the international system of units it is expressed in kelvin, whose symbol is K

The absolute temperature should always be used in the ideal gas state equation as this can only be in kelvin grade scale.

Answer:

461 N

Explanation:

Impulse = change in momentum

F Δt = m Δv

F (0.033 s) = (0.539 kg) (14.4 m/s − (-13.8 m/s))

F = 461 N

Answer:

w = 4,786 rad / s ,  f = 0.76176 Hz

Explanation:

For this problem let's use the concept of angular momentum

       L = I w

The system is formed by the two discs, during the impact the system remains isolated, we have the forces are internal, this implies that the external torque is zero and the angular momentum is conserved

Initial Before sticking

      L₀ = 0 + I₂ w₂

Final after coupling

      [tex]L_{f}[/tex] = (I₁ + I₂) w

The moments of inertia of a disk with an axis of rotation in its center are

      I = ½ M R²

How the moment is preserved

      L₀ = [tex]L_{f}[/tex]

      I₂ w₂ = (I₁ + I₂) w

      w = w₂ I₂ / (I₁ + I₂)

Let's reduce the units to the SI System

      d₁ = 60 cm = 0.60 m

      d₂ = 40 cm = 0.40 m

      f₂ = 200 min-1 (1 min / 60 s) = 3.33 Hz

Angular velocity and frequency are related.

      w₂ = 2 π f₂

      w₂ = 2π 3.33

      w₂ = 20.94 rad / s

Let's replace

       w = w₂ (½ M₂ R₂²) / (½ M₁ R₁² + ½ M₂ R₂²)

       w = w₂ M₂ R₂² / (M₁ R₁² + M₂ R₂²)

Let's calculate

      w = 20.94 8 0.40² / (12 0.60² + 8 0.40²)

      w = 20.94 1.28 / 5.6

      w = 4,786 rad / s

Angular velocity and frequency are related.

      w = 2π f

      f = w / 2π

      f = 4.786 / 2π

      f = 0.76176 Hz

Answer:

The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.

Explanation:

Given that,

Distance d = 2.00 mm

Wavelength = 511 nm

Order number = 3

Order number = 5

We need to calculate the width of third-order bright fringe

Using formula of width

[tex]d\sin\theta=m\lambda[/tex]

[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]

Put the value into the formula

[tex]\theta=\sin^{-1}\dfrac{3\times511\times10^{-9}}{2.00\times10^{-3}}[/tex]

[tex]\theta=0.00076\ rad[/tex]

We need to calculate the width of fifth-order bright fringe

Using formula of width

[tex]d\sin\theta=m\lambda[/tex]

[tex]\theta=\sin^{-1}\dfrac{m\lambda}{d}[/tex]

Put the value into the formula

[tex]\theta=\sin^{-1}\dfrac{5\times511\times10^{-9}}{2.00\times10^{-3}}[/tex]

[tex]\theta=0.00127\ rad[/tex]

Hence, The width of third and fifth order bright fringe is 0.00076 rad and 0.00127 rad.

To solve this problem we will apply the concepts given by the principles of superposition, specifically those described by Bragg's law in constructive interference.

Mathematically this relationship is given as

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

Where,

d = Distance between slits

[tex]\lambda[/tex] = Wavelength

n = Any integer which represent the number of repetition of the spectrum

[tex]\theta = sin^{-1} (\frac{n\lambda}{d})[/tex]

Calculating the value for n, we have

n = 1

[tex]\theta_1 = sin^{-1} (\frac{\lambda}{d})\\\theta_1 = sin^{-1} (\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_1 = 23.3\°[/tex]

n=2

[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (2\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = 52.28\°[/tex]

n =3

[tex]\theta_2 = sin^{-1} (\frac{2\lambda}{d})\\\theta_2 = sin^{-1} (3\frac{632.8*10^{-9}}{1.6*10^{-6}})\\\theta_2 = \text{not possible}[/tex]

Therefore the intensity of light be maximum for angles 23.3° and 52.28°

The work done by the child pulling the sled and dog, with a pulling force of 55 N, over a distance of 7.0 m, considering both pulling work and the work done against friction, is 59.42 J.

The question concerns the work done by a child pulling a sled with a dog. 'Work done' in physics is calculated using the equation, work done = force x distance x cosine of the angle. The force exerted is 55 N, the distance is 7.0 m, and the angle is 20 degrees. Thus, the work done by the child's pulling force as the system moves a distance of 7.0 m, ignoring friction and because cos(20) is approximately 0.94, is calculated as: Work done = 55 N x 7.0 m x cos(20) = 55 N x 7.0 m x 0.94 = 361.3 J.

