An 80.0-g piece of copper, initially at 295°C, is dropped into 250 g of water contained in a 300-g aluminum calorimeter; the water and calorimeter are initially at 10.0°C.

What is the final temperature of the system? (Specific heats of copper and aluminum are 0.092 0 and 0.215 cal/g⋅°C, respectively. cw = 1.00 cal/g°C)
a. 12.8°C
b. 16.5°C
c. 28.4°C
d. 32.1°C

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

Explanation:

Given that,

Mass of the quarterback, m = 80 kg

Mass of the football, m' = 0.43 kg

Speed of the football, v' = 15 m/s

We need to sort the following quantities as known or unknown.

1. The mass of the quarterback = known = 80 kg

2. The mass of the football = known = 0.43 kg

3. The horizontal velocity of quarterback before throwing the ball = known = 0

4. The horizontal velocity of football before being thrown = known = 0

5. The horizontal velocity of quarterback after throwing the ball, = unknown quantity and it can be calculated using the conservation of linear momentum.

6. The horizontal velocity of football after being thrown = known = 15 m/s

Hence, this is the required solution.

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:

[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:

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:

  K = 24.5 keV

Explanation:

The interference phenomenon is described by the equation

       .d sin θ = m λ            m = 1,2,3,…

The pattern is observed on a screen at a distance L = 2.6 m

       tan θ = y / L

As these experiments the angle is very small we can approximate the tangent

        tan θ = sin θ / cos θ

For small angles

        tan θ = sin θ

Let's replace

        d y / L = m λ  

        λ   = d y / m L

Let's reduce the units to the SI system

       d = 51 nm = 51 10⁻⁹ m

       y = 0.4 mm = 0.4 10⁻³ m

Let's calculate the wavelength

Let's use m = 1 for the first interference line

       λ  = 51 10⁻⁹ 0.4 10⁻³ / 2.6

       λ   = 7.846 10⁻¹² m

Let's look for kinetic energy

       K = ½ m v²

       p = mv

       K = ½ m p² / m

       K = p² / 2m

Let's use the wave-particle duality relationship

       p = h /λ  

       K = h² / 2m λ²

Let's calculate

      K = (6.63 10⁻³⁴)² / (2 9.1 10⁻³¹ (7.846 10⁻¹²)²)

      K = 3,923 10⁻¹⁵-15 J

       K = 3.923 10⁻¹⁵ J ( 1 eV / 1.6 10⁻¹⁹ J) =2.452 10⁴ eV

     K = 24.5 keV

Encryption would not likely be the subject of a standard published by the International Electrotechnical Commission (IEC).

Answer: Option C

Explanation:

IEC (International Electrotechnical Commission) is the world's number one organization that develops and releases international standards for all electronic-electrical technologies and its affiliates.

Its standards would cover a wide technology range, from energy generation, transmissions and distributions to house appliances and offices, semiconductors, batteries, fibreglass, solar energy, Nano-technology and sea energy and many more.

IEC standards apply to all electrical technologies, including electronics, magnets and electro-magnetics, electro-acoustics, multi-media, telecommunications, and medical technologies, as well as related general fields such as terminology and symbols, etc.

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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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:

ΔP = P₂ - P₁  = 132. 24 kPa

Explanation:

Given:

ρ = 1370 kg / m³ , v = 9.67 m / s  ,  d₁ = 12.9 cm , d₂ = 16.7 cm , Δy = 9.85 m

Using Bernoulli's equations to determine the difference ΔP  

P₂ + ρ * g * Z₂ + (ρ * V₂²) / 2 = P₁ + ρ * g * Z₁ + (ρ * V₁²) / 2

P₂ - P₁ = ρ * g * (Z₁ - Z₂) + [ ρ * (V₁² - V₂²) ] / 2

P₂ - P₁ = ρ * g * (Z₁ - Z₂) + ¹/₂ * ρ * V₁² * [ ( 1 - (d₁ / d₂) ⁴ ) ]

ΔP = 1370 kg / m³ * 9.8 m/s² * 9.85m  +  0.5 * 1370 kg / m³ * ( 1 - (12.9 cm / 16.7 cm )⁴ )

ΔP = 132247.9364 Pa

ΔP = 132. 24 kPa

Answer:

P₂ - P₁=173.5kPa

Explanation:

The equation of continuity:  

