A block is attached to a spring, with spring constant k, which is attached to a wall. It is initially moved to the left a distance d (at point A) and then released from rest, where the block undergoes harmonic motion. The floor is frictionless.The points labelled A and C are the turning points for the block, and point B is the equilibrium point.1)Which of these quantities are conserved for the spring and block system?mechanical energyx-direction linear momentumneither energy nor momentum

Answer:

(a) Angular acceleration will be [tex]100.53rad/sec^2[/tex]

(b) 50 revolution

Explanation:

We have given time t = 2.5 sec

Initial speed of the drill [tex]\omega _0=0rad/sec[/tex]

Speed after 2.5 sec [tex]=2400rpm=\frac{2\pi \times 2400}{60}=251.327rad/sec[/tex]

From first equation of motion we know that

[tex]\omega =\omega _0+\alpha t[/tex]

[tex]251.327 =0+\alpha \times 2.5[/tex]

[tex]\alpha =100.53rad/sec^2[/tex]

(b) From second equation of motion we know that

[tex]\Theta =\omega _0t+\frac{1}{2}\alpha t^2[/tex]

[tex]\Theta =0\times 2.5+\frac{1}{2}\times 100.53\times  2.5^2=314.16rad=\frac{314.16}{2\pi }=50revolution[/tex]

Answer:

Speed of second block will be 20 m /sec in opposite direction

Explanation:

We have given the mass of two blocks [tex]m_1=40kg[/tex] and [tex]m_2=30kg[/tex]

As both the blocks are initially at rest so [tex]u_1=0m/sec\ and\ u_2=0m/sec[/tex]

Speed of first block [tex]v_1=15m/sec[/tex]

We have to find the speed of second block, that is [tex]v_2[/tex]

From momentum conservation we know that

[tex]m_1u_1+m_2u_2=m_1v_1+m_2v_2[/tex]

[tex]40\times 0+30\times 0=40\times 15+30\times v_2[/tex]

[tex]v_2=-20m/sec[/tex]

So speed of second block will be 20 m /sec in opposite direction.

Answer:

I= 6.292  kg.m²

Explanation:

Given that

m = 1.3 kg

Side of square a= 1.1 m

The distance r

[tex]r=\sqrt{{a^2}+{a^2}}[/tex]

[tex]r={a}{\sqrt 2}[/tex]

[tex]r={1.1}{\sqrt 2}[/tex]

The moment of inertia I

The axis passes through one of the mass then the distance of the that mass from the axis will be zero.

I = m a² + m a² + m r²

By putting the values

I = m a² + m a² + m r²

I =m( 2 a² +  r²)

I =1.3( 2 x 1.1² +  2 x 1.1²)

I = 1.3 x 4  x 1.1² kg.m²

I= 6.292  kg.m²

Explanation:

Part 1.

The force of gravity on the apple, [tex]F_1=9.39\ N[/tex] (downward)

The force of the wind on the apple, [tex]F_2=1.5\ N[/tex] (right)

Let F is the magnitude of the net force acting on it. The resultant of two vectors is given by :

[tex]F=\sqrt{F_1^2+F_2^2}[/tex]

[tex]F=\sqrt{9.39^2+1.5^2}[/tex]

F = 9.509 N

Part 2.

[tex]tan\theta=\dfrac{F_2}{F_1}[/tex]

[tex]tan\theta=\dfrac{1.5}{9.39}[/tex]

[tex]\theta=9.07^{\circ}[/tex]

So, the direction of the net external force is 9.07 degrees wrt vertical. Hence, this is the required solution.

Answer:

The angular acceleration increases.

Explanation:

The relationship between the torque and angular acceleration is :

[tex]\tau=I\times \alpha[/tex]

Where

I is the moment of inertia

[tex]\alpha[/tex] is the angular acceleration

We can see that the torque is directly proportional to the angular acceleration. So, when we have more torque it means angular acceleration increases. Hence, the correct option is (A).

Final answer:

When there is more torque applied to an object with a constant rotational inertia, the angular acceleration of the object increases proportionally. So the correct option is A.

Explanation:

According to the relationship between torque and angular acceleration, when you have more torque (given a constant rotational inertia), the angular acceleration increases. This is because torque (τ) is directly related to angular acceleration (α) through the equation τ = Iα, where I represent the moment of inertia. If the moment of inertia remains constant, an increase in torque results in a proportional increase in angular acceleration, similar to how pushing a merry-go-round harder makes it accelerate faster.

Answer:

I₂ = 25.4 W

Explanation:

Polarization problems can be solved with the malus law

     I = I₀ cos² θ

Let's apply this formula to find the intendant intensity (Gone)

Second and third polarizer, at an angle between them is

    θ₂ = 68.0-22.2 = 45.8º

    I = I₂ cos² θ₂

    I₂ = I / cos₂ θ₂

    I₂ = 75.5 / cos² 45.8

    I₂ = 155.3 W

We repeat for First and second polarizer

   I₂ = I₁ cos² θ₁

   I₁ = I₂ / cos² θ₁

   I₁ = 155.3 / cos² 22.2

   I₁ = 181.2 W

Now we analyze the first polarizer with the incident light is not polarized only half of the light for the first polarized

    I₁ = I₀ / 2

   I₀ = 2 I₁

   I₀ = 2 181.2

   I₀ = 362.4 W

Now we remove the second polarizer the intensity that reaches the third polarizer is

    I₁ = 181.2 W

The intensity at the exit is

    I₂ = I₁ cos² θ₂

    I₂ = 181.2 cos² 68.0

   I₂ = 25.4 W

To solve this exercise it is necessary to take into account the concepts related to the magnetic field and its vector representation through the cross product or vector product.