However, the total work done is reduced due to friction between the sled and the ground. The sled's total weight (15 kg sled + 5.0 kg dog = 20 kg) multiplied by gravity (9.8 m/s²) gives  the normal force (20 kg * 9.8 m/s² = 196 N). Multiplying the normal force by the friction coefficient (0.22), gives the frictional force (196 N * 0.22 = 43.12 N). Hence, the work done against friction is: Work done against friction = frictional force x distance = 43.12 N x 7.0 m = 301.88 J. Therefore, the actual work done by the child equals the pull work minus the work against friction, which is 361.3 J - 301.88 J = 59.42 J.

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To solve this problem it is necessary to apply the concepts related to the magnetic field

the flux density and the magnitude of the magnetization.

Each of these will be tackled as the exercise is carried out, for example for the first part we have to:

Part A) Magnitude of the magnetic field

[tex]H = \frac{NI}{L}[/tex]

Where,

N = Number of loops

I = Current

L = Length

If we replace the given values the value of the magnitude of the magnetic field would be:

[tex]H= \frac{340*13}{0.12}[/tex]

[tex]H = 36833 A\cdot turns/m[/tex]

For the second and third part we will apply the concepts of density both in vacuum and positioned at a point, like this:

PARTE B) Flux density in a vacuum

[tex]B = \mu_0 H[/tex]

Where,

[tex]\mu_0 =[/tex] Permeability constant

[tex]B = (4\pi*10^{-7})(36833)[/tex]

[tex]B = 0.04628T[/tex]

PART C) To find the Flux density inside a bar of metal but the magnetic susceptibility is given

[tex]X_m = 1.9*10^{-4}[/tex]

[tex]\mu_R = 1+X_m[/tex]

[tex]\mu_R = 1.00019[/tex]

Then the flux density would be

[tex]B = \mu_0 \mu_R H[/tex]

[tex]B = (4\pi*10^{-7})(1.00019)(36833)[/tex]

[tex]B = 0.04629T[/tex]

PART D) Finally, the magnetization describes the amount of current per meter, and is given by the magnetic susceptibility, that is:

[tex]M = X_m H[/tex]

[tex]M = 1.9*10^{-4}*36833[/tex]

[tex]M = 6.996A/m[/tex]

This solution calculates the magnetic field strength, flux density and magnetization in a wire coil, as well as the flux density inside a metal bar within the coil, using Ampere's law and the formulas for magnetic field and flux densities.

The calculations are derived from Ampere's law and the formulas for magnetic field strength and flux density. For part (a), to find the magnitude of the magnetic field strength H (in A/m), we consider the given length of the coil and the number of turns. H= nI, where n is the number of turns per unit length and I is the current. In the given case, n= 340/0.12 = 2833.33, so H = 2833.33 turns/m * 13 A = 36833 A/m.

For part (b), we need to calculate the flux density B (in T) inside the coil which is in a vacuum. Using the formula B = μH, where μ = 4π x 10-7 T. m/A is the permeability of vacuum, we find B= 4π x 10^-7 T.m/A * 36833 A/m = 4.6 x 10^-2 T.

For part (c), we need to calculate the flux density B inside a bar of metal with susceptibility χ_m = 1.90 x 10-4. In this case, B = μ(H + M), where M = χ_m H is the magnetization. We find M= 1.90 x 10^-4 * 36833 A/m = 7 A/m, so B in metal = 4π x 10^-7 T.m/A * (36833 A/m + 7 A/m) = 4.6 x 10^-2 T.

For part (d), we've already calculated the magnitude of the magnetization M to be 7 A/m.

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The ideal concept for solving this question is based on the Doppler effect, for which it is indicated that the source's listening frequency changes as the distance and the relative speed between the receiver and the transmitter are also changed. However, if the relative velocity between the two objects is zero as in the particular case presented (since both travel at 75km / h) we have that there will be no change in frequency.

Therefore the frequency that I hear and that my sister would listen would be the same.

Answer:

(B) Resistor only

Explanation:

Alternating Current: These are currents that changes periodically with time.

An LRC  Ac circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, connected in series.

In a purely resistive circuit, current and voltage are in phase.

In a purely capacitive circuit, the current leads  the voltage by π/2

In a purely inductive circuit, the current lags the voltage by π/2.

Therefore when a alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.

The right option is (B) Resistor only.

The circuit elements are the current and voltage in phase in:

B) Resistor only

These are flows that changes occasionally with time. A LRC AC circuit is an AC circuit that contains a Resistor, a capacitor and an inductor, associated in series. In an absolutely resistive circuit, current and voltage are in stage. In an absolutely capacitive circuit, the current leads  the voltage by π/2. In an absolutely inductive circuit, the current slacks the voltage by π/2.

Thus, when an alternating current is set up in LRC circuit, in the resistor, the current and the voltage are in phase.

So, correct option is (B).

Find more information about Alternating current here:

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