A₁v₁=A₂v₂  

where A₁=πd₁²/4 and A₂=πd₂²/4

v₂=(A₁/A₂)v₁

v₂={(πd₁²/4)/(πd₂²/4)}v₁

v₂=(d₁²/d₂²)v₁

Use Bernoulli's equation

P₂+pgz₂+(pv₂²/2)=P₁+pgz₁+(pv₁²/2)

The difference between the fluid pressure at location 2 and the fluid pressure at location 1

P₂ - P₁=pg(z₁-z₂)+{p(v₁²-v₂²)}/2=pg(z₁-z₂)+1/2pv₁²(1-(d₁/d₂)⁴)

P₂ - P₁=(1.370×10³×9.8×9.85)+(1/2)(1.370×10³×(9.67)²){(1-(0.129m/0.167m)⁴}

P₂ - P₁=1.735×10⁵Pa

P₂ - P₁=173.5kPa

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:E

Explanation:

This can be explained by Doppler effect which gives the relation between apparent Frequency and actual frequency when the source of sound is moving

[tex]f'=f\frac{v+v_o}{v-v_s}[/tex]

where [tex]f'=apparent\ frequency [/tex]

[tex]f=actual\ frequency[/tex]

[tex]v=velocity\ of\ sound[/tex]

[tex]v_o=velocity\ of\ observer[/tex]

[tex]v_s=velocity\ of\ source[/tex]

here [tex]v_o =0[/tex] as observer is standing

when Ambulance is approaching then velocity of sound and velocity of ambulance have relative velocity thus denominator is less than and apparent frequency is more while when ambulance is going away then velocity of sound waves and velocity of observer is in opposite direction thus denominator is less than 1 and apparent frequency is less.

Thus [tex]f_2<f<f_1[/tex]

The Doppler effect explains the changes in sound frequency from a moving source as perceived by a stationary observer. The observed frequency increases as the source approaches and decreases when it moves away. Hence, the actual frequency of the source (the siren) is less than the frequency observed when the source is approaching (f1) and greater than the frequency observed than when it is retreating (f2). The correct answer is 'e. f2 < f < f1'.

This question relates to the phenomenon of the Doppler effect, which is the change in frequency or wavelength of a wave in relation to an observer who is moving relative to the wave source. In this case, when an ambulance with a siren is approaching a static observer, the observed frequency f1 will be higher than the actual frequency f (the frequency of the siren when you are sitting inside the ambulance), due to compression of the sound waves.

However, when the ambulance starts moving away from the observer, the observed frequency f2 will be lower than f, due to elongation or stretching out of the sound waves. Hence, the correct answer is 'e. f2 < f < f1'.

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Answer:

The asteroid requires 5.14 years to make one revolution around the Sun.

Explanation:

Kepler's third law establishes that the square of the period of a planet will be proportional to the cube of the semi-major axis of its orbit:

[tex]T^{2} = a^{3}[/tex] (1)

Where T is the period of revolution and a is the semi-major axis.

In the other hand, the distance between the Earth and the Sun has a value of [tex]1.50x10^{8} Km[/tex]. That value can be known as well as an astronomical unit (1AU).

But 1 year is equivalent to 1 AU according with Kepler's third law, since 1 year is the orbital period of the Earth.

For the special case of the asteroid the distance will be:

[tex]a = 2.98(1.50x10^{8}Km)[/tex]

[tex]a = 4.47x10^{8}Km[/tex]

That distance will be expressed in terms of astronomical units:

[tex]4.47x10^{8}Km.\frac{1AU}{1.50x10^{8}Km}[/tex] ⇒ [tex]2.98AU[/tex]

Finally, from equation 1 the period T can be isolated:

[tex]T = \sqrt{a^{3}}[/tex]

[tex]T = \sqrt{(2.98)^{3}}[/tex]  

[tex]T = \sqrt{26.463592}[/tex]

[tex]T = 5.14AU[/tex]

Then, the period can be expressed in years:

[tex]5.14AU.\frac{1yr}{1AU} ⇒ 5.14 yr[/tex]

[tex]T = 5.14 yr[/tex]

Hence, the asteroid requires 5.14 years to make one revolution around the Sun.

Final answer:

Using Kepler's third law, the asteroid that is 2.98 times the mean distance of Earth from the Sun has an orbital period of approximately 5.14 years.