According to the definition the direction of the electromagnetic wave propagation is given by

[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]

Where,

E = Electric Field

B = Magnetic Field

According to the information provided, the direction of propagation of the electromagnetic wave is on the X axis, which for practical purposes we will denote as [tex]\hat {i}[/tex], on the other hand it is also indicated that the magnetic field is in the Y direction, that for practical purposes we will denote it as [tex]\hat {j}[/tex]. In this way using the previous equation we would have to,

[tex]\hat{n} = \vec{E} \times \vec{B}[/tex]

[tex]\hat{i} = \vec{E} \times \hat{j}[/tex]

The cross product identity is

[tex]\hat{i}=-\hat{k} \times \hat{j}[/tex]

From the equation we can notice that the electric field would be given by,

[tex]\vec{E} = -\hat{k}[/tex]

Therefore the direction of electric field is negative z-axis.

Answer:

[tex]V_p = 267.258 m/s[/tex]

[tex]V_n = 38.375 m/s[/tex]      

Explanation:

using the law of the conservation of the linear momentum:

[tex]P_i = P_f[/tex]

where [tex]P_i[/tex] is the inicial momemtum and [tex]P_f[/tex] is the final momentum

the linear momentum is calculated by the next equation

P = MV

where M is the mass and V is the velocity.

so:

[tex]P_i = m(270 m/s)[/tex]

[tex]P_f = mV_P + M_nV_n[/tex]

where m is the mass of the proton and [tex]V_p[/tex] is the velocity of the proton after the collision, [tex]M_n[/tex] is the mass of the nucleus and [tex]V_n[/tex] is the velocity of the nucleus after the collision.

therefore, we can formulate the following equation:

m(270 m/s) = m[tex]V_p[/tex] + 14m[tex]V_n[/tex]

then, m is cancelated and we have:

270 = [tex]V_p[/tex] + [tex]14V_n[/tex]

This is a elastic collision, so the kinetic energy K is conservated. Then:

[tex]K_i = \frac{1}{2}MV^2 = \frac{1}{2}m(270)^2[/tex]

and

Kf = [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]

then,

[tex]\frac{1}{2}m(270)^2[/tex] =  [tex]\frac{1}{2}mV_p^2 +\frac{1}{2}(14m)V_n^2[/tex]

here we can cancel the m and get:

[tex]\frac{1}{2}(270)^2[/tex] =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]

now, we have two equations and two incognites:

270 = [tex]V_p[/tex] + [tex]14V_n[/tex]  (eq. 1)

[tex]\frac{1}{2}(270)^2[/tex] =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]

in the second equation, we have:

36450 =  [tex]\frac{1}{2}V_p^2 +\frac{1}{2}(14)V_n^2[/tex]  (eq. 2)

from this last equation we solve for [tex]V_n[/tex] as:

[tex]V_n = \sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]

and replace in the other equation as:

270 = [tex]V_p +[/tex] 14[tex]\sqrt{\frac{36450-\frac{1}{2}V_p^2 }{\frac{1}{2} } }[/tex]

so,

[tex]V_p = -267.258 m/s[/tex]

Vp is negative because the proton go in the -i hat direction.

Finally, replacing this value on eq. 1 we get:

[tex]V_n = \frac{270+267.258}{14}[/tex]

[tex]V_n = 38.375 m/s[/tex]  

The final Celsius temperature is C. greater by a factor between 1 and 2.

Explanation:

We can solve this problem by applying the pressure law, which states that for an ideal gas kept at constant volume, the ratio between the pressure and the absolute temperature of the gas is constant:

[tex]\frac{p}{T}=constant[/tex]

or

[tex]\frac{p_1}{T_1}=\frac{p_2}{T_2}[/tex]

where in this problem:

[tex]p_1 = p_0[/tex] is the initial pressure of the gas

[tex]T_1[/tex] is the initial absolute temperature (in Kelvin)

[tex]p_2 = 2 p_0[/tex] is the final pressure

[tex]T_2[/tex] is the final temperature in Kelvin

Re-arranging the equation,

[tex]\frac{p_0}{T_1}=\frac{2p_0}{T_2}\\T_2 = 2T_1[/tex]

So, the temperature in Kelvin has doubled. We can now rewrite this relationship by rewriting the Kelvin temperature in Celsius degrees:

[tex]T_{C2}-273 = 2(T_{C1}-273)\\T_{C2}=2T_{C1}-273[/tex]

where [tex]T_{C1}, T_{C2}[/tex] are the initial and final temperatures in Celsius degrees.

This means that the temperature in Celsius increases by a factor between 1 and 2, so the correct answer is C.

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The final Celsius temperature of the gas after being warmed at constant volume to reach a pressure of 2p0 is greater by a factor between 1 and 2, according to Gay-Lussac's law. So the correct option is C.

The question deals with the behavior of gas under the influence of temperature changes at constant volume, which is described by Gay-Lussac's law. This law states that for a given mass of gas at constant volume, the pressure of the gas is directly proportional to its Kelvin (absolute) temperature. Since the initial pressure of the gas is p0 and the final pressure is 2p0, and we know that the volume remains constant, the final Kelvin temperature must also double. To convert Celsius to Kelvin, you add 273.15 to the Celsius temperature. If the initial temperature in Celsius T1 corresponds to the Kelvin temperature K1, and after the change, the final Kelvin temperature is 2K1, then when converted back to Celsius, the final temperature T2 will not be twice T1—because the relationship between Kelvin and Celsius is linear and not proportional. Therefore, the final Celsius temperature will be higher by a certain amount but not double.

Given this, the correct answer is that the final Celsius temperature of the gas is greater by a factor between 1 and 2.

Answer:

Width of central bright fringe will be 0.00228 m

Explanation:

We have given wavelength of light [tex]\lambda =632.8nm=632.8\times 10^{-9}m[/tex]

Distance between the slits [tex]d=0.5mm=0.5\times 10^{-3}m[/tex]

Distance between slit and screen D = 1.8 m

We have to find the width

We know that width is given by

[tex]width=\frac{\lambda D}{d}=\frac{632.8\times 10^{-9}\times 1.8}{0.5\times 10^{-3}}=0.00228m[/tex]

So width of central bright fringe will be 0.00228 m

Applying the given values to the formula for the width of the central bright fringe in a double-slit interference pattern yields a result of approximately 4.556 × 10^(-3) m.

The width of the central bright fringe in a double-slit interference pattern can be calculated using the formula:

w = λL / d

where:

w is the width of the central bright fringe,

λ is the wavelength of the light,

L is the distance from the slits to the screen, and

d is the distance between the slits.

In this case:

λ = 632.8 nm = 6.328 × 10^(-7) m,

L = 1.8 m, and

d = 0.5 mm = 5 × 10^(-4) m.

Plug these values into the formula:

w = (6.328 × 10^(-7) m × 1.8 m) / (5 × 10^(-4) m)

Calculating this expression gives:

w ≈ 4.556 × 10^(-3) m

So, the width of the central bright fringe is indeed approximately 4.556 × 10^(-3) m

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Answer:

Zero

Explanation:

Total energy of the EARTH  is the sum total of the all the kinetic and the gravitation potential energy of the earth.

Which can be written as

[tex]E_{total} = KE+U[/tex]

[tex]= \frac{1}{2}mv^2+ \frac{GMm}{R}[/tex]

At very large distances the kinetic as well as the gravitational potential energies become zero.

therefore, E_total= 0

When an object is launched at its escape velocity, its total mechanical energy at infinite distance from the planet is zero. The object loses all its kinetic energy, and its potential energy also becomes zero as it reaches infinity.

Escape velocity is the minimum initial velocity required for an object to escape the gravitational pull of a planet or other large body and reach an infinite distance where gravitational force becomes zero. When an object is launched at this escape velocity, its total mechanical energy (Etotal) at an infinite distance is zero. This is because the object gives up all its kinetic energy and the potential energy approaches zero as the distance (r) approaches infinity.

Using conservation of energy, we can illustrate this with the following steps:

According to the information presented, it is necessary to take into account the concepts related to mass flow, specific potential energy and the power that will determine the total work done in the system.

By definition we know that the change in mass flow is given by

[tex]\dot{m} = \rho AV[/tex]

[tex]\dot {m} = \rho Q[/tex]

Remember that the Discharge is defined as Q = AV, where A is the Area and V is the speed.

Substituting with the values we have we know that the mass flow is defined by

[tex]\dot{m} = 1000*0.03[/tex]

[tex]\dot{m} = 30kg/s[/tex]

To calculate the power we need to obtain the specific potential energy, which is given by

[tex]\Delta pe = gh[/tex]

[tex]\Delta pe = 9.8*45[/tex]

[tex]\Delta pe = 441m^2/s^2[/tex]

So the power needed to deliver the water into the storage tank would be

[tex]\dot {E} = \dot{m}\Delta pe[/tex]

[tex]\dot {E} = 30*441[/tex]

[tex]\dot{E} = 13230W = 13.23kW[/tex]

Finally the mechanical power that is converted to thermal energy due to friction effects is:

[tex]\dot{W}_f = \dot{W}_s - \dot{E}[/tex]

[tex]\dot{W}_f 20-13.23[/tex]

[tex]\dot{W} = 6.77kW[/tex]

Therefore the mechanical power due to friction effect is 6.77kW

The irreversible head loss in this pump system is 23 meters and the lost mechanical power during the pumping process is 6.8 kW.

In this system, the pump supplies energy to the water and drives it up to the higher reservoir. Out of this energy provided by the pump, part of it is used for lifting the water while some is lost as head loss due to friction and other losses in the system. Using the principle of conservation of energy and the Bernoulli's equation, one can calculate the irreversible head loss and the mechanical power loss.

Given the mechanical power provided by the pump, P = 20 kW, the height difference h = 45 m, the flow rate Q = 0.03 m3/s, and the density of water ρ = 1000 kg/m3, the useful power for lifting the water equals ρghQ=1000x 9.81x 45 x 0.03=13.2 kW. Thus, the lost power due to irreversible head loss equals the total power minus the useful power, or 20 kW - 13.2 kW = 6.8 kW. The corresponding head loss equals the lost power divided by the power used for lifting the water, or h_loss = (6.8 kW)/(ρgQ) = 23 m.