Explanation:

To calculate the orbital period of an asteroid using Kepler's third law of planetary motion, which relates the orbital period of a planet to its average distance from the Sun, we first need to understand that the asteroid's mean distance from the Sun is given as 2.98 times the Earth's mean distance from the Sun.

Since 1 astronomical unit (AU) is the mean distance of Earth from the Sun, we'll denote the asteroid's mean distance as 2.98 AU. Kepler's third law can be formulated as P² = a³, where P is the orbital period in Earth years, and a is the semimajor axis or mean distance from the Sun in astronomical units.

To find the period P, we take the cube of the mean distance (2.98³) and then take the square root of that number to find the orbital period in years.

Calculation:

a = 2.98 AU (mean distance to the Sun)
a³ = 2.98 x 2.98 x 2.98 = 26.45
P² = 26.45
P = √26.45
P ≈ 5.14 years

Therefore, the asteroid takes approximately 5.14 years to make one revolution around the Sun.

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

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]                                    

In order to solve this problem it is necessary to apply the concepts related to Newton's second law and the respective representation of the Forces in their vector components.

The horizontal component of this force is given as

F_x = Fcos(6.7)

While the vertical component of this force would be

F_y = Fsin(6.7)

In the vertical component, the sum of Force indicates that:

[tex]\sum F_y= 0 [/tex]

The Normal Force would therefore be equivalent to the weight and vertical component of the applied force, therefore:

[tex]N = mg+Fsin(6.7)[/tex]

In the horizontal component we have that the Force of tension in its horizontal component is equivalent to the Force of friction:

[tex]\sum F_x = 0[/tex]

[tex]F_x = F_{friction}[/tex]

[tex]Fcos (6.7) = N\mu[/tex]

Using the previously found expression of the Normal Force and replacing it we have to,

[tex]Fcos(6.7)= \mu (mg+Fsin(6.7))[/tex]

Replacing,

[tex]Fcos(6.7)= (0.87) (mg+Fsin(6.7))[/tex]

[tex]Fcos(6.7) = (0.87)(mg) + (0.87)(Fsin(6.7))[/tex]

[tex]Fcos(6.7) -(0.87)(Fsin(6.7)) = 0.87 (mg)[/tex]

[tex]F(cos(6.7)-0.87sin(6.7)) = 0.87 (mg)[/tex]

[tex]F = \frac{0.87 (mg)}{(cos(6.7)-0.87sin(6.7))}[/tex]

[tex]F = \frac{0.87(128000*9.8)}{(cos(6.7)-0.87sin(6.7))}[/tex]

[tex]F = 1.95*10^6N[/tex]

Finally the acceleration would be by Newton's second law:

[tex]F = ma[/tex]

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

[tex]a = \frac{ 1.95*10^6}{128000}[/tex]

[tex]a = 15.234m/s^2[/tex]

Therefore the greatest acceleration the man can give the airplane is [tex]15.234m/s^2[/tex]

the greatest acceleration the man can give the airplane is approximately 7.38 m/s²

The greatest acceleration the man can give the airplane when pulling at an angle of 6.7° above the horizontal is given by the equation:

[tex]\[ a = \frac{\mu (m_m + m_a) g - m_a g \sin(\theta)}{m_m + m_a \cos(\theta)} \][/tex]

where:

- [tex]\( \mu \)[/tex] is the coefficient of static friction between the man's shoes and the runway,

- [tex]\( m_m \)[/tex] is the mass of the man,

- [tex]\( m_a \)[/tex] is the mass of the airplane,

- g is the acceleration due to gravity,

- [tex]\( \theta \)[/tex] is the angle at which the man pulls the cable.

Given:

[tex]- \( \mu = 0.87 \),[/tex]

[tex]- \( m_m = 76 \) kg,[/tex]

[tex]- \( m_a = 128000 \) kg,[/tex]

[tex]- \( g = 9.8 \) m/s²,[/tex]

[tex]- \( \theta = 6.7° \).[/tex]

First, we need to convert the angle from degrees to radians because the sine and cosine functions in trigonometry typically use radians:

[tex]\[ \theta_{\text{radians}} = \theta_{\text{degrees}} \times \frac{\pi}{180} \][/tex]

[tex]\[ \theta_{\text{radians}} = 6.7° \times \frac{\pi}{180} \][/tex]