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Answer:

Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz

Explanation:

Since, the two are in out of phase,

their path difference is

d= nλ

[tex]d_2-d_1= n\lambda[/tex]

Given d1= 2.75 m

D= 3.90 m

[tex]d_2= \sqrt{D^2- d_1^2}[/tex]

[tex]d_2= \sqrt{3.90^2- 2.75^2}[/tex]

d_2= 2.76 m

2.76-2.75= 1×λ

λ= 0.01 m

0.01= 1*λ

λ =0.01

frequency ν = v/λ = 340/0.01

f= 3400 Hz

Hence lowest (nonzero) frequency that gives destructive interference in this case = 3400 Hz

Answer:

45 W/m^2

Explanation:

Intensity of light, Io = 90 W/m^2

According to the law of Malus

[tex]I=I_{0}Cos^{2}\theta[/tex]

The average value of Cos^θ is half

So, I = Io/2

I = 90 /2

I = 45 W/m^2

Unpolarized light, when passed through a polarizer, reduces its intensity by half. So, the intensity if the light that emerges from a vertical filter will be 45 W/m².

Given that the incident intensity of the unpolarized light is 90 W/m², when passed through a vertically oriented optical filter, the emerging light will be polarized and will have its intensity halved as it's the property of a polarizing filter to decrease the intensity of unpolarized light by a factor of 2. The formula used in this process is I = Io cos² θ. In the case of unpolarized light passing through a single polarizer, θ is 0. So, the formula simplifies to I = Io/2.

Therefore, the intensity of the light that emerges from the vertically oriented optical filter is: I = 90 W/m² / 2 = 45 W/m².

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Answer:

The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

Explanation:

Given that,

Energy density = 0.1 J/m³

Separation = 0.2 mm

We need to calculate the potential difference

Using formula of energy density

[tex]J=\dfrac{1}{2}\epsilon_{0}E^2[/tex]

[tex]J=\dfrac{1}{2}\epsilon_{0}\dfrac{V^2}{d^2}[/tex]

[tex]V^2=\dfrac{0.1\times(0.2\times10^{-3})^2\times2}{8.85\times10^{-12}}[/tex]

[tex]V^2=\sqrt{903.95}[/tex]

[tex]V=30.06\ V[/tex]

We need to calculate the speed of electron

Using energy conservation

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

Put the value into the formula

[tex]1.6\times10^{-19}\times30.06=\dfrac{1}{2}\times9.1\times10^{-31}\times v^2[/tex]

[tex]v^2=\dfrac{1.6\times10^{-19}\times30.06\times2}{9.1\times10^{-31}}[/tex]

[tex]v=\sqrt{1.057\times10^{13}}[/tex]

[tex]v=3.2\times10^{6}\ m/s[/tex]

Hence, The speed of electron is [tex]3.2\times10^{6}\ m/s[/tex]

Answer:

17.7 cm^3

Explanation:

depth, h = 120 m

density of water, d = 1000 kg/m^3

V1 = 1.4 cm^3

P1 = P0 + h x d x g

P2 = P0

where, P0 be the atmospheric pressure

Let V2 be the volume of the bubble at the surface of water.

P0 = 1.01 x 10^5 Pa

P1 = 1.01 x 10^5 + 120 x 1000 x 9.8 = 12.77 x 10^5 Pa

Use

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

Thus, the volume of bubble at the surface of water is 17.7 cm^3.

The volume (in cm3) of the bubble just before it pops at the surface of the lake is mathematically given as

V2 = 17.7 cm^3

Question Parameter(s):

An air bubble released by a remotely operated underwater vehicle, 120 m

below the surface of a lake, has a volume of 1.40 cm3.

Generally, the equation for the initial Pressure  is mathematically given as

P1 = P0 + h x d x g

Where,atmospheric pressure

P0 = 1.01 x 10^5 Pa

Therefore

P1 = 1.01 * 10^5 + 120 * 1000 &* 9.8

P1= 12.77 * 10^5 Pa

In conclusion

P1 x V1 = P2 x V2

12.77 x 10^5 x 1.4 = 1.01 x 10^5 x V2

V2 = 17.7 cm^3

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To find the angular acceleration of the cylinder, calculate the moment of inertia of the cylinder, determine the torque due to the falling bucket, and then use Newton's second law for rotation to find the angular acceleration.

The subject involves calculating the angular acceleration of a cylinder when the bucket of water tied to a rope around the cylinder is let go. We assume no friction or air resistance in this case.

First, we need to calculate the moment of inertia (I) of the solid cylinder, which can be calculated using the formula I = 0.5 * m * r^2. Here, m is the mass of the cylinder (50 kg) and r is the radius (half of the diameter, 0.125 m). After calculating I, use Newton's second law for rotation τ = I * α, where τ is torque and α is the angular acceleration.

The torque τ can be calculated as the product of the force exerted by the falling bucket (F) and the radius r of the cylinder. The force F is the weight of the falling bucket, m * g, where m is the mass of the bucket (20 kg) and g is acceleration due to gravity (about 9.8 m/s^2). Once you have calculated τ, you can solve for α by rearranging the equation to α = τ/I.

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The angular acceleration of the 50-kg cylinder is found to be approximately 62.7 rad/s².

To find the angular acceleration of the 50-kg cylinder when the bucket of water falls down to the bottom of the well, we will use the concept of rotational dynamics. The moment of inertia (I) of a solid cylinder is given by:

I = (1/2) [tex]\times[/tex] M[tex]\times[/tex] R²

where M is the mass and R is the radius. Given the mass M is 50 kg and the diameter is 25 cm, so the radius R is 0.125 m:

I = (1/2) * 50 kg[tex]\times[/tex] (0.125 m)² = 0.390625 kg m²

The torque (τ) caused by the falling bucket is:

The force exerted by the bucket of water is its weight, F = m [tex]\times[/tex] g, where m is the mass of the bucket and water (20 kg) and g is the acceleration due to gravity (9.8 m/s²):

So,

Using Newton's second law for rotation:

where α is the angular acceleration:

Solving for α:

Therefore, the angular acceleration of the 50-kg cylinder is approximately 62.7 rad/s².

Answer:

[tex]\lambda=3.20*10^{-7}m[/tex]

Explanation:

The wavelength is inversely proportional to the frequency. The wavelength is equal to the speed of the wave, divided by the frequency. In the case of electromagnetic waves like ultraviolet radiation, the speed of propagation is the speed of light.

[tex]\lambda=\frac{c}{f}\\\lambda=\frac{3*10^8\frac{m}{s}}{9.38*10^{14}Hz}\\\lambda=3.20*10^{-7}m[/tex]

Answer :  The wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]

Explanation : Given,

Frequency = [tex]9.38\times 10^{14}Hz=9.38\times 10^{14}s^{-1}[/tex]

Formula used :

[tex]\nu=\frac{c}{\lambda}[/tex]

where,

[tex]\nu[/tex] = frequency

[tex]\lambda[/tex] = wavelength

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

Now put all the given values in the above formula, we get:

[tex]9.38\times 10^{14}s^{-1}=\frac{3\times 10^8m/s}{\lambda}[/tex]

[tex]\lambda=3.20\times 10^{-7}m[/tex]

Therefore, the wavelength of the radiation absorbed by ozone is, [tex]3.20\times 10^{-7}m[/tex]

Answer:3.51

Explanation:

Given

Coefficient of Friction [tex]\mu =0.4 [/tex]

Consider a small element at an angle \theta having an angle of [tex]d\theta [/tex]

Normal Force[tex]=T\times \frac{d\theta }{2}+(T+dT)\cdot \frac{d\theta }{2}[/tex]

[tex]N=T\cdot d\theta [/tex]

Friction [tex]f=\mu \times Normal\ Reaction[/tex]

[tex]f=\mu \cdot N[/tex]

and [tex]T+dT-T=f=\mu Td\theta [/tex]

[tex]dT=\mu Td\theta [/tex]

[tex]\frac{dT}{T}=\mu d\theta [/tex]

[tex]\int_{T_2}^{T_1}\frac{dT}{T}=\int_{0}^{\pi }\mu d\theta [/tex]

[tex]\frac{T_2}{T_1}=e^{\mu \pi}[/tex]

[tex]\frac{T_2}{T_1}=e^{0.4\times \pi }[/tex]

[tex]\frac{T_2}{T_1}==e^{1.256}[/tex]

[tex]\frac{T_2}{T_1}=3.51[/tex]

Answer:

The speed of the water is 14.68 m/s.

Explanation:

Given that,

Time = 30 minutes

Distance = 11.0 m

Pressure = 101.3 kPa

Density of water = 1000 kg/m³

We need to calculate the speed of the water

Using equation of motion

[tex]v^2=u^2+2gs[/tex]

Where, u = speed of water

g = acceleration due to gravity

h = height

Put the value into the formula

[tex]0=u^2-2\times9.8\times11.0[/tex]

[tex]u=\sqrt{2\times9.8\times11.0}[/tex]

[tex]u=14.68\ m/s[/tex]

Hence, The speed of the water is 14.68 m/s.

Answer:

Velocity of the helium nuleus  = 1.44x10⁴m/s

Velocity of the proton = 2.16x10⁴m/s

Explanation:

From the conservation of linear momentum of the proton collision with the He nucleus:

[tex] P_{1i} + P_{2i} = P_{1f} + P_{2f] [/tex] (1)

where [tex]P_{1i}[/tex]: is the proton linear momentum initial, [tex]P_{2i}[/tex]: is the helium nucleus linear momentum initial, [tex]P_{1f}[/tex]: is the proton linear momentum final, [tex]P_{2f}[/tex]: is the helium nucleus linear momentum final

From (1):

[tex] m_{1}v_{1i} + 0 = m_{1}v_{1f} + m_{2}v_{2f} [/tex] (2)

where m₁ and m₂: are the proton and helium mass, respectively, [tex]v_{1i}[/tex] and [tex]v_{2i}[/tex]: are the proton and helium nucleus velocities, respectively, before the collision, and [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex]: are the proton and helium nucleus velocities, respectively, after the collision

By conservation of energy, we have:

[tex] K_{1i} + K_{2i} = K_{1f} + K_{2f} [/tex] (3)

where [tex]K_{1i}[/tex] and  [tex]K_{2i}[/tex]: are the kinetic energy for the proton and helium, respectively, before the colission, and [tex]K_{1f}[/tex] and  [tex]K_{2f}[/tex]: are the kinetic energy for the proton and helium, respectively, after the colission

From (3):

[tex] \frac{1}{2}m_{1}v_{1i}^{2} + 0 = \frac{1}{2}m_{1}v_{1f}^{2} + \frac{1}{2}m_{2}v_{2f}^{2} [/tex] (4)  

Now we have two equations: (2) ad (4), and two incognits: [tex]v_{1f}[/tex] and [tex]v_{2f}[/tex].