Now, we can plug in the values into the equation:

[tex]\[ a = \frac{0.87 (76 \text{ kg} + 128000 \text{ kg}) \times 9.8 \text{ m/s}^2 - 128000 \text{ kg} \times 9.8 \text{ m/s}^2 \times \sin(6.7°)}{76 \text{ kg} + 128000 \text{ kg} \times \cos(6.7°)} \][/tex]

Calculating the sine and cosine of 6.7°:

[tex]\[ \sin(6.7)\approx 0.117 \][/tex]

[tex]\[ \cos(6.7) \approx 0.993 \][/tex]

Now, we substitute these values into the equation:

[tex]\[ a = \frac{0.87 (76 + 128000) \times 9.8 - 128000 \times 9.8 \times 0.117}{76 + 128000 \times 0.993} \][/tex]

Solving the numerator:

[tex]\[ (0.87 \times 128076 \times 9.8) - (128000 \times 9.8 \times 0.117) \][/tex]

[tex]\[ = (0.87 \times 128076 \times 9.8) - (128000 \times 1.1466) \][/tex]

[tex]\[ = 939608.16 \][/tex]

Solving the denominator:

[tex]\[ 76 + (128000 \times 0.993) \][/tex]

[tex]\[ = 127308 \][/tex]

Finally, we divide the numerator by the denominator to find the acceleration:

[tex]\[ a = \frac{939608.16}{127308} \][/tex]

[tex]\[ a \approx 7.38 \text{ m/s}^2 \][/tex]

The answer is: 7.38.

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

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.

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.

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.

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].

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:

FN₁ = 1146.6N : Force exerted on the ladder by the floor , vertical and upward

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Explanation:

The equilibrium equation are:

∑Fx=0

∑Fy=0

∑M = 0  

M = F*d  

Where:  

∑M : Algebraic sum of moments

M : moment  ( N*m)  

F : Force ( N)  

d :Perpendicular distance of the force to the point  ( meters )

Data

m =45 kg  : mass of the ladder

M =72 kg : mass of the fire fighter

g = 9.8 m/s²: acceleration due to gravity

L = 12 m : ladder length

h =  9.3 m: ladder height

L/3 = 12/3 = 4m Location of the center of mass of the ladder of the way up

L/2 = 12/2 = 6m Location of the center of mass of the  fire fighter

µ = 0 : coefficient of friction between the ladder and the wall

θ  : angle that makes  the  ladder  with the floor

sinθ = h/L = 9.3 m/12 m

θ =sin⁻¹( 9.3 / 12)

θ = 50.8°

Forces acting on the ladder

W₁ =m*g = 45 kg* 9.8 m/s² = 441 N: Weight of the ladder (vertical downward)

W₂ =M*g = 72 kg * 9.8 m/s² = 705.6 N : Weight of the fire fighter(vertical downward)  

FN₁ :Normal force that the floor exerts on the ladder (vertical upward)  (point A)

fs : friction force that the floor exerts on the ladder (horizontal and opposite the movement )(point A)  

FN₂ :  Normal Force that the wall exerts on the ladder ( horizontal and opposite to friction force between the floor and the ladder)

∑Fy=0

FN₁ -W₁ -W₂= 0

FN₁ = W₁ + W₂

FN₁ = 441N+ 705.6N

FN₁ = 1146.6N : Force exerted on the ladder by the wall (vertical and upward)

Calculation of the distances of the forces at the point A (contact point of the ladder on the floor)

d₁ = 4*cos 50.8° (m) = 2.53 m:  Distance from W₁ to the point A

d₂ =6*cos 50.8° (m)= 3.79 m  : Distance from W₂ to the point A

d₃ = 9.3 m : Distance from FN₂ to the point A

The equilibrium equation of the moments at the point A  (contact point of the ladder with the floor)  

∑MA = 0  

FN₂(d₃) - W₁( d₁) - W₂(d₂) = 0

FN₂(d₃) = W₁(d₁) + W₂(d₂)

FN₂(9.3) = (441 )(2.53) + (705.6)( 3.79 )

FN₂(9.3) = 1115.73 + 2674.2

FN₂ = (3790) / (9.3)  

FN₂ = 407.5 N : Force exerted on the ladder by the wall , horizontal and opposite to friction force between the floor and the ladder

Force excreted by floor on ladder is 1146.6 N (vertically upward) and force excreted by wall on ladder is 407.5 N (horizontally towards the normal force).