Solving equation (2) for [tex]v_{1f}[/tex], we have:

[tex] v_{1f} = v_{1i} -\frac{m_{2}}{m_{1}} v_{2f} [/tex] (5)

From getting (5) into (4) we can obtain the [tex]v_{2f}[/tex]:

[tex] v_{2f}^{2} \cdot (\frac{m_{2}^{2}}{m_{1}} + m_{2}) - 2v_{2f}v_{1i}m_{2} = 0 [/tex]

[tex] v_{2f}^{2} \cdot (\frac{(4u)^{2}}{1u} + 4u) - v_{2f}\cdot 2 \cdot 3.6 \cdot 10^{4} \cdot 4u = 0 [/tex]

From solving the quadratic equation, we can calculate the velocity of the helium nucleus after the collision:

[tex] v_{2f} = 1.44 \cdot 10^{4} \frac{m}{s} [/tex] (6)

Now, by introducing (6) into (5) we get the proton velocity after the collision:

[tex] v_{1f} = 3.6 \cdot 10^{4} -\frac{4u}{1u} \cdot 1.44 \cdot 10^{4} [/tex]

[tex] v_{1f} = -2.16 \cdot 10^{4} \frac{m}{s} [/tex]

The negative sign means that the proton is moving in the opposite direction after the collision.

I hope it helps you!

Final answer:

In an elastic collision between a proton and a helium nucleus, their final velocities can be found using the principles of conservation of momentum and kinetic energy.

Explanation:

In an elastic collision, both momentum and kinetic energy are conserved. To find the final velocities of the proton and helium nucleus after the collision, we can use the principles of conservation of momentum and kinetic energy.

Since the helium nucleus is initially at rest, its momentum before the collision is zero. The momentum before the collision can be calculated as:

Initial momentum proton = proton mass × proton velocity

Final momentum proton + Final momentum helium nucleus = proton mass × proton final velocity + helium nucleus mass × helium nucleus final velocity

Using the conservation of momentum, we can solve for the final velocities of the proton and helium nucleus. The final kinetic energy of the helium nucleus can also be calculated using the equation:

Final kinetic energy helium nucleus = 0.5 × helium nucleus mass × helium nucleus final velocity^2

To develop this problem it is necessary to apply the concepts of sum of capacitors in a circuit, either in parallel or in series.

When capacitors are connected in series, they consecutively add their capacitance.

However, when they are connected in parallel, the sum that is made is that of the inverse of the capacitance, that is [tex]\frac {1} {C}[/tex] where, C is the capacitance.

For the given case, it is best to connect two of these capacitors in series and one in parallel, so

We have three 100 pF capacitors, then

[tex]C_1 = 100 pF\\C_2 = 100 pF\\C_3 = 100 pF\\[/tex]

Here we can see how two capacitors 1 and 2 are in series and the third in parallel.

[tex]\frac{1}{C_{serie}} = \frac{1}{100pF}+\frac{1}{100pF}[/tex]

[tex]\frac{1}{C_{serie}} = \frac{1}{50pF}[/tex]

Investing equality

[tex]C_{serie} = 50pF[/tex]

Adding it in parallel, then

[tex]C_{total} = C_{serie} +C_3[/tex]

[tex]C_{total} = 50pF+100pF[/tex]

[tex]C_{total} = 150pF[/tex]

Answer: It should be one that matches the symmetry of the charge distribution

Explanation: Gaussian surfaces are usually carefully chosen to exploit symmetries of a situation to simplify the calculation of the surface integral. If the Gaussian surface is chosen such that for every point on the surface the component of the electric field along the normal vector is constant, then the calculation will not require difficult integration as the constants which arise can be taken out of the integral.It is defined as the closed surface in three dimensional space by which the flux of vector field be calculated.

Final answer:

The best choice for a Gaussian surface is one that matches the symmetry of the charge distribution, for example, a spherical Gaussian surface for a point charge and a cylindrical Gaussian surface for a cylindrical charge distribution, as it simplifies the integral for the flux.

Explanation:

The best choice for the shape of a Gaussian surface is E) It should be one that matches the symmetry of the charge distribution. When dealing with point charges that are spherically symmetric, selecting a spherical Gaussian surface makes sense since the electric field is radial. Likewise, for a cylindrical charge distribution, a cylindrical Gaussian surface is most appropriate. The goal is to make the flux integral easy to evaluate and to find a surface over which the electric field is constant in magnitude and makes the same angle with every element of the surface.

Using a Gaussian surface that matches the symmetry of the charge distribution, such as a spherical shell for a point charge or spherical charge distribution, ensures the electric field (E) is constant in magnitude across the surface, simplifying the calculations. Conversely, using a less ideal surface, like a cylindrical surface to envelop a spherical charge, leads to a more complex integral due to the variations in electric field strength (E) across the surface.

Answer:

N₂/N₁ = 45

Explanation:

The relation of the number of turns with the voltage is given by the EMF equation of transformer:  

[tex] \frac{V_{2}}{V_{1}} = \frac{N_{2}}{N_{1}} [/tex] (1)

where V₁: voltage of the primary winding, V₂: voltage of the secondary winding, N₁: number of turns in the primary winding, and N₂: number of turns in the secondary winding

Assuming that the primary winding is connected to the input voltage supply, and using equation (1), we can calculate the ratio of N₂ to N₁:

[tex] \frac{N_{2}}{N_{1}} = \frac{V_{2}}{V_{1}} = \frac{5.0 \cdot 10^{3}V}{110V} = 45 [/tex]       

So, the ratio N₂/N₁ is 45.