Force is the effect of pull or push due to which the object having a mass changes its velocity.

The force is of two types-

The length of the ladder is 12 meter and the mass of the ladder is 45 kg.Its upper end is a distance h of 9.3 and the mass of fire fighter is 72 kg.

The sine angle for the ladder can be given as,

[tex]\sin\theta=\dfrac{9.3}{12}\\\theta=50.8^o[/tex]

The forces acting on the system is normal force and force due to the weight of the two bodies.

The summation of the vertical forces is equal to the zero to keep it at rest. Therefore,

[tex]F_n-45\times9.8-72\times9.8\\F_n=1146.6 \rm N[/tex]

By the equilibrium of the momentum for the system,

[tex]F_{N2}\times(9.3)+45(9.8)\times(4\cos50.8)+72(9.8)\times(6\cos50.8)\\F_{N2}=407.5\rm N[/tex]

Thus, the force excreted by the floor on ladder is 1146.6 N (vertically upward) and the force excreted by the wall on ladder is 407.5 N (horizontally towards the normal force).

Learn more about the force here;

https://brainly.com/question/388851

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.

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.

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:

a) Please, see the attached figure.

b) The horizontal component of the initial velocity is 8.5 m/s

The vertical component of the initial velocity is 6.0 m/s

c) The clown will return to a height of 1.0 m after 1.2 s of the launch.

d) The clown will land safely on the mattress, 10.2 m from the cannon. If we include air resistance in the calculation, he will surely not reach the mattress because, without air resistance, he lands just 20 cm from the closest edge of the mattress.

Explanation:

Hi!

a) Please, see the attached figure.

b) As shown in the figure, the initial velocity vector is the following:

v0 = (v0x, v0y)

Using trigonomety of right triangles:

cos angle = adjacent side / hypotenuse

In this case:

Adjacent side = v0x

hypotenuse = v0

(see figure)

Then:

cos 35° = v0x / v0

v0 · cos 35° = v0x

v0x = 10.4 m/s · cos 35°

v0x = 8.5 m/s

The horizontal component of the initial velocity is 8.5 m/s

We proceed in the same way to find the vertical component:

sin angle = opposite side / hypotenuse

sin 35° = v0y / v0 (see figure to notice that opposite side = v0y)

v0 · sin 35° = v0y

10.4 m/s · sin 35° = v0y

v0y = 6.0 m/s

The vertical component of the initial velocity is 6.0 m/s

c) The equation of the position vector of the clown at time t is the following:

r = (x0 + v0 · t · cos α, y0 + v0 · t · sin α + 1/2 · g · t²)

Where:

x0 = initial horizontal position.

v0 = initial velocity.

t = time.

α = launching angle.

y0 = initial vertical position.

g = acceleration due to gravity (-9.8 m/s² considering the upward direction as positive).

We have to find the time at which the vertical component of the position vector is 1 m. Let´s place the origin of the system of reference at the point where the cannon is located on the ground so that x0 = 0 and y0 = 1.0 m.

Using the equation of the vertical component of the position:

y = y0 + v0 · t · sin α + 1/2 · g · t²

1.0 m = 1.0 m + 10.4 m/s · t · sin 35° - 1/2 · 9.8 m/s² · t²

0 = 10.4 m/s · t · sin 35° - 4.9 m/s² · t²

0 = t (10.4 m/s · sin 35° - 4.9 m/s² · t) (t = 0, when t = 0 the clown is 1.0 m above the ground, just leaving the cannon).

0 = 10.4 m/s · sin 35° - 4.9 m/s² · t

-10.4 m/s · sin 35° / -4.9 m/s² = t

t = 1.2 s

The clown will return to a height of 1.0 m after 1.2 s of the launch.

d) Now, let´s calculate the horizontal traveled distance after 1.2 s using the equation of the horizontal component of the position vector:

x = x0 + v0 · t · cos α (x0 = 0)

x = 10.4 m/s · 1.2 s · cos 35°

x = 10.2 m

Since the mattress is located at 10 m from the cannon and it is 2.0 m long, the clown will land safely on the mattress. However, the clown almost miss the mattress (he lands just 20 cm from the closest edge), so, if we include air resistance in the calculation, he will surely not reach the mattress.

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.