Have a nice day!

Answer:

They have the same momentum, but the car has more kinetic energy than the truck.

Explanation:

Let [tex]m_c[/tex] is the mass of the car and [tex]m_t[/tex] is the mass of truck such that,

[tex]m_t=2m_c[/tex]

You push both the car and the truck for the same amount of time with the same force. Momentum of an object is given by :

[tex]p=mv=F\times t[/tex]

Since, force and time are same, so they have same momentum. The kinetic energy of an object is given by :

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

Since, the mass of truck is more, it will have maximum kinetic energy. So, the correct option is (a) "They have the same momentum, but the car has more kinetic energy than the truck".

The car has more momentum and more KE than the truck because car mass is lower so it moves fast as compared to truck.

Momentum is the product of its mass and velocity. When we compare car with truck , car has a lower mass so when the mass is lower the velocity will be higher so car has more momentum.

We also know that kinetic energy also depends on mass so less mass have more kinetic energy.

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We can find the distance the sled travels before it comes to rest by equating the work done by the frictional force with the initial kinetic energy of the sled. We first calculate the initial kinetic energy with (1/2)mv². The force of friction is given by μmg. The work done by friction is equal to this force times the distance traveled, and we set this equal to the negative initial kinetic energy to account for the sled stopping.

The subject of this question is about the concept of work and energy in a physics problem involving a sled on a frozen pond. To solve this, we would use the principle of work and energy where the work done on the sled is equal to the change in its kinetic energy according to the work-energy theorem. Since the sled comes to rest, the final kinetic energy is zero.

First, we calculate the initial kinetic energy of the sled using the formula (1/2)mv², where m is the mass (8.54 kg), and v is the velocity (1.87 m/s). Next, we know that the work done by the friction on the sled is equal to the force of friction times the distance traveled which is equal to the change in kinetic energy. The force of friction can be calculated by multiplying the mass of the sled, the acceleration due to gravity (9.8 m/s²), and the coefficient of kinetic friction (0.087). Set this work equal to the negative initial kinetic energy (as it acts to bring the sled to a stop) and solve for the distance.

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To find the distance the sled moves before coming to rest, we can use the work and energy concepts. Work done by the friction force is equal to the change in kinetic energy. By setting the work done by friction equal to the change in kinetic energy and using the given values, we can solve for the distance the sled moves.

To find the distance the sled moves before coming to rest, we will use the work and energy concepts. The work done on an object is equal to the change in its kinetic energy, so we can use the work-energy principle to solve the problem. The work done by the friction force is equal to the force of friction multiplied by the distance the sled moves. Setting the work done by friction equal to the change in kinetic energy, we can solve for the distance the sled moves.

The work done by the friction force is given by the equation:
W = F * d * cos(180°)

where W is the work done, F is the force of friction, d is the distance, and cos(180°) is the angle between the force of friction and the displacement. The work done by the friction force is equal to the negative change in kinetic energy of the sled, which can be calculated as:
ΔKE = -(1/2) * m * v²

where ΔKE is the change in kinetic energy, m is the mass of the sled, and v is the final velocity of the sled.

Setting the work done by friction equal to the negative change in kinetic energy and solving for the distance d:

W = -(1/2) * m * v²
F * d * cos(180°) = -(1/2) * m * v²

Using the given values for the mass of the sled (m = 8.54 kg), the initial velocity (v = 1.87 m/s), and the coefficient of kinetic friction (µ = 0.087), we can solve for the distance d.

First, we need to calculate the magnitude of the friction force using the equation:
F = µ * N

where µ is the coefficient of kinetic friction and N is the normal force. The normal force is equal to the weight of the sled, which can be calculated as:
N = m * g

where g is the acceleration due to gravity (approximately 9.8 m/s²).

Substituting the values into the equations:

F = 0.087 * (8.54 kg * 9.8 m/s²) = 7.94 N

Now, we can substitute the values for the force of friction and the change in kinetic energy into the equation:

F * d * cos(180°) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Using the cosine of 180° as -1:

7.94 N * d * (-1) = -(1/2) * (8.54 kg) * (1.87 m/s)²

Simplifying the equation:

-7.94 N * d = -(1/2) * (8.54 kg) * (1.87 m/s)²

Dividing both sides of the equation by -7.94 N:

d = [(1/2) * (8.54 kg) * (1.87 m/s)²] / 7.94 N

Calculating the distance d:

d = 2.008 m

Therefore, the sled moves a distance of 2.008 meters before coming to rest.

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Answer:

Explanation:

If the acceleration due to gravity on the Moon's surface is [tex]a_M[/tex] and the acceleration due to gravity on the Earth is [tex]a_E[/tex], we can write that:

[tex]a_M=\frac{1}{6}a_E[/tex]

[tex]\frac{a_M}{a_E}=\frac{1}{6}[/tex]

If the radius of the Moon is [tex]r_M[/tex] and the radius of the Earth is [tex]r_E[/tex], we can write that:

[tex]r_M=\frac{1}{4}r_E[/tex]

[tex]\frac{r_M}{r_E}=\frac{1}{4}[/tex]

By Newton's 2nd Law we know that F=ma and using Newton's law of universal gravitation  we can calculate the gravitational force an object with mass m experiments from a planet with mass M being at a distance r from it. We will assume our object is on the surface so this distance will be the radius of the planet.

Since the force the object experiments is the force of gravitation we can write, for Earth:

[tex]F=ma_E=\frac{GM_Em}{r_E^2}[/tex]

which means:

[tex]a_E=\frac{GM_E}{r_E^2}[/tex]

[tex]M_E=\frac{a_Er_E^2}{G}[/tex]

And for the Moon:

[tex]F=ma_M=\frac{GM_Mm}{r_M^2}[/tex]

which means:

[tex]a_M=\frac{GM_M}{r_M^2}[/tex]

[tex]M_M=\frac{a_Mr_M^2}{G}[/tex]

We can then write the fraction:

[tex]\frac{M_M}{M_E}=\frac{a_Mr_M^2}{G}\frac{G}{a_Er_E^2}=\frac{a_M}{a_E}(\frac{r_M}{r_E})^2=\frac{1}{6}(\frac{1}{4})^2=0.01[/tex]

Which means:

[tex]M_M=0.01M_E[/tex]

Final answer:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration and the given ratios between the Moon and Earth's acceleration and radii. Using these values, we can solve for the mass of the Moon in terms of the mass of the Earth.

Explanation:

To find the mass of the Moon in terms of the mass of the Earth, we can use the formula for gravitational acceleration: g = GM/r². Given that the acceleration due to gravity on the Moon is one-sixth that of Earth and the radius of the Moon is one-quarter that of Earth, we can set up the following equation:

(1/6) * (9.8 m/s²) = GM/(1/4 * R)²

Simplifying, we get: 1/6 * 9.8 = GM/(1/16)

Now, we can solve for the mass of the Moon (M) in terms of the mass of the Earth (m):

M = (1/6 * 9.8 * R²)/(1/16 * G)

Substituting the values for R and G, we get:

M = (1/6 * 9.8 * (1/4)²)/(1/16 * 6.67 × 10⁻¹¹)

M ≈ 0.123m

Answer:

[tex]\frac{d\phi}{dt}=\pi r^2\frac{dB}{dt}[/tex]

Explanation:

the rate of the magnetic flux is given by,

[tex]\frac{d\phi}{dt}= \frac{d(BA)}{dt}[/tex]

where B= magnetic strength

A= area of cross section

r= radius of circle

\phi= flux

=[tex]A\frac{dB}{dt}[/tex]

=[tex]\pi r^2\frac{dB}{dt}[/tex]

Answer:

Explanation:

Diameter of the circular coil d = 17.7 cm = 0.177m

Current in the coil, I = 7.4 A

Number of loops in the coil, N = 18

Magnetic field, B = 5.5 x 10-5 T

Angle below line, (θ) = 66°

Write the expression for the torque in the coil

Torque = N x I x B x A x sin (θ)

Where A is the area of the coil and θ° is the angle between the magnetic field and the coil face.

The cross- sectional area of the coil = (pie)(d/2)²

A = (π)(0.177/2)² = 0.0246 m²

The Earth’s magnetic field points into the earth at an angle 66 below the line. The line points towards the north

Hence to find the angle between the magnetic field and the coil face we need to subtract the given angle by 90°

Theta = 90 – 66 = 24°

Substitute the value to find out torque

Torque = 18 x 7.4 x 5.5 x 10⁻⁵ x 0.0246 x sin(24) = 7.33 x 10⁻⁵ N-m

The torque on the coil is 7.33 x 10⁻⁵ N-m

Answer:

[tex]1\times 10^{8}\ Pa[/tex]

Explanation:

k = Boltzmann constant = [tex]1.38\times 10^{-23}\ J/K[/tex]

r = Radius of gas molecule = [tex]2\times 10^{-10}\ m[/tex]

t = Temperature = 300 K

P = Pressure

Volume of gas per molecule is given by

[tex]V=\frac{4}{3}\pi r^3\\\Rightarrow V=\frac{4}{3}\pi (2\times 10^{-10})^3\\\Rightarrow V=3.35103\times 10^{-29}\ m^3[/tex]

From the ideal gas law we have

[tex]PV=kt\\\Rightarrow P=\frac{kt}{V}\\\Rightarrow P=\frac{1.38\times 10^{-23}\times 300}{3.35103\times 10^{-29}}\\\Rightarrow P=123544104.35\ Pa=1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of the molecules should cause noticeable deviations from ideal-gas behavior is [tex]1\times 10^{8}\ Pa[/tex]

The pressure at which the finite volume of oxygen molecules causes deviations from ideal-gas behavior can be estimated using the ideal gas law and considering the volume of a molecule. At ordinary temperatures, the approximate pressure is 2.04 × 10^11 Pa.

To estimate the pressure at which the finite volume of oxygen molecules causes noticeable deviations from ideal-gas behavior, we need to equate the volume of gas per molecule to the volume of a molecule. The volume of the gas per molecule is equivalent to the volume of a sphere with a radius of 2.0 × 10-10 m, which is approximately 4.19 × 10-30 m3. Using the ideal gas law, we can calculate the pressure using the formula PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature in Kelvin. Assuming 1 mole of oxygen gas (which contains 6.022 × 1023 molecules), the number of moles can be calculated by dividing the volume of the gas per molecule by Avogadro's number. The approximate pressure at which deviations from ideal-gas behavior become noticeable is around 2.04 × 1011 Pa (Pascals).

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